The ratio test and the root test, both built by comparison against a geometric series, why both are silent at a limit of exactly one, how a term's shape picks between them, and the strategy assembling every test in the chapter.
Subject: Calculus II · 66 slides · symbolic lesson
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Title
Calculus II · Section 5.6
Tests the series supplies for itself, and how to choose among every test in the chapter
Objectives
Every test so far either needed a second series to compare with, or an integral you could actually do. The two tests in this lesson need neither: the series supplies its own comparison. Then the section closes the chapter by assembling every test into one strategy.
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 509 — learning objectives 5.6.1 to 5.6.3
Look back at the tests you have. The divergence test can only ever say diverges. The integral test needs a function you can integrate. The comparison tests need you to invent a second series with the right behaviour. Each is powerful, and each asks you to bring something from outside the series.
The two tests in this lesson ask for nothing extra. They look at the series' own terms, measure how quickly they shrink, and compare the tail with a geometric series built from that measurement. That makes them the natural tools for factorials and exponentials, where every other test struggles.
The second half of the lesson is the part to keep for the rest of the course: the book's strategy for choosing among all the tests of Chapter 5, and Table 5.3, the summary of every one of them in a single place.
Warm-up
Discussion prompt
A geometric series has nth term a times r to the n minus 1. Write the ratio of each term to the one before it, and state exactly when the series converges. Answer before revealing.
Write your answer before revealing it. Everything in this lesson grows out of the geometric series, so it is worth having its rule sharp in your mind.
The key fact is that in a geometric series the ratio of each term to the one before is the same number, r, at every step. Whether the size of r is below one decides convergence completely, and when it is below one you even get the sum.
Now imagine a series whose ratio is not constant but settles down to a limit as n grows. Far out in the tail, that series looks almost geometric. The ratio test is simply that thought made precise, and the root test is a second way of measuring the same effective ratio.
Section
Part 1
Concept
Figure (svg): Two panels. Left: the terms of one over n, one over n squared, and two to the n over n factorial, for n from 1 to 12, all shrinking toward zero. Right: the ratio of each term to the one before; for one over n and one over n squared the ratios climb toward the dashed line at 1, while for two to the n over n factorial they fall toward zero.
\[ \frac1n \to 0 \text{ (diverges)}, \quad \frac{1}{n^2} \to 0 \text{ (converges)}, \quad \frac{2^n}{n!} \to 0 \text{ (?)} \]
The divergence test only asks whether the terms reach zero. What decides convergence is how fast they get there, and the ratio of each term to the one before is a direct measure of that speed.
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 509 — introduction to the ratio test
Look at the left panel first. All three sequences of terms head down to zero, and by eye it is hard to say which will give a finite total. You already know the answer for two of them: the harmonic series diverges and the reciprocal squares converge. So the limit of the terms, on its own, cannot be what decides.
Now look at the right panel, which plots the ratio of each term to the one before. For one over n and one over n squared the ratios creep up toward one: each term is almost as big as the previous one. For two to the n over n factorial the ratios fall toward zero: each term is a smaller and smaller fraction of the last.
That ratio is a measure of speed. When it settles well below one, the terms collapse like a geometric sequence, and that is the situation the ratio test detects.
Concept
In a geometric series the ratio of neighbours is a constant. In most other series it changes with n, but it often settles to a limit.
\[ \frac{a_{n+1}}{a_n} = \frac{2^{n+1}/(n+1)!}{2^{n}/n!} = \frac{2}{n+1} \]
\[ \frac{2}{n+1} \to 0 \]
If the ratio settles at a number rho, then far out in the series each term is roughly rho times the one before. The tail behaves like a geometric series with ratio rho, and a geometric series is something you can decide.
The first line is the calculation you will do in almost every ratio-test problem: put the next term over the current one and simplify. Here the powers of two leave a single two, and the factorials leave a single factor of n plus one, so the ratio is two over n plus one.
That ratio is not constant, which is why the series is not geometric. But it settles, and its limit is zero. Far out in the series, each term is a tiny fraction of the one before.
Give the limit a name, rho. The plan for the next few slides is to show that a series whose ratio settles at rho behaves, in its tail, like a geometric series with ratio rho. Then the geometric rule transfers: below one converges, above one diverges.
Concept
Ratio test — For a series with nonzero terms, take the limit rho of the absolute value of the next term over the current one. Below one, the series converges absolutely; above one or infinite, it diverges; exactly one, the test gives no information.
\[ \rho = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| \]
| value of ρ | verdict |
|---|---|
| 0 ≤ ρ < 1 | converges absolutely |
| ρ > 1 or ρ = ∞ | diverges |
| ρ = 1 | no information |
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 509 — Theorem 5.16
Read the statement in three parts. First the hypothesis: the terms must be nonzero, because you are dividing by them. Then the quantity: the limit of the absolute value of the next term over the current one. Then the three cases, which the table lays out.
Two details are easy to miss. The convergent verdict is absolute convergence, which is stronger than plain convergence, so the ratio test settles the question from Section 5.5 at the same time. And an infinite limit counts as above one: terms whose ratios grow without bound certainly diverge.
The third row is the one that costs marks. When rho equals one, the test gives no information whatsoever. It does not mean converges, and it does not mean diverges. Part 4 of this lesson shows why it has to be that way.
Notation
Annotate
On: \( \rho = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|, \quad \rho < 1 \;\Longrightarrow\; \sum |a_n| \text{ converges} \)
Step through the annotations one at a time. The first is about orientation: the next term goes on top. It matches the geometric series, where the ratio r is what you multiply by to move forward one step. Flip it and a limit of zero becomes a limit of infinity, reversing the verdict.
The absolute value bars are why the test works for series with any signs. It measures only how the sizes shrink, which is also why its convergent verdict is absolute convergence.
The third annotation is the subtle one. The condition is on the limit of the ratios, not on each ratio. Every ratio of the harmonic series is below one, yet it diverges, because the ratios climb toward one. You will meet this again as a counterexample later.
Worked example
Use the ratio test on the series below.
\[ \sum_{n=1}^{\infty}\frac{2^n}{n!} \]
Write the next term over this term
Why: Replace n by n plus 1 for the top; the bottom is the term itself.
\[ \frac{a_{n+1}}{a_n} = \frac{2^{n+1}/(n+1)!}{2^{n}/n!} \]
Flip and multiply
Why: Dividing by a fraction is multiplying by its reciprocal.
\[ = \frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^{n}} \]
Cancel the powers of two
Why: One factor of two survives.
\[ = 2\cdot\frac{n!}{(n+1)!} \]
Cancel the factorials
Why: The factorial of n plus 1 is n plus 1 times the factorial of n.
\[ (n+1)! = (n+1)\,n! \;\Longrightarrow\; \frac{a_{n+1}}{a_n} = \frac{2}{n+1} \]
Take the limit and compare with one
Why: The denominator grows without bound.
\[ \rho = \lim_{n\to\infty}\frac{2}{n+1} = 0 < 1 \;\Longrightarrow\; \text{converges} \]
Check against the known sum
Why: This is the series for e squared with its first term removed (Chapter 6 proves it). Ten terms already agree to four decimal places.
\[ S_{10} \approx 6.388995, \qquad e^2 - 1 \approx 6.389056 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 511 — Example 5.23a
This is the template for every ratio-test calculation, so follow the lines in order. Write the next term over the current one. Turn the division into multiplication by the reciprocal. Then cancel, one kind of factor at a time: the powers of two first, then the factorials.
The factorial cancellation is the move to remember. The factorial of n plus one is n plus one times the factorial of n, so their quotient is a single factor. That is why factorials, which are awkward in every other test, are easy here.
The limit is zero, well below one, so the series converges absolutely. The check uses a fact you will prove in Chapter 6: this series is the power series for e squared with its first term, one, left out. Ten terms already match e squared minus one to four decimal places, which is what a ratio shrinking toward zero predicts.
Prediction
\[ \sum_{n=1}^{\infty}\frac{10^n}{n!} = 10 + 50 + 166.7 + \cdots + 2755.7 + 2755.7 + \cdots \]
Predict first
The ninth and tenth terms are both about 2756, and the partial sums pass 20,000. Does the series converge?
Correct: Yes: the ratio limit is 0
Why: The ratio of neighbours is 10/(n + 1), which tends to 0. Convergence is about the tail, and a few thousand large terms at the start change the value of the sum, never whether it exists. From n = 10 on, each term is at most 10/11 of the one before, and that factor keeps shrinking.
Commit to an answer before revealing. The numbers are chosen to make convergence look impossible: the terms climb into the thousands, and the partial sums pass twenty thousand.
But convergence is only ever about the tail. However large the first few terms are, they add a fixed finite amount. What matters is what happens far out, and the ratio of neighbours, ten over n plus one, eventually drops below one and keeps falling toward zero.
