A series is the limit of its partial sums, not a magical infinite addition, and this deck starts there. It builds the nth-term test, geometric and telescoping sums, the proof that the harmonic series diverges, the comparison, ratio, and root tests, and the distinction between absolute and conditional convergence. It targets the classic traps: thinking that terms shrinking to zero forces convergence, that |r| may be anything in the geometric formula, that the ratio test settles the case when its limit is 1, and that a conditionally convergent series keeps its sum under rearrangement.
Subject: Foundations of Higher Mathematics · 111 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Define a series as the limit of its sequence of partial sums, and say precisely what it means for a series to converge.
2. Apply the nth-term test, and explain why it can only ever prove divergence.
3. Sum geometric and telescoping series from their closed-form partial sums, and prove the harmonic series diverges.
4. Choose and run the comparison, limit-comparison, ratio, and root tests.
5. Distinguish absolute from conditional convergence, use the alternating series test, and state why rearranging a conditional series can change its sum.
Warm-up
Discussion prompt
Before we open Infinite Series: without looking back, what was the main idea of Monotone Convergence, Bolzano-Weierstrass & Cauchy, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers the three theorems that make "the iterates settle down" rigorous: the Monotone Convergence Theorem, the Bolzano-Weierstrass theorem, and the Cauchy criterion. It targets the misconceptions that monotonicity alone gives convergence, that every bounded sequence converges, that Cauchy implies convergent in any space, and that a subsequential limit is THE limit.
Concept
You cannot literally perform infinitely many additions. So an infinite sum is not an operation we carry out; it is a value we assign by a limiting process.
The whole subject rests on one move: turn the infinite sum into a sequence of ordinary finite sums, then ask whether that sequence has a limit.
\[ \sum_{n=1}^{\infty} a_n \;\;\text{means}\;\; \lim_{N \to \infty} \sum_{n=1}^{N} a_n \]
Counterexample
Discussion prompt
You cannot literally perform infinitely many additions. So an infinite sum is not an operation we carry out; it is a value we assign by a limiting process.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The whole subject rests on one move: turn the infinite sum into a sequence of ordinary finite sums, then ask whether that sequence has a limit.
Concept
Given a sequence of terms, its partial sums are the running totals: add the first term, then the first two, then the first three, and so on.
\[ S_N \;=\; \sum_{n=1}^{N} a_n \;=\; a_1 + a_2 + \cdots + a_N \]
partial sum — The finite sum of the first N terms of a series. The partial sums form their own sequence, and the behavior of THAT sequence is the whole game.
Analogy
Discussion prompt
Explain The partial sum sequence by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Given a sequence of terms, its partial sums are the running totals: add the first term, then the first two, then the first three, and so on.
Picture it
Figure (svg): A number line with partial sums S1 at 0.5, S2 at 0.75, S3 at 0.875 marching toward the value 1.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Picture each partial sum as a dot on the number line. Adding the next term nudges you along by that term's size.
Intuition
Picture each partial sum as a dot on the number line. Adding the next term nudges you along by that term's size.
Figure (svg): A number line with partial sums S1 at 0.5, S2 at 0.75, S3 at 0.875 marching toward the value 1.
If the dots crowd in toward a single point, the series converges to that point. If they wander off or oscillate forever, it diverges.
Explain it
Discussion prompt
Explain A series is a running total on the line to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Picture each partial sum as a dot on the number line. Adding the next term nudges you along by that term's size.
Concept
There is nothing more to the definition than this: a series converges exactly when its partial sum sequence converges, and the sum is that limit.
\[ \sum_{n=1}^{\infty} a_n = S \quad\Longleftrightarrow\quad \lim_{N \to \infty} S_N = S \]
converges — A series converges when its sequence of partial sums has a finite limit S. That limit S is called the sum of the series.
Definition probe
Sort into buckets
Every line below is part of the definition of partial sum or of converges — one or the other, never both. Put each where it belongs.
Concept
Keep two objects separate. The series is the whole infinite construction. The sum is a single number the series may or may not point at.
Writing an equals sign between a series and a number is a claim that the limit exists and equals that number. It is never automatic.
Concept
Changing, deleting, or inserting finitely many terms shifts every later partial sum by a fixed constant. A fixed shift cannot create or destroy a limit.
So convergence is a property of the tail of the series. Where you start the index is irrelevant to the yes-or-no question; it only changes the value of the sum.
Ranking
Put in order
Put the moves of Worked example: a telescoping series into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Write the term as a difference so consecutive pieces will cancel.
Worked example
Evaluate the series whose general term splits into a difference.
\[ \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \]
Split the term by partial fractions
Why: Write the term as a difference so consecutive pieces will cancel.
\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]
Write out the partial sum and cancel
Why: Each minus piece kills the next plus piece; only the first and last survive.
\[ S_N = \Big(1 - \tfrac{1}{2}\Big) + \Big(\tfrac{1}{2} - \tfrac{1}{3}\Big) + \cdots + \Big(\tfrac{1}{N} - \tfrac{1}{N+1}\Big) = 1 - \frac{1}{N+1} \]
Take the limit of the closed form
Why: The leftover tail term goes to zero, leaving the sum.
\[ \lim_{N \to \infty} \Big(1 - \tfrac{1}{N+1}\Big) = 1 \]
Verify the closed form at the first three partial sums
Why: Check S_N against direct addition for N equal to 1, 2, 3.
\[ S_1 = 1-\tfrac12 = \tfrac12,\quad S_2 = 1-\tfrac13 = \tfrac23,\quad S_3 = 1-\tfrac14 = \tfrac34 \]
| N | direct sum | formula 1-1/(N+1) |
|---|---|---|
| 1 | 1/2 | 1/2 |
| 2 | 1/2+1/6 = 2/3 | 2/3 |
| 3 | 2/3+1/12 = 3/4 | 3/4 |
Picture it
Animation
Shows: Each line of the worked example "a telescoping series", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check S_N against direct addition for N equal to 1, 2, 3.
