5.2 Infinite Series

1. Infinite Series

Title

Calculus II · Section 5.2

Adding forever, made precise

2. What this lesson gives you

Objectives

You cannot perform infinitely many additions, so an infinite sum has to be defined. This lesson gives the definition and the two families of series whose sums you can compute exactly.

• Define the sum of a series as the limit of its partial sums, and use that definition directly
• Prove that the harmonic series diverges, even though its terms shrink to zero
• Use the sum, difference and constant-multiple rules, and know when they are licensed
• Decide when a geometric series converges and compute its exact sum
• Evaluate a telescoping series by cancelling inside the partial sum

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, pp. 450-470 — learning objectives 5.2.1 to 5.2.3

Everything in Chapter 5 rests on one definition, and this lesson gives it to you. An infinite series looks like an addition problem that never ends, and the whole difficulty is that you cannot actually do infinitely many additions. The way out is to add finitely many terms, which you can do, and then ask what those finite totals approach.

With that definition in hand you will meet the three series that the rest of the chapter keeps coming back to. The harmonic series is the great warning: its terms shrink to nothing and yet its total is infinite. Geometric series and telescoping series are the two families whose sums you can compute exactly, and later sections use them as yardsticks for everything else.

Along the way you will also learn which algebraic moves are allowed with infinite sums, and, just as important, which familiar moves quietly break.

3. Before anything new: two sequences from Section 5.1

Warm-up

Commit to an answer for each sequence before you reveal. These two are not random: they are about to reappear as the running totals of two series in this lesson, and recognising them will save you work.

The first sequence, k over k plus 1, creeps up toward 1: divide the top and bottom by k and the fraction becomes one over one plus one over k. The second, minus one to the k, jumps back and forth between minus one and one forever. A sequence converges only if its terms settle down near a single number, and this one settles nowhere.

Keep the Section 5.1 definition of a limit of a sequence in mind. In a few slides you will see that deciding whether a series converges is nothing more than deciding whether one particular sequence converges.

4. What an infinite sum means

Section

Part 1

5. Oil seeping into a lake

Concept

A thousand gallons of oil enter a lake in the first week, five hundred in the second, and each week half as much as the week before. Measure in thousands of gallons and keep a running total.

\[ S_1 = 1, \quad S_2 = 1 + \tfrac12, \quad S_3 = 1 + \tfrac12 + \tfrac14, \quad S_4 = 1 + \tfrac12 + \tfrac14 + \tfrac18 \]

The running totals creep up on 2 and never pass it. The amount of oil in the lake approaches two thousand gallons.

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, pp. 450-451 — the oil example and Figure 5.10

This is the book's opening example, and it is worth slowing down on because it shows the two things a series involves. Each week a new amount of oil arrives: that is a term. The amount in the lake so far is the running total: that is a partial sum.

Look at the figure. The orange bars are the weekly amounts, and they shrink by half each week. The blue dots are the running totals, and they rise quickly at first, then flatten out just below the dashed line at 2. The table tells the same story in numbers: after twenty weeks the total is 1.999998 thousand gallons.

So the honest answer to how much oil ends up in the lake is: the totals get as close to two thousand gallons as you like, and never exceed it. That sentence is exactly what it will mean to say the series equals 2.

6. Adding forever is not an operation

Intuition

Addition takes two numbers and returns one. Repeat it and you can add a million numbers, but never infinitely many: there is always a next term waiting.

So an infinite sum cannot be computed; it has to be defined. The definition uses the one tool you have for infinity, a limit, applied to the only things you can compute, the finite running totals.

\[ \underbrace{a_1 + a_2 + \cdots + a_k}_{\text{a finite sum you can compute}} \quad \xrightarrow{\;k\to\infty\;} \quad \text{the value of the series} \]

It is easy to read an infinite sum as an instruction, add all of these, and assume it must have an answer. But addition is defined for two numbers at a time, and by repeating it you can only ever reach finitely many terms. There is no last step at which the adding is finished.

So mathematicians do not compute infinite sums; they define them. The definition uses the finite totals, which you can always compute, and then asks for their limit, a notion you already made precise for sequences in Section 5.1.

The payoff of treating it as a definition is that some infinite sums will simply fail to have a value, and that is not a paradox. It just means the running totals have no limit. You will see several examples in the next few slides.