If you chose one of the no answers, notice what you were reacting to: the size of the early terms. The ratio test ignores them completely, and rightly so. The next slide shows the whole picture.
Worked example
\[ \sum_{n=1}^{\infty}\frac{10^n}{n!} \]
Form the ratio
Why: Same shape as Example 5.23(a), with 10 in place of 2.
\[ \frac{a_{n+1}}{a_n} = \frac{10^{n+1}}{(n+1)!}\cdot\frac{n!}{10^{n}} \]
Cancel
Why: One factor of 10 and one factor of n plus 1 survive.
\[ \frac{a_{n+1}}{a_n} = \frac{10}{n+1} \]
Take the limit
Why: The ratio eventually drops below any positive number.
\[ \rho = \lim_{n\to\infty}\frac{10}{n+1} = 0 < 1 \;\Longrightarrow\; \text{converges} \]
Figure (svg): Bars showing ten to the n over n factorial for n from 1 to 30. They rise to a peak of about 2756 at n equals 9 and 10, then fall steeply and are almost invisible by n equals 25. A dashed vertical line at n equals 10 marks where the ratio of neighbours drops below 1.
Locate the peak
Why: The ratio is at least 1 exactly while n is at most 9, so the terms climb until then and fall after.
\[ \frac{10}{n+1} \ge 1 \iff n \le 9, \qquad a_9 = a_{10} \approx 2755.7 \]
Check with partial sums
Why: The sum is e to the 10 minus 1; twenty terms are still well short, forty are correct to eight figures.
\[ S_{20} \approx 21990.48, \quad S_{40} \approx 22025.46579, \quad e^{10} - 1 \approx 22025.46579 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 522 — Exercise 318
The algebra is identical to the previous example with ten in place of two, and the limit is again zero. The interesting part is the picture.
Each bar is ten over n plus one times the one before it. While that factor is at least one, which happens up to n equal to nine, the bars grow. At n equal to nine the factor is exactly one, so the ninth and tenth terms are equal. From then on the factor is below one and keeps shrinking, so the bars fall away faster and faster.
The check shows how the sum behaves. Twenty terms still fall short by about thirty-five, because the terms around fifteen and twenty are still sizeable. Forty terms match e to the ten minus one to eight figures. A large early hump delays the convergence; it never prevents it.
Worked example
\[ \sum_{n=1}^{\infty}\frac{n^n}{n!} \]
Form the ratio
Why: Next term over this term, flipped into a product.
\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n} \]
Cancel the factorials
Why: The factorial ratio leaves one over n plus 1.
\[ = \frac{(n+1)^{n+1}}{(n+1)\,n^n} \]
Cancel one factor of n plus 1
Why: The exponent drops from n plus 1 to n.
\[ = \frac{(n+1)^{n}}{n^{n}} = \left(\frac{n+1}{n}\right)^{n} \]
Split the fraction
Why: n plus 1 over n is 1 plus one over n.
\[ = \left(1 + \frac1n\right)^{n} \]
Take the limit
Why: This is the limit that defines e.
\[ \rho = \lim_{n\to\infty}\left(1 + \frac1n\right)^{n} = e \approx 2.718 > 1 \;\Longrightarrow\; \text{diverges} \]
Figure (svg): Two panels. Left: the terms n to the n over n factorial for n from 1 to 10 on a logarithmic scale, rising in a nearly straight line from 1 past 1000. Right: the ratios, one plus one over n all to the n, rising from 2 toward a dashed line at e, about 2.718, well above the line at 1.
Check with the divergence test
Why: A ratio above one means the terms grow, so they cannot tend to zero.
\[ a_1, \dots, a_5 = 1,\; 2,\; 4.5,\; 10.67,\; 26.04 \;\Longrightarrow\; a_n \not\to 0 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, pp. 511-512 — Example 5.23b
Here both the top and the bottom grow very quickly, and it is not obvious which wins. The ratio test settles the contest. After the factorials cancel and one factor of n plus one is removed, the ratio becomes n plus one over n, all to the power n.
Rewrite that base as one plus one over n and you have the limit that defines e, from Section 5.1. The ratio tends to e, about 2.718, which is above one, so the series diverges.
The left panel uses a log scale, where multiplying by the same factor each step shows up as a straight line: the terms are growing by nearly a factor of e every time. The check is the divergence test: terms that grow cannot possibly tend to zero. Whenever the ratio test says diverges, this check is always available.
Worked example
\[ \sum_{n=1}^{\infty}\frac{(-1)^n(n!)^2}{(2n)!} \]
Take the absolute value of the ratio
Why: The sign alternates, and the ratio test only looks at sizes.
\[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{((n+1)!)^2}{(2n+2)!}\cdot\frac{(2n)!}{(n!)^2} \]
Cancel the squared factorials
Why: Each factorial of n plus 1 over the factorial of n leaves n plus 1.
\[ \frac{((n+1)!)^2}{(n!)^2} = (n+1)^2 \]
Cancel the doubled factorials
Why: Going from 2n to 2n plus 2 adds two new factors.
\[ \frac{(2n)!}{(2n+2)!} = \frac{1}{(2n+2)(2n+1)} \]
Combine
Why: Put the two pieces together.
\[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)(n+1)}{(2n+2)(2n+1)} \]
Take the limit
Why: Expand; the leading coefficients are 1 on top and 4 below.
\[ \rho = \lim_{n\to\infty}\frac{n^2+2n+1}{4n^2+6n+2} = \frac14 < 1 \;\Longrightarrow\; \text{converges absolutely} \]
Figure (svg): The partial sums of the series minus one to the n times n factorial squared over 2n factorial, for k from 1 to 10, joined by a zigzag. They alternate above and below a dashed line at about minus 0.3722 and close in on it within a few steps.
Check the rate
Why: Consecutive step sizes shrink by the ratio formula's values, one third, then 0.3, 0.29, 0.28, heading for one quarter.
\[ \frac{0.1667}{0.5} \approx 0.33, \quad \frac{0.05}{0.1667} = 0.30, \quad \frac{0.00397}{0.01429} \approx 0.28 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 512 — Example 5.23c
This series alternates in sign, but the ratio test does not care: it works with the absolute value of the ratio, and the sign disappears in the first line.
The factorials need care. The squared factorials cancel to n plus one, squared. For the other pair, going from the factorial of 2n to the factorial of 2n plus 2 adds two new factors, 2n plus 1 and 2n plus 2, so their quotient is one over their product. Put the pieces together, expand, and compare leading coefficients: one on top, four below.
A limit of one quarter means absolute convergence. The picture shows what that looks like. The partial sums zigzag because of the alternating sign, and each step is roughly a quarter to a third of the one before, so the zigzag closes in on the sum very quickly.
Worked example
\[ \sum_{n=1}^{\infty}\frac{n^3}{3^n} \]
Form the ratio
Why: Next over current, flipped into a product.
\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^3}{3^{n+1}}\cdot\frac{3^n}{n^3} \]
Separate the power of three from the cube
Why: Three to the n over three to the n plus 1 is one third.
\[ = \frac13\left(\frac{n+1}{n}\right)^{3} = \frac13\left(1 + \frac1n\right)^{3} \]
Take the limit
Why: The one over n vanishes.
\[ \rho = \frac13(1 + 0)^3 = \frac13 < 1 \;\Longrightarrow\; \text{converges} \]
Check numerically
Why: The ratio is still above 1 at n equal to 2, drops under one half by n equal to 7, and is 0.343 at n equal to 100; the partial sums settle at 4.125.
\[ \frac{a_8}{a_7} \approx 0.498, \quad \frac{a_{101}}{a_{100}} \approx 0.343, \quad S_{200} \approx 4.1250 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 512 — Checkpoint 5.21
A power of n over an exponential. The ratio splits cleanly into two parts: the exponential contributes one third, and the cube contributes one plus one over n, cubed, which tends to one.
That second part is worth noticing, because it happens every time. Any fixed power of n contributes a factor tending to one in the ratio. Only exponentials and factorials change the limit. So here rho is one third, and the series converges.
The check shows that the ratio need not be below one from the start. At n equal to two the ratio is still above one, since the terms are still growing. By n equal to seven it is below one half, and at one hundred it is close to one third. The partial sums settle at 4.125.
Trap
Testing the sum of two to the n over n factorial:
\[ \frac{a_n}{a_{n+1}} = \frac{n+1}{2} \to \infty \]
\[ \Longrightarrow\; \text{diverges?} \]
Wrong. The ratio is upside down.
The ratio test divides the next term by the current one, like the r of a geometric series. Inverting the ratio inverts every verdict: infinity turns into zero.
\[ \frac{a_{n+1}}{a_n} = \frac{2}{n+1} \to 0 < 1 \]
This slip is easy to make because both orders are ratios of neighbouring terms, and the algebra looks equally tidy either way. The wrong line divides the current term by the next one and gets a ratio that grows without bound.