Step zero
Discussion prompt
Worked example: a second telescoping sum — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Split into half-differences
Answer:
Worked example
Evaluate a sum over odd factors.
\[ \sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)} \]
Split into half-differences
Why: Partial fractions again, with a factor of one half out front.
\[ \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) \]
Telescope the partial sum
Why: Interior terms cancel in pairs, leaving the two ends.
\[ S_N = \frac{1}{2}\left(1 - \frac{1}{2N+1}\right) \]
Take the limit
Why: The tail vanishes; the one half survives.
\[ \lim_{N \to \infty} S_N = \frac{1}{2} \]
Verify at N equal to 1, 2, 3
Why: Compare the closed form with direct addition of the first terms.
| N | direct sum | formula |
|---|---|---|
| 1 | 1/3 | 1/3 |
| 2 | 1/3+1/15 = 2/5 | 2/5 |
| 3 | 2/5+1/35 = 3/7 | 3/7 |
Picture it
Animation
Shows: Each line of the worked example "a second telescoping sum", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Compare the closed form with direct addition of the first terms.
Trap
Claiming that because the individual terms settle down to a limit, the series must settle down too.
For the harmonic series the terms march to zero, yet the running total climbs without bound. The terms converging says nothing final about the sum.
\[ a_n = \frac{1}{n} \to 0, \qquad \text{but} \qquad \sum \frac{1}{n} = \infty \]
Always ask the question about the partial sums, never about the terms alone.
The convergence of the term sequence and the convergence of the series are two different questions about two different sequences.
\[ \text{term sequence: } (a_n) \qquad\text{vs.}\qquad \text{partial-sum sequence: } (S_N) \]
Concept
There is a one-directional rule that follows immediately from the definition.
\[ \sum a_n \text{ converges} \;\Longrightarrow\; \lim_{n \to \infty} a_n = 0 \]
Read as a divergence test by contraposition: if the terms do not go to zero, the series diverges. It can never confirm convergence.
Intuition
Each term is the gap between two consecutive partial sums.
\[ a_n = S_n - S_{n-1} \]
If the partial sums close in on a single limit, then two consecutive ones get arbitrarily near each other, so their difference is squeezed to zero.
\[ S_n \to S,\; S_{n-1} \to S \;\Longrightarrow\; a_n = S_n - S_{n-1} \to S - S = 0 \]
Estimation
Predict first
Decide whether this series can converge.
Commit before you compute: what does Worked example: nth-term test kills a series come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the intuition with partial sums
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Each term is near one half, so the running total grows by roughly one half per step and cannot settle.
Worked example
Decide whether this series can converge.
\[ \sum_{n=1}^{\infty} \frac{n}{2n+1} \]
Compute the limit of the general term
Why: Divide numerator and denominator by n and let n grow.
\[ \lim_{n \to \infty} \frac{n}{2n+1} = \lim_{n \to \infty} \frac{1}{2 + 1/n} = \frac{1}{2} \]
Apply the nth-term test
Why: The limit is not zero, so convergence is impossible.
\[ \lim_{n\to\infty} a_n = \tfrac{1}{2} \neq 0 \;\Longrightarrow\; \text{diverges} \]
Verify the intuition with partial sums
Why: Each term is near one half, so the running total grows by roughly one half per step and cannot settle.
| n | term n/(2n+1) | roughly |
|---|---|---|
| 10 | 10/21 | 0.476 |
| 100 | 100/201 | 0.498 |
| 1000 | 1000/2001 | 0.4998 |
Picture it
Animation
Shows: Each line of the worked example "nth-term test kills a series", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Each term is near one half, so the running total grows by roughly one half per step and cannot settle.
Missing information
Discussion prompt
Test this series, where the term looks like it might vanish.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
This is the defining limit of the number e, not something tending to zero.
Worked example
Test this series, where the term looks like it might vanish.
\[ \sum_{n=1}^{\infty} \left(1 + \frac{1}{n}\right)^{n} \]
Recall the classic limit of the base-and-exponent
Why: This is the defining limit of the number e, not something tending to zero.
\[ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n} = e \approx 2.718 \]
Apply the nth-term test
Why: The terms approach e, which is far from zero, so the series diverges.
\[ \lim_{n\to\infty} a_n = e \neq 0 \;\Longrightarrow\; \text{diverges} \]
Verify numerically at small n
Why: The terms are already above 2 and rising toward e, never near zero.
| n | term | value |
|---|---|---|
| 1 | (2)^1 | 2.000 |
| 2 | (3/2)^2 | 2.250 |
| 5 | (6/5)^5 | 2.488 |
Picture it
Animation
Shows: Each line of the worked example "a sneakier nth-term divergence", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The terms are already above 2 and rising toward e, never near zero.
Trap
The single most common error in the subject: reading the nth-term test backwards.
Asserting that because the terms shrink to zero, the harmonic series must converge.
\[ \frac{1}{n} \to 0 \;\;\overset{?}{\Longrightarrow}\;\; \sum \frac{1}{n} \text{ converges} \]
The implication only runs one way. Terms going to zero is necessary, never sufficient.