7. Partial sums and the sum of a series

Concept

partial sum — The kth partial sum of a series is the total of its first k terms. The partial sums form a sequence of their own.

\[ S_k = \sum_{n=1}^{k} a_n = a_1 + a_2 + \cdots + a_k \]

convergence of a series — The series converges if its sequence of partial sums converges to a real number S, called the sum of the series. Otherwise the series diverges.

\[ \sum_{n=1}^{\infty} a_n = S \quad\text{means}\quad \lim_{k\to\infty} S_k = S \]

A series therefore involves two sequences: the terms, which are the steps, and the partial sums, which are where you stand after each step. Convergence is decided entirely by the second.

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 452 — definition of an infinite series

Read the two definitions slowly, because every later test depends on them. The kth partial sum is an ordinary finite sum: stop after k terms and add. As k runs through 1, 2, 3 and onward, the partial sums form a brand-new sequence.

The series converges exactly when that new sequence converges, and its sum is the limit. If the partial sums grow without bound, or keep oscillating, the series diverges and has no sum at all.

The last line on the slide is the one to remember. A series carries two sequences: the terms, which are the individual steps, and the partial sums, which record where you are after each step. Convergence is a property of the second one. Almost every mistake in this chapter comes from looking at the first sequence when the question is about the second.

8. Reading the definition piece by piece

Notation

Step through the annotations one at a time. The key one is the first: the sigma with infinity on top is not a calculation you perform. It is a name for a limit, and it only refers to an actual number when that limit exists.

Notice that two different letters are doing two different jobs. The index n labels the terms inside a sum, and it is a dummy variable, like the variable of integration. The letter k says where you stop, and each value of k gives you one partial sum.

Once the definition is written this way, the whole toolkit of Section 5.1 applies. The limit laws, the squeeze theorem, and the Monotone Convergence Theorem are all statements about sequences, and a series is decided by a sequence.

9. Halving the distance to two

Picture it

Each week's oil covers exactly half of the gap between the running total and 2. After k weeks the gap is one over two to the k minus 1, which shrinks to zero, so the partial sums converge to 2.

\[ 2 - S_k = \frac{1}{2^{k-1}} \to 0 \quad\Longrightarrow\quad \sum_{n=1}^{\infty}\left(\frac12\right)^{n-1} = 2 \]

The number line shows why the oil totals approach 2 and never pass it. The first jump takes you from 0 to 1, halfway to 2. The next jump is one half, which again covers half of the remaining gap, landing you at one and a half. Each jump covers half of what is left.

After k jumps the gap that remains is one over two to the power k minus 1. That gap shrinks to zero, which is precisely the statement that the partial sums have limit 2. No single partial sum ever equals 2; the limit is where they are heading, not a place they arrive.

This picture is the ancient paradox of walking halfway to a wall, then halfway again, and so on. The resolution is exactly the definition you just learned: infinitely many steps can have a finite total, because the total is defined as a limit.

10. The index can start anywhere

Concept

The same series can be written with its index starting at any whole number; only the list of terms matters.

\[ \sum_{n=1}^{\infty}\left(\frac12\right)^{n-1} = \sum_{n=0}^{\infty}\left(\frac12\right)^{n} = \sum_{n=5}^{\infty}\left(\frac12\right)^{n-5} \]

To move an index, substitute. Starting at n equal to 2 and putting m equal to n minus 1 shifts the start to 1:

\[ \sum_{n=2}^{\infty}\frac{1}{n^2} = \sum_{m=1}^{\infty}\frac{1}{(m+1)^2} \qquad (n = m+1) \]

Both sides list one quarter, one ninth, one sixteenth, and so on. Changing the starting point of a series without substituting changes which terms are added.

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 452 — reindexing

Series in the wild do not always start at n equal to 1. The book writes the oil series three different ways, with the index starting at 1, at 0 and at 5, and all three list the same numbers: 1, one half, one quarter and so on.

To move the starting point without changing the series, substitute, exactly as you would change the variable in an integral. If the series starts at n equal to 2 and you want it to start at 1, set m equal to n minus 1. Then n equal to 2 becomes m equal to 1, and every n in the formula becomes m plus 1.

The check is always the same: write the first two or three terms of each version and make sure they agree. Changing the lower limit without changing the formula adds or removes terms, and that changes the sum.