The protection is to remember where the test comes from. It imitates a geometric series, where you multiply by r to get the next term. So the ratio is next over current. If you are ever unsure, test your formula on a geometric series with ratio one half: the correct orientation gives one half, the wrong one gives two.
Sorting
Sort into buckets
Compute each ratio limit in your head, then sort the series by the ratio test's verdict.
Work out each ratio limit before dragging. Look for the parts of the term that change the limit, the exponentials and factorials, and remember that any power of n contributes a factor of one.
Three to the n over n factorial gives zero. The factorial of n over n to the n gives one over e, the same limit as in Example 5.23(b) turned upside down. Both converge absolutely. The factorial over five to the n gives a ratio that grows without bound, and minus three to the n over n squared has ratios of size tending to three, so both diverge.
The last two are built only from powers of n, so their ratio limits are both one and the test is silent. That is not a failure of your calculation; it is the test telling you to use a different tool. The first of them converges by comparison with the p-series for p equal to two.
Section
Part 2
Concept
Suppose rho is less than 1. There is room between them, so choose a number R strictly in between, and let epsilon be the gap from rho up to R.
\[ 0 \le \rho < R < 1, \qquad \varepsilon = R - \rho > 0 \]
The ratios converge to rho, so past some index N every ratio is within epsilon of rho, and none of them can reach R.
\[ \left|\frac{a_{n+1}}{a_n}\right| < \rho + \varepsilon = R \quad \text{for all } n \ge N \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 509 — proof of Theorem 5.16, part i
Now for why the test works. The proof is short, and every step uses something you already have, so it is worth following once properly.
The whole idea is to make room. If rho is below one, there is a gap between them, and you can choose a number R inside the gap. Think of R as a safety margin: a fixed ratio that is still below one, but slightly bigger than rho.
The definition of a limit then does the work. Because the ratios approach rho, they eventually stay within any distance you like of it. Choose that distance to be the gap from rho up to R, and past some index N every ratio is below R. From that point on, every step multiplies the size of the term by less than R.
Concept
Multiply out one step at a time, starting from the term at N.
\[ |a_{N+1}| < R\,|a_N| \]
\[ |a_{N+2}| < R\,|a_{N+1}| < R^2|a_N| \]
\[ |a_{N+3}| < R\,|a_{N+2}| < R^3|a_N| \]
\[ |a_{N+k}| < R^k|a_N| \quad \text{for every } k \ge 1 \]
Each step multiplies by less than R, so after k steps the term is below R to the k times where it started.
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 510 — the chain of inequalities
Apply the bound one step at a time. The term after N is less than R times the term at N. The next one is less than R times that, so less than R squared times the term at N. Keep going and after k steps you are below R to the k times the term at N.
Look at the last line and notice what it says. The tail of the series, term by term, is smaller than a geometric sequence with ratio R. Since R was chosen below one, that geometric sequence shrinks to zero, and its series converges.
This is the moment the comparison is manufactured. You did not have to guess a comparison series, as in Section 5.4. The series built it for you out of its own ratios.
Picture it
Figure (svg): On a logarithmic scale, dots show the terms n cubed over 3 to the n for n from 1 to 20: they rise to 1 at n equals 3, then fall along a nearly straight line. From n equals 7 a dashed straight line, the geometric sequence a sub 7 times one half to the k, runs above every later dot.
Checkpoint 5.21's terms, on a log scale. Its ratio limit is one third; choose R equal to one half, and from N equal to 7 every ratio is below it. Every later term sits under the dashed geometric bound.
Here is the proof drawn with real numbers, using Checkpoint 5.21. The vertical scale is logarithmic, so a geometric sequence appears as a straight line, falling by the same amount at every step.
The ratio limit for this series is one third. Choose R equal to one half, safely between one third and one. The ratios fall below one half from n equal to seven, so from there the dashed geometric bound, the seventh term times one half to the k, lies above every later dot.
Notice that the early terms do whatever they like: they rise at first and peak at n equal to three. The proof simply ignores them. Only the tail needs to sit under the line.
Concept
\[ \sum_{k=1}^{\infty}|a_{N+k}| < \sum_{k=1}^{\infty}R^k|a_N| = \frac{R\,|a_N|}{1-R} \]
The right side is a geometric series with ratio R below 1, so it converges, and the comparison test makes the tail of absolute values converge.
\[ \sum_{n=1}^{\infty}|a_n| = \underbrace{\sum_{n=1}^{N}|a_n|}_{\text{finite}} + \underbrace{\sum_{n=N+1}^{\infty}|a_n|}_{\text{converges}} \]
A finite head plus a convergent tail converges, so the series converges absolutely. Nothing new was used: a limit, a geometric series and the comparison test.
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 510 — end of the proof of part i
The first line adds up the chain of inequalities. The tail of absolute values is term by term smaller than a geometric series whose first term is R times the term at N, with ratio R. That geometric series has a finite sum, so by the comparison test the tail converges.
The second line puts the head back on. The first N terms are just a finite sum, a number, and adding a number to a convergent series gives a convergent series. So the series of absolute values converges, which is what absolute convergence means.
Step back and notice that nothing new was needed: the definition of a limit from Section 5.1, the geometric series from Section 5.2, and the comparison test from Section 5.4. The ratio test is those three tools packaged together.
Intuition
Now pick R between 1 and rho. The same chain runs the other way.
\[ |a_{N+k}| > R^k|a_N| \to \infty \]
Terms that grow without bound cannot tend to zero, so the series itself diverges, not merely its absolute values. The divergence half of the ratio test is the divergence test in disguise.
And at rho equal to 1 there is no room for an R on either side, so the proof has nothing to compare with. That is where the silence comes from.
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 510 — proof of part ii
For rho above one, choose R between one and rho instead. The same argument, with every inequality reversed, shows that the terms eventually grow at least as fast as a geometric sequence with ratio R, which is above one. So their size runs off to infinity.
That has a useful consequence. Terms that do not tend to zero fail the divergence test, so the series itself diverges, not merely the series of absolute values. The ratio test's divergent verdict is therefore full divergence, with no conditional-convergence loophole.
At rho equal to one, the argument breaks: there is no room for an R strictly between rho and one on either side. The proof has nothing to compare with, and that, not some oversight, is the origin of the silent case.
Matching
Match the pairs
Why: Each line uses one earlier tool: the gap below one, the definition of a limit (Section 5.1), repeated multiplication, the geometric series (Section 5.2) with the comparison test (Section 5.4), and the fact that finitely many terms never affect convergence.
Try to match each line before checking. If you can say why each step is true, you can rebuild the proof on your own and you will never misremember what the test concludes.
The chain goes: room below one, then the definition of the limit to trap the ratios below R, then multiplying the inequalities together, then comparing with a geometric series, then adding back the finite head. Five short steps, each one a tool from earlier in the chapter.
Section
Part 3
Concept
Suppose the nth root of the nth term settles at a number rho. Then for large n each term is roughly rho to the n.
\[ \sqrt[n]{|a_n|} \to \rho \;\Longrightarrow\; |a_n| \approx \rho^{n} \]
\[ |a_N| + |a_{N+1}| + \cdots \approx \rho^{N} + \rho^{N+1} + \cdots \]
The right side is a geometric series with ratio rho. So the verdicts are the same as before: below 1 converges absolutely, above 1 diverges. The root test reads the ratio off a single term instead of two neighbours.
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 512 — the idea of the root test
The root test starts from a slightly different question. Instead of comparing each term with its neighbour, ask what number, raised to the power n, gives the nth term. That number is the nth root of the term.
If those roots settle at rho, then far out each term is roughly rho to the n, and the tail looks like a geometric series with ratio rho. The same conclusions follow as before: below one converges absolutely, above one diverges.
So the two tests measure the same thing, the effective geometric ratio of the tail. They just extract it differently: the ratio test from two neighbouring terms, the root test from a single term. When both limits exist, they are in fact equal.
Concept
Root test — Take the limit rho of the nth root of the absolute value of the nth term. Below one, the series converges absolutely; above one or infinite, it diverges; exactly one, the test gives no information.
\[ \rho = \lim_{n\to\infty}\sqrt[n]{|a_n|} \]
It is made for terms that are something raised to the power n, because then the nth root simply removes the power:
\[ |a_n| = (b_n)^{n} \;\Longrightarrow\; \sqrt[n]{|a_n|} = b_n \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 513 — Theorem 5.17
The statement mirrors the ratio test exactly, with the nth root of the absolute value of the nth term in place of the ratio. The same three cases, the same absolute convergence, and the same silence at one.
The second half of the slide tells you when to use it. If the term is some expression raised to the power n, the nth root simply removes that power, and you are left with the limit of the expression. That is usually a limit you can do in one line.
By contrast, if you tried the ratio test on such a term, you would get a ratio of two expressions raised to different powers, which is usually much messier. Choosing the right test is mostly a matter of looking for a power n on the whole term.