The harmonic series is the standing counterexample: its terms vanish and yet it diverges. We prove this next.
\[ \frac{1}{n} \to 0 \quad\text{and yet}\quad \sum_{n=1}^{\infty}\frac{1}{n} = \infty \]
Concept
A geometric series is one where each term is a fixed multiple of the one before. That fixed multiple is the ratio.
\[ \sum_{n=0}^{\infty} a\,r^{n} = a + ar + ar^2 + ar^3 + \cdots \]
common ratio — The constant r by which each term is multiplied to get the next. The convergence of the whole series is decided entirely by the size of r.
Intuition
There is a slick way to collapse the finite geometric sum. Multiply the whole partial sum by the ratio, line it up shifted by one, and subtract.
\[ S_N = a + ar + \cdots + ar^{N}, \qquad rS_N = ar + ar^2 + \cdots + ar^{N+1} \]
Almost everything cancels in the subtraction. Only the very first and very last pieces remain, which is why a clean closed form exists.
Estimation
Predict first
Derive the partial sum, then the infinite sum, of a geometric series.
Commit before you compute: what does Worked example: closed form of the geometric sum come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the partial-sum formula at N equal to 1, 2, 3
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Test with a equal to 1 and r equal to one half against direct addition.
Worked example
Derive the partial sum, then the infinite sum, of a geometric series.
Subtract the shifted copy
Why: S minus rS cancels every interior term, leaving two ends.
\[ S_N - rS_N = a - ar^{N+1} \]
Solve for the partial sum
Why: Factor out S and divide, valid whenever r is not 1.
\[ S_N = a\,\frac{1 - r^{N+1}}{1 - r} \]
Take the limit when the ratio is small
Why: When the size of r is below 1, its powers die, so the numerator settles at 1.
\[ |r| < 1 \;\Longrightarrow\; r^{N+1} \to 0 \;\Longrightarrow\; \sum_{n=0}^{\infty} a r^n = \frac{a}{1-r} \]
Verify the partial-sum formula at N equal to 1, 2, 3
Why: Test with a equal to 1 and r equal to one half against direct addition.
| N | direct sum (a=1, r=1/2) | formula (1-r^(N+1))/(1-r) |
|---|---|---|
| 1 | 1+1/2 = 3/2 | 3/2 |
| 2 | 3/2+1/4 = 7/4 | 7/4 |
| 3 | 7/4+1/8 = 15/8 | 15/8 |
Picture it
Animation
Shows: Each line of the worked example "closed form of the geometric sum", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Test with a equal to 1 and r equal to one half against direct addition.
Pattern
Predict first
The table runs: 0 | 5 | 5.000 · 1 | 5+10/3 | 8.333
In Worked example: evaluate a geometric series, given the rows so far: what is the next one — the row where N is 3?
Correct: 3 | 5+10/3+20/9+40/27 | 11.48
| N | partial sum | value |
|---|---|---|
| 0 | 5 | 5.000 |
| 1 | 5+10/3 | 8.333 |
| 3 | 5+10/3+20/9+40/27 | 11.48 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. The index starts at zero, so the first term is a itself; the ratio is two thirds.
Worked example
Sum the series exactly.
\[ \sum_{n=0}^{\infty} 5\left(\frac{2}{3}\right)^{n} \]
Read off the first term and the ratio
Why: The index starts at zero, so the first term is a itself; the ratio is two thirds.
\[ a = 5, \qquad r = \frac{2}{3}, \qquad |r| < 1 \]
Apply the closed form
Why: The size of r is below 1, so the formula applies.
\[ \frac{a}{1-r} = \frac{5}{1 - \tfrac{2}{3}} = \frac{5}{\tfrac{1}{3}} = 15 \]
Verify with the first partial sums
Why: The running totals should climb toward 15 from below.
| N | partial sum | value |
|---|---|---|
| 0 | 5 | 5.000 |
| 1 | 5+10/3 | 8.333 |
| 3 | 5+10/3+20/9+40/27 | 11.48 |
Picture it
Animation
Shows: Each line of the worked example "evaluate a geometric series", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The running totals should climb toward 15 from below.
Step zero
Discussion prompt
Worked example: a repeating decimal is a geometric series — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Identify the geometric data
Answer:
Worked example
Express the repeating decimal 0.474747... as an exact fraction.
\[ 0.474747\ldots = \sum_{n=0}^{\infty} \frac{47}{100}\left(\frac{1}{100}\right)^{n} \]
Identify the geometric data
Why: Each repeat block shifts two decimal places, so the ratio is one hundredth.
\[ a = \frac{47}{100}, \qquad r = \frac{1}{100} \]
Apply the closed form
Why: Sum with the a over one minus r rule.
\[ \frac{47/100}{1 - 1/100} = \frac{47/100}{99/100} = \frac{47}{99} \]
Verify by dividing back
Why: Long division of the fraction reproduces the repeating block.
\[ \frac{47}{99} = 0.474747\ldots \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "a repeating decimal is a geometric series", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Long division of the fraction reproduces the repeating block.
Trap
Plugging any ratio into the closed form, even one that is 1 or larger in size.