11. Example 5.7(a): every term is at least one half

Worked example

Use the sequence of partial sums to decide whether this series converges.

\[ \sum_{n=1}^{\infty}\frac{n}{n+1} \]

\[ S_1 = \tfrac12, \quad S_2 = \tfrac12 + \tfrac23, \quad S_3 = \tfrac12 + \tfrac23 + \tfrac34 \]

\[ \frac{n}{n+1} \ge \frac12 \iff 2n \ge n + 1 \iff n \ge 1 \]

\[ S_k \ge \underbrace{\tfrac12 + \tfrac12 + \cdots + \tfrac12}_{k} = \frac{k}{2} \]

\[ \frac{k}{2} \to \infty \;\Longrightarrow\; \{S_k\} \text{ unbounded} \;\Longrightarrow\; \text{diverges} \]

\[ S_4 \approx 2.717 > 2, \qquad S_{10} \approx 7.980 > 5 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 453 — Example 5.7a

The question is about the partial sums, so the first move is to write a few out. They quickly get messy to add exactly, and that is a hint: you do not need their exact values, only whether they are bounded.

The key observation is that no term is smaller than one half. Cross-multiplying turns that into a statement you can see at once, that n is at least 1. Then k terms, each at least one half, add up to at least k over 2.

In the figure the dots sit on or above the dashed line k over 2, and since that line climbs forever, so do the dots. An unbounded sequence cannot converge, because convergent sequences are bounded, a fact from Section 5.1. So the series diverges.

The check uses real numbers: after four terms the total is about 2.717 and after ten about 7.980, both comfortably above the bound.

12. Example 5.7(b): a running total that flips

Worked example

\[ \sum_{n=1}^{\infty}(-1)^n = -1 + 1 - 1 + 1 - \cdots \]

\[ S_1 = -1 \]

\[ S_2 = -1 + 1 = 0 \]

\[ S_3 = -1 + 1 - 1 = -1 \]

\[ S_k = \begin{cases} -1 & k \text{ odd} \\ 0 & k \text{ even} \end{cases} \]

\[ \{S_k\} = \{-1, 0, -1, 0, \dots\} \;\Longrightarrow\; \sum (-1)^n \text{ diverges} \]

\[ S_{2j-1} = -1 \to -1, \qquad S_{2j} = 0 \to 0, \qquad -1 \ne 0 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 453 — Example 5.7b

This series adds and subtracts one forever. Writing out the partial sums one term at a time shows the pattern immediately: minus one, zero, minus one, zero. Each pair of terms cancels, so the running total depends only on whether you have added an odd or an even number of terms.

That is the warm-up's bouncing sequence again, just shifted down, and it has no limit. So the series diverges. It does not diverge to infinity; its partial sums are perfectly bounded. It diverges because they never settle, which the figure shows as a seesaw between two levels.

The check makes the reason precise. If the partial sums had a limit, then the partial sums at odd positions and at even positions would both approach that same limit. They approach minus one and zero, which are different, so there can be no limit.

13. Example 5.7(c): partial sums with a pattern

Worked example

\[ \sum_{n=1}^{\infty}\frac{1}{n(n+1)} \]

\[ S_1 = \frac{1}{1\cdot 2} = \frac12 \]

\[ S_2 = \frac12 + \frac16 = \frac46 = \frac23 \]

\[ S_3 = \frac23 + \frac1{12} = \frac9{12} = \frac34 \]

\[ S_4 = \frac34 + \frac1{20} = \frac{16}{20} = \frac45 \]

\[ S_k = \frac{k}{k+1} \]

\[ \lim_{k\to\infty}\frac{k}{k+1} = 1 \;\Longrightarrow\; \sum_{n=1}^{\infty}\frac{1}{n(n+1)} = 1 \]

\[ S_{10} = \sum_{n=1}^{10}\frac{1}{n(n+1)} \approx 0.909091 = \frac{10}{11} \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, pp. 453-454 — Example 5.7c

Here the partial sums can be added exactly, and doing so carefully, one term at a time over a common denominator, reveals a striking pattern: one half, two thirds, three quarters, four fifths. Every partial sum is k over k plus 1.

That is the first warm-up sequence, whose limit is 1, so the series converges and its sum is 1. The figure shows both sequences at once: the orange terms dwindle toward zero, while the blue partial sums rise toward the dashed line at 1. The sum is where the blue dots go.