Notation
Annotate
On: \( \rho = \lim_{n\to\infty}\sqrt[n]{|a_n|}, \quad \rho < 1 \;\Longrightarrow\; \sum|a_n| \text{ converges} \)
The first two annotations are about what the root applies to. The index of the root changes with n, so each term gets its own root. And the root is of the whole term, with the sign removed, not just of the part that happens to carry the power n.
The last annotation is the fact you will use more than any other. The nth root of n tends to one, and so does the nth root of n squared, or n cubed, or any fixed power of n. So powers of n are invisible to the root test, just as they are to the ratio test.
That explains at once why every p-series gives rho equal to one under both tests, which is the subject of Part 4.
Worked example
\[ \sum_{n=1}^{\infty}\left(\frac{n^2+3n}{4n^2+5}\right)^{n} \]
Spot the shape
Why: The whole term is a positive fraction raised to the power n.
\[ a_n = (b_n)^{n}, \quad b_n = \frac{n^2+3n}{4n^2+5} \]
Take the nth root
Why: The root cancels the power exactly.
\[ \sqrt[n]{a_n} = \frac{n^2+3n}{4n^2+5} \]
Divide top and bottom by n squared
Why: The highest power in the denominator.
\[ \sqrt[n]{a_n} = \frac{1 + 3/n}{4 + 5/n^2} \]
Take the limit
Why: Both small fractions vanish.
\[ \rho = \frac{1+0}{4+0} = \frac14 < 1 \;\Longrightarrow\; \text{converges absolutely} \]
Figure (svg): On a logarithmic scale, dots show the terms of the series with nth term n squared plus 3n over 4n squared plus 5, all to the n, for n from 1 to 20, and a dashed line shows one quarter to the n. After the first few terms the dots run parallel to the line, a constant distance above it.
Check the root sequence numerically
Why: At n equal to 10, 100 and 1000 the root closes in on one quarter from above.
\[ \sqrt[n]{a_n}: \quad 0.3210,\; 0.2575,\; 0.2507 \;\to\; 0.25 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 513 — Example 5.24a
The term is a single fraction raised to the power n, which is exactly the shape the root test is built for. Take the nth root and the power vanishes, leaving just the fraction.
What remains is a limit of a rational function, which you handle by dividing top and bottom by the highest power in the denominator. The leading coefficients give one quarter, below one, so the series converges absolutely.
The figure explains what the root test is really measuring. On a log scale, one quarter to the n is a straight line, and after the first few terms the dots run parallel to it. The terms are roughly a constant multiple of one quarter to the n, so they fall at exactly that rate. The numerical check shows the roots approaching one quarter from above.
Worked example
The book prints this sum from n equal to 1, but ln 1 is 0 and that term is undefined, so start at 2.
\[ \sum_{n=2}^{\infty}\frac{n^n}{(\ln n)^n} \]
Write the term as one power
Why: Both numerator and denominator are raised to the n.
\[ \frac{n^n}{(\ln n)^n} = \left(\frac{n}{\ln n}\right)^{n} \]
Take the nth root
Why: The power disappears.
\[ \sqrt[n]{a_n} = \frac{n}{\ln n} \]
Apply L'Hôpital's rule
Why: Top and bottom both grow without bound.
\[ \lim_{x\to\infty}\frac{x}{\ln x} = \lim_{x\to\infty}\frac{1}{1/x} = \lim_{x\to\infty}x = \infty \]
Conclude
Why: An infinite root limit is in the divergent case.
\[ \rho = \infty \;\Longrightarrow\; \text{diverges} \]
Check with the terms
Why: They grow from the start, so the divergence test agrees.
\[ a_2 \approx 8.33, \quad a_3 \approx 20.36, \quad a_4 \approx 69.31, \quad a_5 \approx 289.4 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 513 — Example 5.24b
Before starting, notice the misprint in the book. It starts this sum at n equal to one, but the natural log of one is zero, so the first term would divide by zero. Starting at two fixes it and does not change whether the series converges.
Both parts of the term are raised to the n, so combine them into one power and take the nth root. You are left with n over ln n, a limit of the form infinity over infinity, which L'Hôpital's rule turns into the limit of x, which is infinite.
An infinite root limit is in the divergent case. The check is immediate: the terms are 8.3, 20.4, 69.3 and 289.4 for n from two to five, growing fast, so they cannot tend to zero.
Worked example
\[ \sum_{n=1}^{\infty}\frac{1}{n^n} \]
Take the nth root
Why: The term is one over n, all raised to the n.
\[ \sqrt[n]{\frac{1}{n^n}} = \sqrt[n]{\left(\frac1n\right)^{n}} = \frac1n \]
Take the limit
Why: The root sequence goes all the way to zero.
\[ \rho = \lim_{n\to\infty}\frac1n = 0 < 1 \;\Longrightarrow\; \text{converges} \]
Figure (svg): The terms one over n to the n drawn as bars stacked into a staircase of partial sums for k from 1 to 7: the first step reaches 1, the second 1.25, and by the fourth step the staircase is flat against a dashed line at about 1.2913.
Check with a geometric comparison
Why: For n at least 2, n to the n is at least 2 to the n, so the terms sit under a convergent geometric series; five terms already give 1.29126.
\[ \frac{1}{n^n} \le \left(\frac12\right)^{n} \;(n \ge 2), \qquad S_5 \approx 1.29126 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 514 — Checkpoint 5.22
The term is one over n, raised to the power n, so the nth root is simply one over n. Its limit is zero, and the series converges.
A root limit of zero means the terms shrink faster than any geometric sequence at all, and the picture shows it. The first riser is a quarter, the next one twenty-seventh, then one two hundred fifty-sixth, and after that the staircase is flat to the eye.
The check gives an independent argument. From n equal to two on, n to the n is at least two to the n, so each term is at most one half to the n, and a geometric series with ratio one half converges. The sum is about 1.29129, and five terms already agree with it to four decimal places.
Worked example
\[ \sum_{n=1}^{\infty}\left(1 - \frac1n\right)^{n^2} \]
Split the exponent
Why: n squared is n times n, so the term is a power of a power.
\[ \left(1 - \frac1n\right)^{n^2} = \left[\left(1 - \frac1n\right)^{n}\right]^{n} \]
Take the nth root
Why: The outer power disappears.
\[ \sqrt[n]{a_n} = \left(1 - \frac1n\right)^{n} \]
Use the companion of the limit for e
Why: With a minus sign inside, the limit is e to the minus 1.
\[ \lim_{n\to\infty}\left(1 - \frac1n\right)^{n} = e^{-1} \approx 0.368 \]
Conclude
Why: One over e is below one.
\[ \rho = \frac1e < 1 \;\Longrightarrow\; \text{converges} \]
Figure (svg): Dots showing one minus one over n, all to the n, for n from 1 to 40: they start at 0, rise quickly, and level off at a dashed line at one over e, about 0.368, far below the dashed line at 1.
Check numerically
Why: The roots at n equal to 10, 100 and 1000 approach 0.3679.
\[ \sqrt[n]{a_n}: \quad 0.3487,\; 0.3660,\; 0.3677 \;\to\; 0.3679 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 522 — Exercise 341
The exponent is n squared, not n, and that is the whole trick. Write n squared as n times n, and the term becomes a power of a power: one minus one over n raised to the n, and all of that raised to the n.
Now the nth root removes the outer power, leaving one minus one over n, to the n. That is the partner of the limit that defines e: with a minus sign inside, it tends to one over e, about 0.368.
The picture shows the roots rising from zero and levelling off at one over e, far below the dividing line at one. So rho is one over e and the series converges absolutely. Try the ratio test on this series and you will see why the root test was the better choice.
Trap
Testing the sum of n squared over two to the n:
\[ \sqrt[n]{\frac{n^2}{2^n}} = \frac{n^2}{2} \]
\[ \to \infty \;\Longrightarrow\; \text{diverges?} \]
Wrong. Only the denominator was rooted.
The nth root applies to the whole term. The nth root of n tends to 1, so the power of n contributes nothing in the limit.
\[ \sqrt[n]{\frac{n^2}{2^n}} = \frac{(\sqrt[n]{n})^2}{2} \]
\[ \to \frac{1^2}{2} = \frac12 < 1 \]
The wrong line looks natural because the two to the n is the obvious target of an nth root. But the root applies to the whole fraction, so the n squared in the numerator must be rooted too.
Rooted properly, the numerator becomes the nth root of n, squared. The nth root of n tends to one, so the numerator tends to one and the root limit is one half. The series converges.
Keep the rule from the notation slide close: any fixed power of n contributes a factor of one in the limit. So if a power of n ever seems to be deciding a root-test problem, you have almost certainly forgotten to root it.
Step zero
\[ (a)\; \sum_{n=1}^{\infty}\frac{n!}{n^n} \qquad (b)\; \sum_{n=1}^{\infty}\left(\frac{2n}{3n+1}\right)^{n} \]
Discussion prompt
Before any algebra: which test suits (a), which suits (b), and what feature of the term tells you?