Applying the formula to a ratio of 2 gives a finite negative number for a sum of exploding positive terms. Nonsense.
\[ \sum_{n=0}^{\infty} 2^{n} \;\overset{?}{=}\; \frac{1}{1-2} = -1 \]
The closed form is only valid when the size of the ratio is strictly less than 1. Otherwise the powers do not die and the series diverges.
\[ \sum_{n=0}^{\infty} a r^n = \frac{a}{1-r} \quad\text{only when}\quad |r| < 1; \quad |r| \ge 1 \Rightarrow \text{diverges} \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
The closed form is only valid when the size of the ratio is strictly less than 1. Otherwise the powers do not die and the series diverges.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Concept
The harmonic series adds the reciprocals of the whole numbers. Its terms shrink to zero, which makes its divergence genuinely surprising.
\[ \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots \]
Intuition
The terms shrink, but not fast enough. The trick is to group the terms into blocks of doubling length and show each block contributes at least one half.
There are infinitely many blocks, each worth at least one half, so the total is unbounded.
Hypothesis
Predict first
Worked example: the harmonic series diverges is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Group terms into blocks of length 1, 2, 4, 8, ...
Why: Each block runs from index one past a power of two up to the next power of two.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Prove divergence by the grouping argument of Oresme.
Group terms into blocks of length 1, 2, 4, 8, ...
Why: Each block runs from index one past a power of two up to the next power of two.
\[ 1 + \frac{1}{2} + \underbrace{\Big(\frac{1}{3}+\frac{1}{4}\Big)}_{2\text{ terms}} + \underbrace{\Big(\frac{1}{5}+\cdots+\frac{1}{8}\Big)}_{4\text{ terms}} + \cdots \]
Bound each block below by one half
Why: A block of length two-to-the-k has each term at least one over two-to-the-(k+1), so the block sums to at least one half.
\[ \frac{1}{3}+\frac{1}{4} \ge \frac{1}{4}+\frac{1}{4} = \frac{1}{2}, \qquad \frac{1}{5}+\cdots+\frac{1}{8} \ge 4\cdot\frac{1}{8} = \frac{1}{2} \]
Add up the block bounds
Why: Infinitely many halves overwhelm any target, so the partial sums are unbounded.
\[ S_{2^k} \ge 1 + \frac{k}{2} \longrightarrow \infty \]
Verify the block bound for the first blocks
Why: Confirm the growth-by-one-half pattern on concrete partial sums.
| k | partial sum S at 2^k | lower bound 1+k/2 |
|---|---|---|
| 1 | 1.500 | 1.5 |
| 2 | 2.083 | 2.0 |
| 3 | 2.718 | 2.5 |
Picture it
Animation
Shows: Each line of the worked example "the harmonic series diverges", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Confirm the growth-by-one-half pattern on concrete partial sums.
Concept
The reciprocals of powers form the single most useful family of benchmark series. Everything hinges on one threshold.
\[ \sum_{n=1}^{\infty} \frac{1}{n^{p}} \quad\text{converges} \iff p > 1 \]
The harmonic series is the boundary case at exactly 1, where it diverges. Anything with power above 1 converges; anything at or below 1 diverges.
Estimation
Predict first
Classify two series by the p-series rule.
Commit before you compute: what does Worked example: reading off two p-series come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against the threshold
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Both classifications sit on the correct side of the boundary at 1.
Worked example
Classify two series by the p-series rule.
Identify the power in the first series
Why: A square root in the denominator is a power of one half.
\[ \sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}}, \qquad p = \tfrac{1}{2} \le 1 \]
Conclude divergence
Why: The power does not exceed 1, so it diverges.
\[ p = \tfrac{1}{2} \le 1 \;\Longrightarrow\; \text{diverges} \]
Do the same for the square case
Why: Power two exceeds 1, so this one converges.
\[ \sum \frac{1}{n^{2}}, \qquad p = 2 > 1 \;\Longrightarrow\; \text{converges} \]
Verify against the threshold
Why: Both classifications sit on the correct side of the boundary at 1.
| series | power p | verdict |
|---|---|---|
| sum 1/sqrt(n) | 1/2 | diverges |
| sum 1/n^2 | 2 | converges |
Picture it
Animation
Shows: Each line of the worked example "reading off two p-series", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both classifications sit on the correct side of the boundary at 1.
Concept
For series with nonnegative terms, you can pin an unknown series against a known one, term by term.
\[ 0 \le a_n \le b_n: \quad \sum b_n \text{ converges} \Rightarrow \sum a_n \text{ converges} \]
\[ 0 \le b_n \le a_n: \quad \sum b_n \text{ diverges} \Rightarrow \sum a_n \text{ diverges} \]
Intuition
If your terms sit under the terms of a series with a finite total, your running total is capped, so it must converge.
If your terms sit above the terms of a series that already blows up, you blow up at least as fast. Direction is everything.
Ranking
Put in order
Put the moves of Worked example: comparison from above into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Adding one to the denominator only shrinks the fraction.
Worked example
Test the series for convergence.
\[ \sum_{n=1}^{\infty} \frac{1}{n^{2}+1} \]
Bound each term above by a known convergent term
Why: Adding one to the denominator only shrinks the fraction.
\[ \frac{1}{n^{2}+1} \le \frac{1}{n^{2}} \]
Cite the reference p-series
Why: The bounding series is a p-series with power two, which converges.
\[ \sum \frac{1}{n^{2}} \text{ converges} \;(p = 2 > 1) \]
Conclude by comparison
Why: Smaller nonnegative terms under a convergent series converge.
\[ 0 \le \frac{1}{n^2+1} \le \frac{1}{n^2} \;\Longrightarrow\; \sum \frac{1}{n^2+1} \text{ converges} \]
Verify the term inequality at small n
Why: Confirm the bound really holds so the comparison is legitimate.
| n | 1/(n^2+1) | 1/n^2 |
|---|---|---|
| 1 | 0.500 | 1.000 |
| 2 | 0.200 | 0.250 |
| 3 | 0.100 | 0.111 |
Picture it
Animation
Shows: Each line of the worked example "comparison from above", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Confirm the bound really holds so the comparison is legitimate.