Be honest about what has been proved. Spotting a pattern in five cases is not a proof that it holds for every k. The check confirms it at k equal to 10, and Part 5 of this lesson turns the pattern into a proof with partial fractions and telescoping.

14. Checkpoint 5.7: n plus 1 over n

Prediction

\[ \sum_{n=1}^{\infty}\frac{n+1}{n} = 2 + \frac32 + \frac43 + \frac54 + \cdots \]

\[ \frac{n+1}{n} = 1 + \frac1n > 1 \;\Longrightarrow\; S_k > k \to \infty \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 454 — Checkpoint 5.7

Choose an answer before revealing. The tempting wrong answer is that the series converges to 1, because the terms approach 1. That is the limit of the terms, which is the wrong sequence.

Rewrite each term as one plus one over n. Every term is bigger than one, so after k terms the running total is bigger than k. The partial sums are unbounded, so the series diverges, exactly as in Example 5.7(a).

A useful habit is emerging here: when the terms approach a nonzero number, the running total gains about that much at every step, and it cannot level off. Section 5.3 turns this habit into a theorem, the divergence test.

15. Trap: reporting the limit of the terms

Trap

16. The harmonic series

Section

Part 2

17. The harmonic series climbs, but slowly

Concept

\[ \sum_{n=1}^{\infty}\frac1n = 1 + \frac12 + \frac13 + \frac14 + \cdots \]

The terms shrink to zero and the totals grow at a crawl. From the table alone you could not tell whether they level off. Only an argument can decide.

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 454 — equation 5.5 and the table of partial sums

The harmonic series is the sum of the reciprocals of the whole numbers. Its terms shrink to zero, and the table in the book, reproduced here, shows its partial sums growing astonishingly slowly: after a million terms the total is still under fifteen.

The figure puts k on a logarithmic scale, so each tick is ten times the one before. On that scale the dots rise in an almost straight line, gaining about 2.3 for every factor of ten. That is what logarithmic growth looks like: unbounded, but so slow that no amount of computation will show you the growth continuing forever.

So the numbers alone are genuinely ambiguous. They are consistent with a series that levels off at twenty, or at a hundred, or never. Deciding it needs an argument, and the next few slides give you one.

18. How many terms to pass ten?

Estimation

\[ S_{83} > 5, \qquad S_{12\,367} > 10 \]

Make your guess before revealing. Most people guess a few hundred, reasoning that doubling the target should roughly double the work.

The real answer, found by adding terms one at a time on a computer, is 12,367 terms. Passing five took 83 terms; passing ten takes about a hundred and fifty times as many. Each extra unit of height costs roughly e times as many terms as the one before, which is the signature of growth like a logarithm.

Keep this in mind whenever a computer seems to show a series settling down. Slow divergence and convergence can look identical over any range of terms you could ever compute.

19. Each block is worth at least a half

Picture it

Split the terms into blocks that double in length: one term, one term, two, four, eight, and so on. Every term in a block is at least the block's last term, and a block of that many terms each at least that big adds up to at least one half.

This picture is the idea behind the proof on the next slide, so study it before the algebra. The bars have heights one, one half, one third and so on. They are coloured in blocks, and each block is twice as long as the one before it: two bars, then four, then eight.

Inside each block, every bar is at least as tall as the block's last bar, marked by the dashed line. The block from 3 to 4 has two bars, each at least a quarter. The block from 5 to 8 has four bars, each at least an eighth. The block from 9 to 16 has eight bars, each at least a sixteenth. In every case the block adds up to more than one half.

Twice as many bars, each half as tall: the doubling exactly compensates for the shrinking. Since there are infinitely many blocks, each worth more than a half, the total cannot be finite.