Write your choice and your reason before revealing. The point of this slide is the thirty seconds of reading before any algebra, which is where the right test is chosen.
Series (a) has a factorial. Factorials cancel beautifully in a ratio, down to a single factor, so the ratio test is the natural choice. It leaves n over n plus one, to the n, which tends to one over e. Series (b) is a single fraction raised to the n, so the root test removes the power and leaves two n over three n plus one, which tends to two thirds.
Could you do it the other way round? In principle, yes, but the nth root of a factorial is hard to handle, and the ratio for (b) involves different powers of different fractions. The term's shape tells you which test will be easy.
Comparison
Comparison matrix
| ratio test | root test | |
|---|---|---|
| what you compute | |next term / this term| | nth root of |aₙ| |
| best when the term has | factorials | a power n on the whole term |
| ρ < 1 | converges absolutely | converges absolutely |
| ρ > 1 | diverges | diverges |
| ρ = 1 | no information | no information |
| ρ for any p-series | 1 | 1 |
Fill in each blank before checking. This table is the comparison you should be able to reproduce from memory, because the two tests are so alike that they are easily confused.
The verdict rows are identical: below one converges absolutely, above one diverges, exactly one says nothing. The differences are all in the first two rows, what you compute and when it is convenient. Factorials favour the ratio test; a power n on the whole term favours the root test.
The last row is the one to remember when choosing. On any p-series, and on anything built only from powers of n, both tests give one. So for those series, reach for a comparison with a p-series instead.
Section
Part 4
Concept
Apply the ratio test to the p-series, for any real p.
\[ \frac{a_{n+1}}{a_n} = \frac{1/(n+1)^p}{1/n^p} = \frac{n^p}{(n+1)^p} \]
\[ = \left(\frac{n}{n+1}\right)^{p} = \left(\frac{1}{1 + 1/n}\right)^{p} \to 1^p = 1 \]
The limit is 1 for every p. Yet the p-series converges when p is above 1 and diverges when p is at most 1. A ratio limit of one is compatible with both verdicts, so the test must say nothing.
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 511 — proof of Theorem 5.16, part iii
This is the book's proof that the silent case really is silent. Apply the ratio test to the p-series. The ratio of neighbours is n over n plus one, raised to the power p, and as n grows n over n plus one tends to one, so the ratio tends to one for every p.
Now combine that with the p-series rule from Section 5.3. When p is two, the series converges. When p is one, it diverges. Both give exactly the same ratio limit. So if the test gave any verdict at rho equal to one, it would be wrong for one of them.
This is not a weakness that a cleverer version of the test could fix. It is a proof that the limit of the ratios simply does not carry enough information when it equals one.
Concept
\[ \ln\rho = \lim_{n\to\infty}\ln\left(\frac1n\right)^{p/n} = \lim_{n\to\infty}\frac{-p\ln n}{n} \]
\[ \lim_{x\to\infty}\frac{-p\ln x}{x} = \lim_{x\to\infty}\frac{-p/x}{1} = 0 \;\Longrightarrow\; \rho = e^{0} = 1 \]
Figure (svg): Two curves for x from 1 to 100: x to the minus one over x and x to the minus two over x. Both dip below 1 near x equals 3, then creep slowly back up toward the dashed line at 1, which they approach but never cross.
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, pp. 512-513 — the root test on the p-series
The root test runs into the same wall. The nth root of one over n to the p is one over n, to the power p over n, and that is hard to evaluate directly because both the base and the exponent change. Taking the natural log turns the power into a product: minus p times ln n, over n.
That quotient has the form infinity over infinity, and L'Hôpital's rule gives minus p over x, which tends to zero. So the log of rho is zero, and rho is e to the zero, which is one, for every p.
The picture shows the root sequences for p equal to one and two. Both dip a little and then creep back up toward one, very slowly. The curves look alike, and yet one series diverges and the other converges.
Picture it
Figure (svg): Two panels. Left: the ratios n over n plus 1 and n squared over n plus 1 squared, for n from 1 to 30, both rising toward the dashed line at 1. Right: the partial sums of one over n climbing past 3.9 with no ceiling, and of one over n squared levelling off near 1.6.
On the left, the ratios for one over n and for one over n squared both climb to 1. On the right, their partial sums: one runs away, the other levels off. The limit of the ratios cannot tell them apart.
The left panel shows the ratios of the harmonic series and of the reciprocal squares. Both sets of dots climb toward one, and by n equal to thirty they are nearly indistinguishable.
The right panel shows what those series actually do. The harmonic partial sums keep climbing, past four already, and will never stop. The reciprocal squares level off near 1.6. The same ratio limit leads to opposite verdicts.
Keep this pair in mind as your permanent counterexample. Whenever you are tempted to read a verdict from a ratio or root limit of one, remember that it would have to be right for both of these series at once, and it cannot be.
Error analysis
Annotate
On: \( \sum_{n=1}^{\infty}\frac{1}{n^2}: \quad \rho = \lim\frac{n^2}{(n+1)^2} = 1 \;\Longrightarrow\; \text{diverges} \)
Read the argument and try to spot the error before revealing the annotations. The algebra is fine: the ratio limit of the reciprocal squares really is one.
The error is the final arrow. At rho equal to one the ratio test gives no verdict, so writing diverges is an invention. And it is the wrong invention: this is a p-series with p equal to two, and it converges, famously to pi squared over six.
The correct write-up is one line: the ratio limit is one, so the ratio test is inconclusive. Then switch tools. For anything built from powers of n, the p-series rule or a limit comparison with a p-series is almost always the right next step.
Counterexample
Discussion prompt
A classmate argues: if every ratio of consecutive terms is less than 1, the terms keep shrinking, so the series converges. Find a counterexample, and say which part of the ratio test the claim ignores.
Write your counterexample before revealing. The claim sounds convincing: if each term is smaller than the one before, the terms keep shrinking, so surely the total settles.
The harmonic series breaks it. Each ratio, n over n plus one, is below one, and the terms do keep shrinking. But the ratios creep up toward one, so the terms shrink by less and less each time, and the total still runs off to infinity.
The claim ignored the word limit. The ratio test needs the limit of the ratios to be below one, because only then is there room for a fixed R below one that bounds every ratio from some point on. Ratios that approach one leave no such room, and without a fixed R there is no geometric series to compare with.
Prediction
Predict first
For exactly one of these series the ratio test returns ρ = 1. Which?
Correct: sum of (n² + 1)/(n⁴ + 3)
Why: A term built only from powers of n (a rational function, roots included) always has ratio limit 1, because (n + 1)/n → 1. The others contain an exponential or a factorial: n⁵/3ⁿ gives 1/3, 5ⁿ/n! gives 0 and n!/(2n)! gives 0. The silent one converges, but by limit comparison with 1/n², not by the ratio test.
Commit to one option before revealing. You can decide this without computing anything, just by looking at what kind of pieces each term is built from.
The first term is built only from powers of n. Any power of n gives a ratio factor tending to one, so the whole ratio tends to one and the test is silent. Every other option contains an exponential or a factorial, and those are exactly the pieces that move the ratio away from one.
This is the habit to take into the strategy section: before reaching for the ratio test, ask whether the term has an exponential or a factorial in it. If it has neither, the ratio test will almost certainly return one, and a comparison with a p-series is the better choice.
Tweak it
Parameter explorer
The curve is the base-ten logarithm of the term nᵖ rⁿ (Exercise 369). Its ratio limit is r, whatever p is. Push p up: the hump grows, but does the curve still turn down while r is below 1? Then set r to exactly 1, and above it.
\[ \log_{10}\left(n^{{p}}\,({r})^{n}\right) \]
The graph is on a log scale, so the geometric factor r to the n shows up as a straight line with slope the log of r, and the power n to the p shows up as p times the log of n, which rises steeply at first and then flattens.
Set r below one and push p up. The hump grows taller and moves right, but the straight-line decline always wins in the end, and the curve turns down. That is the ratio test's verdict made visible: the ratio limit is r, whatever p is, so every such series converges.
Now set r to exactly one. The geometric factor disappears and you are left with a pure power of n, where the ratio test is silent and the p-series rule takes over. Push r above one and the terms grow without bound. This is Exercise 369, and the moral is that exponentials always beat powers.
Section
Part 5
Concept
| series or test | conclusion | comment |
|---|---|---|
| divergence test: lim aₙ | lim aₙ ≠ 0 ⟹ diverges | lim aₙ = 0: inconclusive; never proves convergence |
| geometric Σ a rⁿ⁻¹ | |r| < 1: converges to a/(1 − r); |r| ≥ 1: diverges | reindex to the form a + ar + ar² + ⋯ |
| p-series Σ 1/nᵖ | p > 1 converges; p ≤ 1 diverges | p = 1 is the harmonic series |
| alternating Σ (−1)ⁿ⁺¹ bₙ | bₙ decreasing to 0 ⟹ converges | only for alternating series |
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, pp. 517-518 — Table 5.3
This is the first half of the book's summary table, and it holds the series and tests that need the least work. The divergence test costs one limit and can only prove divergence. Geometric series and p-series are the reference families: you decide them by reading off r or p, and you compare everything else against them.