Fill the middle
Fill in the blanks
From Worked example: comparison against a geometric series — finish the line. Write what belongs on the right of the equals sign before you look.
\sum_1}^{\infty} \frac{1}{2^{n}+1}}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Adding one to the denominator only decreases the fraction.
Worked example
Test this series.
\[ \sum_{n=1}^{\infty} \frac{1}{2^{n}+1} \]
Bound above by a geometric term
Why: Adding one to the denominator only decreases the fraction.
\[ \frac{1}{2^{n}+1} \le \frac{1}{2^{n}} \]
Cite the reference geometric series
Why: The bounding series has ratio one half, which converges.
\[ \sum_{n=1}^{\infty} \frac{1}{2^{n}} = 1 \quad (|r| = \tfrac12 < 1) \]
Conclude by comparison
Why: Bounded above by a convergent series with nonnegative terms.
\[ \sum \frac{1}{2^{n}+1} \text{ converges} \]
Verify the bound at small n
Why: The inequality holds every term, so the comparison is valid.
| n | 1/(2^n+1) | 1/2^n |
|---|---|---|
| 1 | 0.333 | 0.500 |
| 2 | 0.200 | 0.250 |
| 3 | 0.111 | 0.125 |
Picture it
Animation
Shows: Each line of the worked example "comparison against a geometric series", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The inequality holds every term, so the comparison is valid.
Concept
Term-by-term inequalities can be fiddly. The limit comparison test only needs the terms to be the same size asymptotically.
\[ \lim_{n \to \infty} \frac{a_n}{b_n} = L, \quad 0 < L < \infty \;\Longrightarrow\; \sum a_n \text{ and } \sum b_n \text{ share their fate} \]
If the ratio of terms tends to a finite positive number, both series converge or both diverge together.
Step zero
Discussion prompt
Worked example: limit comparison — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Guess the dominant behavior
Answer:
Worked example
Test the series.
\[ \sum_{n=1}^{\infty} \frac{n+1}{n^{3}+2} \]
Guess the dominant behavior
Why: For large n the term behaves like n over n-cubed, that is one over n-squared.
\[ \frac{n+1}{n^3+2} \approx \frac{n}{n^3} = \frac{1}{n^2}, \qquad b_n = \frac{1}{n^2} \]
Compute the ratio limit
Why: Divide and take the limit; a finite positive value lets the test fire.
\[ \lim_{n\to\infty} \frac{(n+1)/(n^3+2)}{1/n^2} = \lim_{n\to\infty} \frac{n^3+n^2}{n^3+2} = 1 \]
Conclude
Why: The benchmark converges and the limit is finite and positive, so the given series converges too.
\[ 0 < L = 1 < \infty, \;\; \sum \frac{1}{n^2} \text{ converges} \;\Longrightarrow\; \text{converges} \]
Verify the ratio trend numerically
Why: The ratio should sit near 1 and tighten as n grows.
| n | ratio a_n/b_n | value |
|---|---|---|
| 1 | (2)(1)/(3) | 0.667 |
| 10 | (11)(100)/(1002) | 1.098 |
| 100 | approx | 1.010 |
Picture it
Animation
Shows: Each line of the worked example "limit comparison", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The ratio should sit near 1 and tighten as n grows.
Trap
Bounding your terms below a convergent series and thinking you have proved anything.
Knowing a term is smaller than something convergent-or-larger tells you nothing. Here the term is below a divergent series, so the bound is empty.
\[ \frac{1}{2n} \le \frac{1}{n}, \quad \text{but } \sum \frac{1}{n} \text{ diverges} \Rightarrow \text{no conclusion} \]
Match the inequality to the goal. To prove convergence bound above by a convergent series; to prove divergence bound below by a divergent one.
\[ \frac{1}{2n} \ge \frac{1}{2}\cdot\frac{1}{n}, \quad \sum \frac{1}{n} \text{ diverges} \;\Longrightarrow\; \sum \frac{1}{2n} \text{ diverges} \]
Concept
The ratio test measures how fast the terms shrink by looking at the ratio of consecutive terms in the limit.
\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]
\[ L < 1 \Rightarrow \text{converges (absolutely)}, \quad L > 1 \Rightarrow \text{diverges}, \quad L = 1 \Rightarrow \text{inconclusive} \]
Intuition
If the ratio of consecutive terms settles near a number below 1, the tail of the series behaves like a geometric series with that ratio, and geometric series with a small ratio converge.
This is why the ratio test is the reflex for factorials and exponentials, where consecutive terms have a clean ratio.
Worked example
Test the series.
\[ \sum_{n=1}^{\infty} \frac{n}{2^{n}} \]
Form the ratio of consecutive terms
Why: Write term n-plus-one over term n and simplify.
\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)/2^{n+1}}{n/2^{n}} = \frac{n+1}{2n} \]
Take the limit
Why: The ratio tends to one half as n grows.
\[ L = \lim_{n\to\infty} \frac{n+1}{2n} = \frac{1}{2} \]
Apply the test
Why: The limit is below 1, so the series converges.
\[ L = \tfrac{1}{2} < 1 \;\Longrightarrow\; \text{converges} \]
Verify the ratio at small n
Why: The consecutive ratios should approach one half from above.
| n | ratio (n+1)/(2n) | value |
|---|---|---|
| 1 | 2/2 | 1.000 |
| 2 | 3/4 | 0.750 |
| 10 | 11/20 | 0.550 |
Picture it
Animation
Shows: Each line of the worked example "ratio test on a polynomial-over-exponential", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The consecutive ratios should approach one half from above.