20. Grouping proves the harmonic series diverges

Worked example

Show the partial sums are unbounded by grouping terms.

\[ S_2 = 1 + \frac12 \]

\[ S_4 = S_2 + \left(\frac13 + \frac14\right) > S_2 + \left(\frac14 + \frac14\right) \]

\[ S_4 > 1 + \frac12 + \frac12 = 1 + 2\left(\frac12\right) \]

\[ S_8 = S_4 + \left(\frac15 + \frac16 + \frac17 + \frac18\right) > S_4 + 4\cdot\frac18 \]

\[ S_8 > 1 + 3\left(\frac12\right) \]

\[ \frac{1}{2^{j-1}+1} + \cdots + \frac{1}{2^{j}} > 2^{j-1}\cdot\frac{1}{2^{j}} = \frac12 \]

\[ S_{2^j} > 1 + \frac{j}{2} \]

\[ 1 + \frac{j}{2} \to \infty \;\Longrightarrow\; \sum_{n=1}^{\infty}\frac1n \text{ diverges} \]

\[ S_{16} \approx 3.381 > 1 + \tfrac42, \qquad S_{32} \approx 4.058 > 1 + \tfrac52 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, pp. 454-455 — the harmonic series diverges

Follow the argument block by block. After the first two terms, group one third with one quarter; since one third is more than one quarter, that pair is worth more than two quarters, which is a half. Then group the next four terms, each at least an eighth, and they are worth more than four eighths, again a half.

The general step is the heart of it. The jth block runs from just past 2 to the j minus 1 up to 2 to the j, so it contains 2 to the j minus 1 terms, each at least one over 2 to the j. Multiply, and the block is worth more than one half. After j blocks, the partial sum at 2 to the j terms is more than 1 plus j halves.

That lower bound grows without limit, so the partial sums are unbounded and the harmonic series diverges. The check compares the bound with real partial sums at sixteen and thirty-two terms, and both clear it. Section 5.3 proves the same fact a second way, by comparing with an integral.

21. Why must the blocks double?

Socratic

Think about this before revealing. It is worth understanding why the proof is built the way it is, because the same design appears again in later tests.

If every block had the same length, say ten terms, the early blocks would be worth a lot, but the later ones would be worth less and less, because their terms keep getting smaller. The block of ten terms just before a thousand is worth only about a hundredth. Adding infinitely many shrinking contributions might, as far as this argument knows, still give a finite total.

Doubling the length of each block is exactly what cancels the shrinking. The terms in a block are roughly half the size of those in the previous block, but there are twice as many of them, so every block keeps the same guaranteed value. That fixed value, repeated forever, is what forces the total to infinity.

22. Find the error: a million terms is enough

Error analysis

Read the claim first and decide where the argument breaks before you open the annotations. The number itself is not the problem: the millionth partial sum really is about 14.39.

The error is the arrow. Using a partial sum to approximate a series only makes sense when the series has a sum to approximate. Here it does not: the grouping proof shows the partial sums exceed every bound eventually, so there is no sum near 14.4 or anywhere else.

This is the reason proofs matter in this chapter. Computers are wonderful for approximating sums of series that are known to converge, and they are useless for deciding whether a series converges in the first place.

23. How far can a stack of blocks lean?

Real world

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 468 — Exercise 116

This is one of the book's exercises, and it is a real experiment you can try with books or playing cards. Place the top block so its centre of mass sits right at the edge of the block below; then place the stack of two so its combined centre of mass sits at the edge of the next one, and so on down to the table.

The balance condition in the exercise gives the overhang of the kth block from the top as one over 2k. So the total overhang is one half, plus one quarter, plus one sixth, and so on: exactly half of a harmonic partial sum. The figure draws five blocks, whose top block sticks out 1.14 block lengths past the table.

Now use what you know. Harmonic partial sums are unbounded, so half of them are unbounded too, and with enough blocks the top one can hang as far out as you like. It just takes a lot of blocks: thirty-one to get past two block lengths.

24. Algebra with convergent series

Section

Part 3

25. Sums, differences and constant multiples

Concept

Partial sums are finite sums, so ordinary algebra applies to them. For every k:

\[ \sum_{n=1}^{k}(a_n + b_n) = \sum_{n=1}^{k}a_n + \sum_{n=1}^{k}b_n, \qquad \sum_{n=1}^{k}c\,a_n = c\sum_{n=1}^{k}a_n \]

If both series converge, the limit laws for sequences let k go to infinity on each side:

\[ \sum_{n=1}^{\infty}(a_n \pm b_n) = \sum_{n=1}^{\infty}a_n \pm \sum_{n=1}^{\infty}b_n, \qquad \sum_{n=1}^{\infty}c\,a_n = c\sum_{n=1}^{\infty}a_n \]

Theorem 5.7 — If both series converge, then the sum, difference and any constant multiple converge, with the sums behaving as above.