The alternating series test is the odd one out. It is the only test in the chapter that can prove convergence without proving absolute convergence, and it applies only when the signs strictly alternate and the sizes shrink steadily to zero.
Read the comment column as carefully as the conclusions. Each comment is a limitation, and knowing a test's limitations is how you avoid using it where it does not apply.
Concept
| test | conclusion | comment |
|---|---|---|
| comparison | aₙ ≤ bₙ, Σbₙ converges ⟹ Σaₙ converges (and the reverse for divergence) | needs a suitable known series |
| limit comparison | L = lim aₙ/bₙ, 0 < L < ∞ ⟹ same fate | often easier than direct comparison |
| integral | Σaₙ and the integral of f from N share a fate | f positive, continuous, decreasing; must integrate |
| ratio | ρ < 1 converges absolutely; ρ > 1 diverges | factorials and exponentials; silent at 1 |
| root | ρ < 1 converges absolutely; ρ > 1 diverges | terms of the form (bₙ)ⁿ; silent at 1 |
Comparison, limit comparison and the integral test need nonnegative terms. For a series with negative terms, apply them to the absolute values to test absolute convergence.
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, pp. 516-519 — Table 5.3 and the note before it
The second half holds the tests that do real work. The comparison tests need a known series to compare with, usually a p-series or a geometric series. The integral test needs a positive, continuous, decreasing function you can integrate. The ratio and root tests need nothing extra, but they go silent at one.
Notice the sentence below the table. Comparison, limit comparison and the integral test all require terms that are not negative. That does not make them useless for other series: apply them to the absolute values, and a convergent result proves absolute convergence.
Together, the two halves are the whole toolkit of Chapter 5. The strategy on the next slide is the order in which to pick them up.
Picture it
Figure (svg): A flow chart of five question boxes in a row joined by arrows labelled no: familiar series; alternating; like a p-series or geometric series; factorial or power; terms tend to zero. Each box has an arrow labelled yes pointing down to the tool to use: read off p or r; the alternating series test, or continue with the absolute values; comparison or limit comparison; the ratio test, or the root test when the term is something to the n; and for the last box, if the terms do not tend to zero the series diverges, otherwise try the integral test.
Ask the questions in the book's order. The first yes names the tool; a no passes you to the next question. The divergence test and the integral test wait at the end.
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 514 — Problem-Solving Strategy: Choosing a Convergence Test
The book's strategy is five questions, asked in order, and the flow chart lays them out left to right. At each box, a yes sends you down to the tool to use, and a no passes you on to the next question.
The order is not arbitrary. The first question costs nothing: you either recognise the series or you do not. The second decides which question you are answering, convergence or absolute convergence. The third and fourth are quick whenever the term's shape invites them.
The fifth box holds the last resorts. If the terms do not tend to zero, the series diverges in one line. If they do, the integral test is left, which is powerful but only works when you can actually do the integral.
Intuition
| the term looks like | reach for | because |
|---|---|---|
| a ratio of powers of n, roots allowed | limit comparison with a p-series | only the leading powers matter |
| a constant to the n, plus smaller pieces | limit comparison with a geometric series | the exponential dominates |
| contains n! | ratio test | consecutive factorials cancel |
| one expression raised to the n | root test | the nth root removes the power |
| (−1)ⁿ times a positive bₙ | alternating series test | then decide absolute convergence |
| terms plainly not tending to 0 | divergence test | one limit ends the problem |
The strategy is really a habit of reading the term before computing anything. Ten seconds of looking saves ten minutes of the wrong algebra.
The strategy becomes fast once you learn to read the shape of a term at a glance. Each row of the table is a shape, the tool it calls for, and the reason.
Powers of n only, including roots, mean a limit comparison with the p-series of the leading powers. A constant to the n means a comparison with a geometric series. A factorial calls for the ratio test, a whole term to the n for the root test, an alternating sign for the alternating series test, and terms that obviously do not reach zero for the divergence test.
Practise this by covering the right-hand columns of the table and naming the tool from the shape alone. The worked examples that follow are the book's four, plus one of its checkpoints and one series where only the last resort works.
Worked example
\[ \sum_{n=1}^{\infty}\frac{n^2+2n}{n^3+3n^2+1} \]
Steps 1 and 2: not familiar, not alternating
Why: Every term is positive; it is neither a p-series nor geometric.
\[ a_n > 0 \quad \text{for all } n \]
Step 3: keep the leading powers
Why: For large n only the top power above and below matters.
\[ \frac{n^2+2n}{n^3+3n^2+1} \approx \frac{n^2}{n^3} = \frac1n \]
Divide the term by one over n
Why: The limit comparison quotient.
\[ \frac{a_n}{1/n} = \frac{n^3+2n^2}{n^3+3n^2+1} \]
Take the limit
Why: Divide top and bottom by n cubed.
\[ L = \lim_{n\to\infty}\frac{1 + 2/n}{1 + 3/n + 1/n^3} = 1 \]
Conclude
Why: L is positive and finite, and the harmonic series diverges.
\[ 0 < L < \infty, \quad \sum\frac1n = \infty \;\Longrightarrow\; \text{diverges} \]
Check the limit numerically
Why: The quotient approaches 1 as it should.
\[ \frac{a_n}{1/n}: \quad 0.922 \;(n = 10), \quad 0.990 \;(n = 100) \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, pp. 514-515 — Example 5.25a
Run the strategy aloud. It is not a p-series or geometric, and it is not alternating. Step three asks whether it resembles one of the reference families, and for a rational function it always does: keep only the leading powers, and n squared over n cubed is one over n.
That suggests a limit comparison with the harmonic series. Divide the term by one over n, simplify, and the limit is one. A positive, finite limit means the two series share a fate, and the harmonic series diverges, so this one does too.
Notice what would have happened with the ratio test: a rational function has ratio limit one, so you would have done the algebra for nothing. The strategy places comparison before ratio precisely to save you that. The check simply confirms numerically that the quotient approaches one.
Worked example
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}(3n+1)}{n!} \]
Step 2: alternating, and absolute convergence is wanted
Why: Drop the sign and test the absolute values.
\[ \sum_{n=1}^{\infty}|a_n| = \sum_{n=1}^{\infty}\frac{3n+1}{n!} \]
Step 4: a factorial, so the ratio test
Why: The next term replaces n by n plus 1 everywhere, including inside 3n plus 1.
\[ \frac{|a_{n+1}|}{|a_n|} = \frac{3(n+1)+1}{(n+1)!}\cdot\frac{n!}{3n+1} \]
Simplify
Why: Three times n plus 1, plus 1, is 3n plus 4; the factorials leave n plus 1.
\[ = \frac{3n+4}{(n+1)(3n+1)} \]
Take the limit
Why: Degree one on top, degree two below.
\[ \rho = \lim_{n\to\infty}\frac{3n+4}{3n^2+4n+1} = 0 < 1 \]
Conclude
Why: Absolute convergence implies convergence.
\[ \text{converges absolutely, hence converges} \]
Check, and correct the book
Why: The book prints the new numerator as 3n plus 3; it is 3n plus 4, and the limit is 0 either way. Both sums are known exactly.
\[ \sum\frac{3n+1}{n!} = 4e - 1 \approx 9.8731 \]
\[ \sum\frac{(-1)^{n+1}(3n+1)}{n!} = 1 + \frac2e \approx 1.7358 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 515 — Example 5.25b
The series alternates, and the question asks about absolute convergence, so step two tells you to test the absolute values. There is a factorial, so step four says ratio test.
Be careful forming the next term. Every n becomes n plus one, including the one inside three n plus one, which becomes three n plus four. The book prints three n plus three here, a misprint; fortunately the limit is zero either way, because the numerator has degree one and the denominator degree two.
A ratio limit of zero means the absolute values converge, so the series converges absolutely, and therefore converges. The check uses Chapter 6 results: the absolute values sum to four e minus one, about 9.873, and the alternating series to one plus two over e, about 1.736. Both are finite, as the test promised.