Fill the middle
Fill in the blanks
From Worked example: ratio test tames a factorial — finish the line. Write what belongs on the right of the equals sign before you look.
\sum_0}^{\infty} \frac{2^{n}}{n!}}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Dividing consecutive terms cancels almost everything in the factorial.
Worked example
Test the series.
\[ \sum_{n=0}^{\infty} \frac{2^{n}}{n!} \]
Form the ratio
Why: Dividing consecutive terms cancels almost everything in the factorial.
\[ \frac{a_{n+1}}{a_n} = \frac{2^{n+1}/(n+1)!}{2^{n}/n!} = \frac{2}{n+1} \]
Take the limit
Why: The denominator grows without bound, driving the ratio to zero.
\[ L = \lim_{n\to\infty} \frac{2}{n+1} = 0 \]
Apply the test
Why: A limit of zero is below 1, so the series converges; in fact this sum is e-squared.
\[ L = 0 < 1 \;\Longrightarrow\; \text{converges} \]
Verify the ratio shrinks
Why: Consecutive ratios should collapse toward zero.
| n | ratio 2/(n+1) | value |
|---|---|---|
| 0 | 2/1 | 2.000 |
| 3 | 2/4 | 0.500 |
| 9 | 2/10 | 0.200 |
Picture it
Animation
Shows: Each line of the worked example "ratio test tames a factorial", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Consecutive ratios should collapse toward zero.
Concept
When the terms are themselves an n-th power, take the n-th root instead of a ratio.
\[ L = \lim_{n \to \infty} \sqrt[n]{|a_n|}, \qquad L < 1 \Rightarrow \text{converges}, \; L > 1 \Rightarrow \text{diverges}, \; L = 1 \Rightarrow \text{inconclusive} \]
Fill the middle
Fill in the blanks
From Worked example: root test on an n-th power — finish the line. Write what belongs on the right of the equals sign before you look.
\sqrt[n]\frac{n}{2n+1}___\right)^___} = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The outer n-th power cancels cleanly against the n-th root.
Worked example
Test the series.
\[ \sum_{n=1}^{\infty} \left( \frac{n}{2n+1} \right)^{n} \]
Take the n-th root of the term
Why: The outer n-th power cancels cleanly against the n-th root.
\[ \sqrt[n]{\left(\frac{n}{2n+1}\right)^{n}} = \frac{n}{2n+1} \]
Take the limit
Why: The rational expression tends to one half.
\[ L = \lim_{n\to\infty} \frac{n}{2n+1} = \frac{1}{2} \]
Apply the test
Why: The root limit is below 1, so the series converges.
\[ L = \tfrac{1}{2} < 1 \;\Longrightarrow\; \text{converges} \]
Verify the root at small n
Why: The n-th roots should approach one half.
| n | n/(2n+1) | value |
|---|---|---|
| 1 | 1/3 | 0.333 |
| 5 | 5/11 | 0.455 |
| 50 | 50/101 | 0.495 |
Picture it
Animation
Shows: Each line of the worked example "root test on an n-th power", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The n-th roots should approach one half.
Trap
Concluding convergence or divergence when the ratio limit comes out exactly 1.
Both of these give ratio limit 1, yet one converges and one diverges. The test simply cannot see the difference.
\[ \sum \frac{1}{n}: \; L = \lim \frac{n}{n+1} = 1; \qquad \sum \frac{1}{n^2}: \; L = \lim \left(\frac{n}{n+1}\right)^2 = 1 \]
When the ratio limit is 1 the test is silent. Switch tools: use a p-series comparison, limit comparison, or another test.
\[ L = 1 \Rightarrow \text{inconclusive}; \quad \sum \tfrac1n \text{ diverges}, \;\; \sum \tfrac{1}{n^2} \text{ converges} \]
Concept
So far the terms were nonnegative. Once signs appear, there are two grades of convergence. The stronger grade ignores the signs entirely.
\[ \sum a_n \text{ converges absolutely} \iff \sum |a_n| \text{ converges} \]
absolute convergence — A series converges absolutely when the series of absolute values converges. Absolute convergence always implies ordinary convergence.
Intuition
If the total distance travelled by the terms is finite, then letting some steps go backward can only produce cancellation, never escape. The signed sum is trapped inside the unsigned one.
\[ \left| \sum a_n \right| \le \sum |a_n| < \infty \]
Pattern
Predict first
The table runs: 2 | -1+1/4 | -0.750 · 3 | -1+1/4-1/9 | -0.861
In Worked example: an absolutely convergent series, given the rows so far: what is the next one — the row where N is 4?
Correct: 4 | prev+1/16 | -0.799
| N | partial sum | value |
|---|---|---|
| 2 | -1+1/4 | -0.750 |
| 3 | -1+1/4-1/9 | -0.861 |
| 4 | prev+1/16 | -0.799 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. The unsigned series converges, so the signed one converges absolutely, hence converges.