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 455 — Theorem 5.7, algebraic properties of convergent series

The proof of Theorem 5.7 is two lines, and it is worth seeing why it works, because that tells you when it does not. For any finite k the partial sums obey ordinary algebra: you can rearrange a finite sum however you like.

Now let k go to infinity. The limit of a sum of two sequences is the sum of their limits, but only when both limits exist. That is the limit law from Section 5.1, and it is where the hypothesis comes from: both series must already converge.

So the rules are exactly what you would hope, with one condition attached. You can split a series into two pieces, or pull out a constant, provided the pieces you end up with converge. The trap at the end of this part shows what happens when that condition is ignored.

26. Example 5.8: combining two known sums

Worked example

\[ \sum_{n=1}^{\infty}\left[\frac{3}{n(n+1)} + \left(\frac12\right)^{n-2}\right] \]

\[ \sum_{n=1}^{\infty}\frac{1}{n(n+1)} = 1, \qquad \sum_{n=1}^{\infty}\left(\frac12\right)^{n-1} = 2 \]

\[ = \sum_{n=1}^{\infty}\frac{3}{n(n+1)} + \sum_{n=1}^{\infty}\left(\frac12\right)^{n-2} \]

\[ \left(\frac12\right)^{n-2} = \left(\frac12\right)^{-1}\left(\frac12\right)^{n-1} = 2\left(\frac12\right)^{n-1} \]

\[ = 3\sum_{n=1}^{\infty}\frac{1}{n(n+1)} + 2\sum_{n=1}^{\infty}\left(\frac12\right)^{n-1} \]

\[ = 3(1) + 2(2) = 7 \]

\[ S_k = 3\left(1 - \tfrac{1}{k+1}\right) + 4\left(1 - \tfrac{1}{2^k}\right) \to 7, \quad S_{30} \approx 6.9032 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, pp. 455-456 — Example 5.8

This example shows the rules at work. The series is built from two you already know: three times the series from Example 5.7(c), and a geometric series closely related to the oil series.

Splitting is allowed because both pieces converge, and that justification belongs in your answer. The only real work is matching the exponent: one half to the n minus 2 is one half to the minus 1 times one half to the n minus 1, and one half to the minus 1 is simply 2. So the second piece is twice the oil series.

With constants pulled out, you substitute the known sums, 1 and 2, and get 3 plus 4, which is 7. The check writes the kth partial sum using both known partial-sum formulas and confirms that it tends to 7; at thirty terms it is already 6.9032.

27. Checkpoint 5.8: five over two to the n minus 1

Prediction

\[ \sum_{n=1}^{\infty}\frac{5}{2^{n-1}} \]

\[ \sum_{n=1}^{\infty}\frac{5}{2^{n-1}} = 5\sum_{n=1}^{\infty}\left(\frac12\right)^{n-1} = 5(2) = 10 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 456 — Checkpoint 5.8

Choose before revealing. The idea is to recognise the series as a constant multiple of one you already know, rather than working from scratch.

Five over two to the n minus 1 is five times one half to the n minus 1. The oil series, one half to the n minus 1, sums to 2, and the constant multiple rule says multiplying every term by 5 multiplies the sum by 5. So the sum is 10.

If you chose five halves or five, check which series you were actually summing. The first term here is 5, and the partial sums run 5, 7.5, 8.75, and so on, which already rules out any answer below 8.75.

28. Exercises 83 to 86: sums from given sums

Comparison

Suppose the series of a sub n sums to 1 with first term 2, and the series of b sub n sums to minus 1 with first term minus 3. Fill in the blanks.

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 466 — Exercises 83-86

Fill the blanks before checking. You are told only the sums of two series and their first terms, and the algebraic rules let you find the sums of many related series without knowing any other terms.

The first two rows are straight applications of the sum and constant multiple rules. The last two need one more idea: removing the first term from a convergent series subtracts that term from its sum. So the series of a sub n from n equal to 2 sums to 1 minus 2, and the series of b sub n from 2 sums to minus 1 minus minus 3, which is 2.

Shifting the index from n to n plus 1, as in the last row, is the same move in disguise: the series of a sub n plus 1 starts at the second term. Then 3 times minus 1, minus 4 times 2, gives minus 11.

29. Licensed or not?

Sorting

Sort each move by whether it is always justified. The deciding question every time is whether the argument uses a limit law on sequences whose limits actually exist.