Worked example
\[ \sum_{n=1}^{\infty}\frac{e^n}{n^3} \]
Steps 1 to 4
Why: Not familiar, not alternating, no obvious comparison, no factorial; the power is not the whole term.
\[ a_n = \frac{e^n}{n^3} > 0 \]
Step 5: the divergence test
Why: Find the limit of the terms, as a function of x.
\[ \lim_{x\to\infty}\frac{e^x}{x^3} \]
L'Hôpital's rule, three times
Why: Each round lowers the power of x by one; e to the x never changes.
\[ = \lim\frac{e^x}{3x^2} = \lim\frac{e^x}{6x} = \lim\frac{e^x}{6} = \infty \]
Conclude
Why: The terms do not tend to zero.
\[ a_n \to \infty \;\Longrightarrow\; \text{diverges} \]
Figure (svg): Dots showing e to the n over n cubed for n from 1 to 10, over the faint curve e to the x over x cubed. The values dip from 2.72 at n equals 1 to a minimum of about 0.744 at n equals 3, then climb steeply past 22 at n equals 10.
Check with the root test
Why: The nth root of n cubed tends to 1, so the root limit is e, which is above 1: the same verdict.
\[ \sqrt[n]{\frac{e^n}{n^3}} = \frac{e}{(\sqrt[n]{n})^3} \to e > 1 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 515 — Example 5.25c
The first four questions all come up empty. It is not familiar, it is not alternating, there is no obvious comparison, and there is no factorial. There is an exponential, but only in part of the term, so the book moves on to step five: the divergence test.
The limit of e to the x over x cubed is infinity over infinity, and L'Hôpital's rule needs three rounds, each lowering the power of x by one while the exponential stays the same. The limit is infinite, so the terms do not tend to zero and the series diverges.
The picture shows the contest. The cube wins briefly, pulling the terms down to about 0.744 at n equal to three, and then the exponential takes over for good. The check uses the root test instead: the nth root of n cubed tends to one, so the root limit is e, above one. Two tests, one verdict.
Worked example
\[ \sum_{n=1}^{\infty}\frac{3^n}{(n+1)^n} \]
Step 4: the whole term is a power
Why: Both parts are raised to the n.
\[ \frac{3^n}{(n+1)^n} = \left(\frac{3}{n+1}\right)^{n} \]
Take the nth root
Why: The power disappears.
\[ \sqrt[n]{a_n} = \frac{3}{n+1} \]
Take the limit
Why: The denominator grows.
\[ \rho = \lim_{n\to\infty}\frac{3}{n+1} = 0 < 1 \;\Longrightarrow\; \text{converges} \]
Check with a geometric bound
Why: From n equal to 5 the base is at most one half, so the tail sits under a convergent geometric series; the sum is about 3.0901.
\[ n \ge 5: \; \frac{3}{n+1} \le \frac12 \;\Longrightarrow\; a_n \le \left(\frac12\right)^{n} \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, pp. 515-516 — Example 5.25d
Both the numerator and the denominator are raised to the n, so the whole term is one fraction to the n. That is the root test's signal in step four.
The nth root removes the power and leaves three over n plus one, which tends to zero. So the root limit is zero and the series converges, very quickly.
The check makes the geometric comparison explicit. From n equal to five on, three over n plus one is at most one half, so each term is at most one half to the n. The tail sits under a convergent geometric series, which is exactly what the proof of the root test does in general. The sum works out to about 3.090.
Worked example
\[ \sum_{n=1}^{\infty}\frac{2^n}{3^n+n} \]
Step 3: it resembles a geometric series
Why: For large n the lone n is negligible beside 3 to the n.
\[ \frac{2^n}{3^n+n} \approx \frac{2^n}{3^n} = \left(\frac23\right)^{n} \]
Form the limit comparison quotient
Why: Divide by the geometric term.
\[ \frac{a_n}{(2/3)^n} = \frac{2^n}{3^n+n}\cdot\frac{3^n}{2^n} = \frac{3^n}{3^n+n} \]
Take the limit
Why: Divide by 3 to the n; n over 3 to the n tends to 0.
\[ L = \lim_{n\to\infty}\frac{1}{1 + n/3^n} = 1 \]
Conclude
Why: L is positive and finite, and the geometric series converges.
\[ \sum\left(\frac23\right)^{n} \text{ converges} \;\Longrightarrow\; \sum\frac{2^n}{3^n+n} \text{ converges} \]
Figure (svg): Dots showing two to the n over three to the n plus n, for n from 1 to 12, lying almost exactly on the curve two thirds to the x; at n equals 1 the dot is at 0.5 and the curve at 0.667, and by n equals 5 they are indistinguishable.
Check the sizes
Why: The quotient is 0.75, 0.980 and 0.99983 at n equal to 1, 5 and 10; the sum sits just under the geometric series' total.
\[ \sum_{n=1}^{\infty}\frac{2^n}{3^n+n} \approx 1.7100 < 2 = \sum_{n=1}^{\infty}\left(\frac23\right)^{n} \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 516 — Checkpoint 5.23
The checkpoint asks which test is best and why. The term looks almost geometric: the lone n in the denominator is tiny beside three to the n, so the term behaves like two thirds to the n. That points to step three and a limit comparison with a geometric series.
Dividing by two thirds to the n leaves three to the n over three to the n plus n. Divide through by three to the n, and since n over three to the n tends to zero, the limit is one. The geometric series with ratio two thirds converges, so this series does too.
The picture shows how closely the terms hug the geometric curve: by n equal to five they are indistinguishable. The ratio test would also work here, with a ratio limit of two thirds, but the algebra is messier. The sum is about 1.710, just under the geometric total of two.
Worked example
One of this section's problem seeds; no test in Steps 1 to 4 fits.
\[ \sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2} \]
Run the strategy to the end
Why: Not familiar, not alternating, no clean comparison, no factorial or power to the n; the terms tend to 0 and the ratio is silent.
\[ a_n \to 0, \qquad \frac{a_{n+1}}{a_n} \to 1 \]
Check the integral test's hypotheses
Why: Positive, continuous and decreasing for x at least 2, since x and ln x both increase.
\[ f(x) = \frac{1}{x(\ln x)^2}, \quad x \ge 2 \]
Substitute u equal to ln x
Why: Then du is dx over x.
\[ \int\frac{dx}{x(\ln x)^2} = \int\frac{du}{u^2} = -\frac1u = -\frac{1}{\ln x} \]
Evaluate the improper integral
Why: One over ln b tends to 0.
\[ \int_2^{\infty}\frac{dx}{x(\ln x)^2} = \lim_{b\to\infty}\left(\frac{1}{\ln 2} - \frac{1}{\ln b}\right) = \frac{1}{\ln 2} \]
Figure (svg): The curve y equals one over x times ln x squared, for x from 2 to 20, falling from about 1.04 at x equals 2 toward zero, with the region beneath it shaded from x equals 2 onward; the label gives the total area from 2 to infinity as one over ln 2, about 1.443.
Verify with the remainder estimate
Why: The partial sum to 1000 is 1.964988, and the tail lies between one over ln 1001 and one over ln 1000.
\[ 2.10973 < \sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2} < 2.10975 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.3, p. 475 — the integral test, as Step 5 of the strategy
This is one of the section's problem seeds, chosen because every earlier step of the strategy fails. It is not familiar, not alternating, it has no clean comparison, there is no factorial and no power to the n, and both the ratio and root limits are one. The terms do tend to zero, so the divergence test is silent too.
That leaves the integral test. The function is positive and decreasing from two onward, and the substitution u equals ln x turns the integral into the integral of one over u squared. The improper integral comes out to one over ln 2, finite, so the series converges.
The final line uses the remainder estimate from Section 5.3 as a genuine check. A thousand terms, plus the two tail integrals, trap the sum between 2.10973 and 2.10975. The integral found the verdict; the remainder estimate even pins down the value.
Discrimination
Sort into buckets
Read each term and sort the series by the test you would reach for FIRST.
Sort each series by the test you would try first, and give yourself only a few seconds per item. The skill being practised is reading the shape of the term, not doing the algebra.
Factorials go to the ratio test. Terms that are one expression raised to the n go to the root test. Terms built only from powers of n go to a limit comparison with a p-series. And two of these have terms that plainly do not tend to zero, n over two n plus five and one plus one over n to the n, so the divergence test ends them at once.
If you put one plus one over n to the n in the root test bucket, look again: the root is one plus one over n, which tends to one, so the root test would be silent. Checking the limit of the terms is cheaper and decisive.
Ranking
Put in order
Arrange the book's five questions in the order the strategy asks them.
Why: Recognition costs nothing, so it comes first. The alternating question fixes WHICH question you are answering. Comparison and the ratio or root tests are quick when the term's shape invites them. The divergence test and the integral test close the list: the one can only ever prove divergence, and the other needs an integral you can actually evaluate.
Arrange the questions before checking. The reasoning behind the order matters more than memorising it.
Recognition comes first because it costs nothing. The alternating question comes second because it decides which question you are answering. The comparison and the ratio or root questions come next, and each is quick when the term's shape invites it. The divergence test and integral test come last: the first can only ever prove divergence, and the second needs an integral you can evaluate.
Trap
Testing the alternating harmonic series:
\[ \sum\left|\frac{(-1)^{n+1}}{n}\right| = \sum\frac1n = \infty \]
\[ \Longrightarrow\; \text{diverges?} \]
Wrong. Only the absolute values diverge.