Worked example
Test the signed series.
\[ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \]
Strip the signs
Why: Take absolute values of every term.
\[ \left| \frac{(-1)^{n}}{n^{2}} \right| = \frac{1}{n^{2}} \]
Test the unsigned series
Why: This is a p-series with power two.
\[ \sum \frac{1}{n^{2}} \text{ converges} \;(p = 2 > 1) \]
Conclude absolute convergence
Why: The unsigned series converges, so the signed one converges absolutely, hence converges.
\[ \sum \frac{(-1)^n}{n^2} \text{ converges absolutely} \]
Verify the partial sums stay bounded
Why: The signed partial sums hover near the true sum, which is about negative 0.822.
| N | partial sum | value |
|---|---|---|
| 2 | -1+1/4 | -0.750 |
| 3 | -1+1/4-1/9 | -0.861 |
| 4 | prev+1/16 | -0.799 |
Concept
Some series converge only because of the cancellation from alternating signs. Leibniz gives a clean sufficient condition.
\[ \sum (-1)^{n+1} b_n \text{ converges if } b_n \ge 0,\; b_n \text{ decreasing},\; b_n \to 0 \]
Three boxes to check: nonnegative, eventually decreasing, and tending to zero. All three are needed.
Intuition
Alternating partial sums step forward, then back a smaller amount, then forward less, then back less. They bracket the limit from both sides with ever-tighter jumps.
Figure (svg): Partial sums of an alternating series overshooting and undershooting the limit L with shrinking gaps.
The gap between a partial sum and the limit is never larger than the next term, which gives a free error bound.
Step zero
Discussion prompt
Worked example: the alternating harmonic series — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Check the three Leibniz boxes
Answer:
Worked example
Classify the series.
\[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots \]
Check the three Leibniz boxes
Why: The sizes are one over n: nonnegative, decreasing, and tending to zero.
\[ b_n = \frac{1}{n} \ge 0, \quad b_{n+1} < b_n, \quad b_n \to 0 \]
Conclude convergence
Why: All three conditions hold, so the alternating series test gives convergence; the sum is the natural log of two.
\[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \ln 2 \approx 0.693 \]
Test for absolute convergence
Why: The unsigned series is the harmonic series, which diverges, so convergence here is only conditional.
\[ \sum \left| \frac{(-1)^{n+1}}{n} \right| = \sum \frac{1}{n} = \infty \]
Verify by bracketing partial sums
Why: Odd partial sums sit above the limit, even ones below, closing in on ln 2.
| N | partial sum | value |
|---|---|---|
| 1 | 1 | 1.000 |
| 2 | 1-1/2 | 0.500 |
| 3 | 1-1/2+1/3 | 0.833 |
Picture it
Animation
Shows: Each line of the worked example "the alternating harmonic series", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The sizes are one over n: nonnegative, decreasing, and tending to zero.
Ranking
Put in order
Put the moves of Worked example: a second conditional series into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. One over root n is nonnegative, decreasing, and tends to zero.
Worked example
Classify the series.
\[ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}} \]
Check the Leibniz conditions
Why: One over root n is nonnegative, decreasing, and tends to zero.
\[ b_n = \frac{1}{\sqrt{n}} \ge 0, \quad \text{decreasing}, \quad b_n \to 0 \]
Conclude convergence
Why: The alternating series test applies, so the series converges.
\[ \text{alternating test} \;\Longrightarrow\; \text{converges} \]
Test absolute convergence
Why: The unsigned series is a p-series with power one half, which diverges.
\[ \sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}} \text{ diverges} \;(p = \tfrac12 \le 1) \]
Verify the classification
Why: Converges but not absolutely, so it is conditionally convergent.
| property | result |
|---|---|
| series of terms | converges |
| series of absolute values | diverges |
| verdict | conditional |
Picture it
Animation
Shows: Each line of the worked example "a second conditional series", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: One over root n is nonnegative, decreasing, and tends to zero.
Concept
A series that converges, but whose absolute-value series diverges, is called conditionally convergent. Its convergence depends on cancellation, not on the terms being small enough on their own.
conditional convergence — The series converges but the series of absolute values diverges. The sum exists only thanks to sign cancellation, which makes it fragile.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of partial sum, converges, common ratio, absolute convergence, conditional convergence as Infinite Series uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
For a convergent alternating series, stopping at any point leaves an error no bigger than the first term you dropped. This is one of the most useful facts in numerical work.
\[ \left| S - S_N \right| \le b_{N+1} \]
Concept
Here is the shock. For a conditionally convergent series, the sum is not an intrinsic property of the collection of terms. It depends on the order.
By reordering the terms alone, you can make a conditionally convergent series add up to any real number you like, or make it diverge.
\[ \text{conditionally convergent} \;\Longrightarrow\; \forall\, L \in \mathbb{R},\; \exists\ \text{a rearrangement summing to } L \]
Explain it
Discussion prompt
Explain The Riemann rearrangement theorem to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
By reordering the terms alone, you can make a conditionally convergent series add up to any real number you like, or make it diverge.
Intuition
In a conditionally convergent series the positive terms alone add to infinity and the negative terms alone add to negative infinity. You have an unlimited bank of both.
To hit a target, pour in positive terms until you pass it, then negative terms until you drop below, and repeat. Since both reservoirs are infinite and the terms shrink to zero, the running total can be steered to any value.
Analogy
Discussion prompt
Explain Why conditional sums are fragile by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
In a conditionally convergent series the positive terms alone add to infinity and the negative terms alone add to negative infinity. You have an unlimited bank of both.
Trap
Treating an infinite sum like a finite one, where order and grouping never matter.