Step 2 of the strategy asks which question you are answering. Divergent absolute values rule out absolute convergence and nothing more; the alternating series test still applies, so the series converges conditionally.
\[ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} = \ln 2 \]
This mistake comes from running step two of the strategy halfway. The wrong line takes absolute values, finds the harmonic series, correctly concludes that it diverges, and then transfers that verdict to the original series.
Divergent absolute values only rule out absolute convergence. The original series might still converge thanks to cancellation between positive and negative terms, and the alternating harmonic series does, to ln 2. So its correct classification is conditionally convergent.
Contrast this with the ratio and root tests, where a divergent verdict really is divergence of the series itself, because the terms fail to tend to zero. The comparison tests applied to absolute values do not carry that guarantee.
Real world
\[ \frac{1}{\pi} = \frac{2\sqrt2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!\,(1103+26390n)}{(n!)^4\,396^{4n}} \]
Discussion prompt
Ramanujan found this series in the early 1900s, and in the 1980s it was used to compute pi to more than 17 million digits. Use the ratio test to explain why each extra term adds about eight correct digits.
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 520 — student project: series converging to pi and one over pi
This series appears in the book's student project at the end of the section. Try the ratio calculation before revealing: the factorial of 4n plus 4 over the factorial of 4n leaves four factors, and each is about 4n for large n, so together they behave like 256 times n to the fourth.
The factorials of n, raised to the fourth power, contribute n plus one to the fourth, and the power of 396 contributes 396 to the fourth. The linear factor in the numerator changes by a ratio tending to one. So the ratio limit is 256 over 396 to the fourth, about one hundred-millionth.
A ratio that small means each term is roughly a hundred-millionth of the one before, so each new term adds about eight correct digits. That is why the first term alone already matches pi to six decimal places, and two terms to about fifteen.
Estimation
\[ \pi = 4 - \frac43 + \frac45 - \frac47 + \cdots = 4\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1} \]
Predict first
Here the ratio test is silent: the ratio (2n − 1)/(2n + 1) tends to 1. The alternating remainder bound says the error after N terms is below 4/(2N + 1). Roughly how many terms guarantee an error below 0.01?
Correct: About 200
Why: You need 4/(2N + 1) < 0.01, so 2N + 1 > 400 and N = 200: at 200 terms the bound is 4/401 ≈ 0.00998, at 199 it is 4/399 ≈ 0.01003. Two hundred terms for two decimals, against eight digits per term for Ramanujan. A ratio limit of 1 is a warning of slow convergence; a tiny ratio limit is a promise of fast convergence.
\[ \frac{4}{2N+1} < 0.01 \iff 2N + 1 > 400 \iff N \ge 200 \]
OpenStax Calculus Volume 2, §5.6 Ratio and Root Tests §5.6, p. 520 — student project, part 1
Commit to a guess before revealing. This series also converges to pi, but it sits at the opposite extreme from Ramanujan's.
Its ratio of neighbours tends to one, so the ratio test is silent: the series converges only conditionally, by the alternating series test. The alternating remainder bound says the error after N terms is below the next term, four over 2N plus 1. Setting that below one hundredth needs N to be two hundred.
Two hundred terms for two decimal places, against eight digits per term. The comparison teaches you to read the ratio limit as a speed as well as a verdict. A limit near zero promises very fast convergence; a limit of one, even when the series does converge, is a warning that it will be slow.
Section
Part 6
Pattern
\[ \rho < 1: \text{ converges absolutely}, \quad \rho > 1: \text{ diverges}, \quad \rho = 1: \text{ switch tests} \]
This is the whole chapter on one slide, in the order you should work. Read the term, recognise it if you can, settle the alternating question, then match the shape of the term to a tool.
Two rules deserve special attention. When the ratio or root test returns one, write inconclusive and change tools; never read a verdict from it. And keep the divergence test in mind throughout: a quick glance at whether the terms tend to zero can end a problem before any heavier test is needed.
With practice the whole procedure takes a few seconds per series, because most of the work is recognition. The checks that follow test exactly that.
Check
Check your understanding
What does the ratio test conclude about the sum of (−2)ⁿ/n³?
Answer: C
Why: The size of the ratio is 2(n/(n + 1))³, which tends to 2. Since ρ = 2 > 1, the terms grow in size and the series diverges; the alternating sign cannot rescue terms that do not tend to zero.
The sign alternates, but the ratio test only looks at size, so the sign plays no part in the calculation. The cube of n contributes a factor tending to one, and the two to the n contributes a factor of two, so the ratio limit is two.
A limit above one means the terms grow in size, so the series diverges. The alternating sign cannot help, because conditional convergence still needs terms tending to zero.
Check
Check your understanding
Exercise 364: for which x does the ratio test guarantee that the sum of xᵏ/k² converges?
Answer: A
Why: The ratio is |x| times (k/(k + 1))², which tends to |x|. The test guarantees convergence when |x| < 1. At |x| = 1 the test is silent (those endpoints need another test), and for |x| > 1 it diverges. This is exactly how Chapter 6 finds where a power series converges.
This is a preview of Chapter 6. The series has a variable x in it, and you want to know for which x it converges. Treat x as a fixed number and apply the ratio test: the ratio is the size of x times k over k plus one, squared, which tends to the size of x.
So the test guarantees convergence when the size of x is below one and divergence when it is above one. At exactly one the test is silent, and those endpoints need a different tool, here the p-series rule. Every power series in the next chapter is handled exactly this way.
Check
Check your understanding
Exercise 352: which test settles the sum of (1 + 1/n²)ⁿ most quickly?
Answer: B
Why: Since (1 + 1/n²)ⁿ is roughly e to the power n/n², which tends to e⁰ = 1, the terms tend to 1, not 0: the series diverges in one line. The terms are 2, then 1.105 at n = 10 and 1.010 at n = 100.
The power n on the whole term makes the root test look like the obvious choice, and that is the trap. The nth root is one plus one over n squared, which tends to one, so the root test is silent, and the ratio test is silent too.
The divergence test is the right first move. The term is close to e raised to n over n squared, and that exponent tends to zero, so the terms tend to one, not zero. The series diverges in a single line. Always glance at the limit of the terms before reaching for the heavier tests.
Explain it to yourself
Discussion prompt
In two or three sentences, explain why the ratio and root tests can say nothing when ρ = 1. Use what the proof compares the series with, and name a pair of series that proves the silence is forced.
Write your sentences before revealing. If you can explain this, you understand both tests rather than just the procedure.
The two halves of a good answer are the mechanism and the evidence. The mechanism is that both tests work by squeezing a geometric ratio strictly between rho and one, and at rho equal to one there is no room to do that. The evidence is a pair of series with rho equal to one and opposite fates, which proves that no verdict at one could ever be correct in general. The harmonic series and the reciprocal squares are the standard pair.
Exit ticket
\[ \sum_{n=1}^{\infty}\frac{2^n\,n!}{n^n} \]
Discussion prompt
Decide whether this converges. Name the test you chose, say what in the term made you choose it, and compute the limit.
This one problem uses the whole lesson: read the term, choose a test, and carry out the limit. There is a factorial, which points to the ratio test. The n to the n in the denominator is not the whole term raised to the n, so the root test would be awkward.
In the ratio, the powers of two leave a two, the factorials leave n plus one, and one factor of n plus one cancels from n plus one to the n plus one, leaving n over n plus one, all to the n. That tends to one over e, so the ratio limit is two over e, about 0.736. Below one, so the series converges absolutely.
Recap
| tool | compute | verdict |
|---|---|---|
| ratio test | ρ = lim |aₙ₊₁/aₙ| | ρ < 1 converges absolutely; ρ > 1 diverges; ρ = 1 silent |
| root test | ρ = lim ⁿ√|aₙ| | the same three cases |
| why they work | a geometric series with ratio R, ρ < R < 1 | comparison with that series |
| the strategy | familiar → alternating → compare → ratio/root → divergence, integral | the first yes names the tool |
Next, Chapter 6: power series. The ratio test is the tool that finds where each one converges.
Stewart, Calculus: Early Transcendentals 8e, §11.6 Absolute Convergence and the Ratio and Root Tests §11.6 and §11.7, pp. 737-745 — the ratio and root tests and the strategy in Stewart
Two tests and a strategy. The ratio test compares neighbouring terms and the root test takes the nth root of a single term. Both measure the effective geometric ratio of the tail, and both give the same three verdicts: below one converges absolutely, above one diverges, and exactly one tells you nothing.
Both work because they manufacture a geometric series to compare with, which is why they are silent at one: there is no geometric ratio strictly between one and one. The p-series, where both limits are one for every p, proves that the silence is forced.
The strategy puts every test of Chapter 5 in order, cheapest first. In Chapter 6 you will use the ratio test constantly, because it is the tool that finds the interval where a power series converges.
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