The alternating harmonic series sums to the natural log of two, but a rearrangement taking two positives per negative sums to something different.
\[ 1 + \tfrac{1}{3} - \tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{7} - \tfrac{1}{4} + \cdots = \tfrac{3}{2}\ln 2 \neq \ln 2 \]
Reordering is only always safe for absolutely convergent series. For those, every rearrangement gives the same sum.
\[ \text{absolutely convergent} \;\Longrightarrow\; \text{every rearrangement has the same sum} \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
Partial sums are exactly what a running accumulator computes in a loop. Asking whether a series converges is asking whether that accumulator settles or overflows conceptually.
The growth rate of the partial sums is a big-O question: the harmonic partial sums grow like the logarithm, which is why a loop adding one over the index has cost tied to that growth.
\[ \sum_{n=1}^{N} \frac{1}{n} = \ln N + \gamma + O\!\left(\frac{1}{N}\right) \]
Absolutely convergent series are the ones safe to reorder, which matters the moment you sum in parallel and the order of accumulation is no longer fixed.
Counterexample
Discussion prompt
Partial sums are exactly what a running accumulator computes in a loop. Asking whether a series converges is asking whether that accumulator settles or overflows conceptually.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The growth rate of the partial sums is a big-O question: the harmonic partial sums grow like the logarithm, which is why a loop adding one over the index has cost tied to that growth.
Pattern
1. Check the terms first
Why: If the terms do not go to zero, stop: the series diverges by the nth-term test.
2. Recognize a known form
Why: Geometric and telescoping series can be summed exactly; p-series are read off from the power.
3. For factorials or n-th powers, reach for ratio or root
Why: Consecutive-ratio or n-th-root behavior exposes eventual geometric decay.
4. Otherwise compare
Why: Bound above by a convergent series, or below by a divergent one, or use limit comparison against a p-series.
5. If signs alternate, test absolute first, then Leibniz
Why: Absolute convergence is stronger and order-proof; if it fails, the alternating test may still give conditional convergence.
Real world
Discussion prompt
Outside this lesson: where does Infinite Series actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of A decision recipe for testing a series is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
A series is the limit of its partial sums, not a magical infinite addition. Builds the nth-term test, geometric and telescoping sums, the harmonic divergence proof, comparison / ratio / root tests, and absolute vs conditional convergence.
Elimination
Eliminate the wrong options
The terms of this series tend to zero. What may we correctly conclude?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Terms tending to zero is necessary but not sufficient for convergence. The harmonic series has terms going to zero yet diverges, shown by grouping blocks each summing to at least one half. So the limit of the terms alone tells us nothing here.
Check
Consider the harmonic series, whose terms shrink to zero.
\[ \sum_{n=1}^{\infty} \frac{1}{n} \]
Check your understanding
The terms of this series tend to zero. What may we correctly conclude?
Answer: A
Why: Terms tending to zero is necessary but not sufficient for convergence. The harmonic series has terms going to zero yet diverges, shown by grouping blocks each summing to at least one half. So the limit of the terms alone tells us nothing here.
Check
Evaluate the geometric series.
\[ \sum_{n=0}^{\infty} 3\left(\frac{1}{2}\right)^{n} \]
Check your understanding
What is the exact sum?
Answer: A
Why: The first term is a equal to 3 and the ratio r is one half, whose size is below 1, so the sum is a divided by one minus r, that is 3 divided by one half, which equals 6.
Elimination
Eliminate the wrong options
What is the ratio-test limit L, and the conclusion?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The ratio of consecutive terms is (n+1) over (3n), which tends to one third. Since one third is below 1, the ratio test gives convergence.
Check
Apply the ratio test to the series.
\[ \sum_{n=1}^{\infty} \frac{n}{3^{n}} \]
Check your understanding
What is the ratio-test limit L, and the conclusion?
Answer: A
Why: The ratio of consecutive terms is (n+1) over (3n), which tends to one third. Since one third is below 1, the ratio test gives convergence.
Check
Classify the alternating harmonic series.
\[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \]
Check your understanding
Which description is correct?
Answer: A
Why: The alternating series test gives convergence because one over n decreases to zero. But the series of absolute values is the harmonic series, which diverges, so the convergence is conditional, not absolute.
Check
Decide whether the series converges.
\[ \sum_{n=1}^{\infty} \frac{1}{n^{2}+5} \]
Check your understanding
Which reasoning correctly settles it?
Answer: A
Why: Each term is bounded above by one over n-squared, which is a p-series with power two and therefore converges. By the comparison test the smaller nonnegative terms also converge.
Check
Evaluate the telescoping series.
\[ \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \]
Check your understanding
What is the sum?
Answer: A
Why: Splitting the term as one over n minus one over n-plus-one telescopes the partial sum to one minus one over N-plus-one, which tends to 1 as N grows without bound.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — A decision recipe for testing a series · What could adding infinitely many numbers mean · The partial sum sequence · A series is a running total on the line · Convergence is convergence of the partial sums. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A series is nothing more than the limit of its partial sums. Convergence of the series means convergence of that sequence, and the sum is the limit.
The nth-term test only ever proves divergence. Geometric and telescoping series can be summed exactly; the harmonic series diverges even though its terms vanish.
Comparison, limit comparison, ratio, and root tests each fit a shape of series; the ratio and root tests fall silent when their limit is 1.
Absolute convergence is the robust, reorder-proof kind. Conditional convergence rests on cancellation, and the Riemann rearrangement theorem shows it can be steered to any sum at all.
| Situation | First move |
|---|---|
| terms do not go to zero | diverges (nth-term test) |
| geometric or telescoping | sum the closed form |
| factorials or n-th powers | ratio or root test |
| alternating signs | test absolute, then Leibniz |
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