The sum and constant multiple rules are licensed when the pieces converge. Dropping finitely many terms never changes whether a series converges, because every partial sum from then on shifts by the same fixed amount. But it does change the value, by exactly the terms you dropped.

The unlicensed moves are unlicensed for real reasons. A convergent series can split into two divergent ones, as the trap on the next slide shows. And series do not multiply term by term: the geometric series of one half to the n minus 1 sums to 2, so the product of two copies is 4, but the series of their termwise products sums to four thirds.

30. Trap: splitting into infinity minus infinity

Trap

31. Geometric series

Section

Part 4

32. A fixed ratio between terms

Concept

geometric series — A series in which each term is the previous term multiplied by the same number r, the ratio. The first term a is the initial term.

\[ \sum_{n=1}^{\infty}a r^{n-1} = a + ar + ar^2 + ar^3 + \cdots \]

To find a and r, write out a few terms: a is the first one, and r is any term divided by the one before it. The oil series has first term 1 and ratio one half.

\[ 1 + \frac12 + \frac14 + \frac18 + \cdots: \quad a = 1, \quad r = \frac{1/2}{1} = \frac12 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 456 — equation 5.6

A geometric series is one where you get from each term to the next by multiplying by the same number. That number is the ratio r, and the first term is called a. The oil series is the standard example, with first term 1 and ratio one half.

Finding a and r reliably is a skill in itself, because series are often written in forms that disguise them. The safe method is always the same: write out the first two or three terms, take the first one as a, and divide the second by the first to get r. Do not try to read them off the formula, because an exponent like n plus 2 or a starting index of 0 changes the first term.

Geometric series matter well beyond this section. In Chapter 6 they are the starting point for writing functions as infinite polynomials.

33. Deriving the geometric partial sum

Worked example

Find a formula for the kth partial sum, then let k grow. Take a nonzero.

\[ S_k = a + ar + ar^2 + \cdots + ar^{k-1} \]

\[ r = 1: \quad S_k = a + a + \cdots + a = ak \to \pm\infty \]

\[ rS_k = ar + ar^2 + \cdots + ar^{k-1} + ar^k \]

\[ S_k - rS_k = a - ar^k \]

\[ (1-r)S_k = a(1-r^k) \]

\[ S_k = \frac{a(1-r^k)}{1-r}, \quad r \ne 1 \]

\[ S_5 = \frac{1 - (1/2)^5}{1 - 1/2} = \frac{31/32}{1/2} = \frac{31}{16} = 1.9375 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 457 — the partial sums of a geometric series

This derivation is short, and it is worth being able to reproduce it, because it is the only reason the geometric sum formula is true. First, deal with the ratio 1 separately: every term is a, so the partial sums are a times k and grow without bound.

For any other ratio, write the partial sum, then write r times it underneath. Multiplying by r shifts every term one place to the right, so when you subtract, all but two terms cancel: the first term of the top line and the last term of the bottom line. Factoring and dividing by one minus r gives the closed form.

The check plugs in the oil series with five terms and recovers 1.9375, the first entry in the book's table. Notice that nothing so far required the ratio to be small; this formula for the partial sum holds for every ratio except 1.

34. Everything depends on r to the k

Concept

\[ S_k = \frac{a}{1-r} - \frac{a}{1-r}\,r^k \]

The first piece is a fixed number. Only r to the k changes with k, and Section 5.1 settled what that geometric sequence does.

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 457 — convergence of a geometric series

Rewriting the partial sum as a fixed number minus a multiple of r to the k shows exactly what matters: the only part that changes with k is r to the k. So the fate of the series is the fate of a geometric sequence, which Section 5.1 already settled.

Look at the four panels. With ratio one half, the dots rise steadily to 2. With ratio minus one half, the terms alternate in sign, so the partial sums zigzag, overshooting and undershooting two thirds, but the zigzags shrink and they still converge. With ratio 1.1, the partial sums take off. With ratio minus one, they flip between 1 and 0 forever, as in Example 5.7(b).

The table summarises it: convergence exactly when the ratio is less than one in size. The sign of the ratio affects how the partial sums approach the limit, not whether they do.

35. Reading the geometric sum formula

Notation

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 457 — equation 5.7

Go through the annotations one at a time; each one prevents a specific mistake. The first is about a. It is the first term of the series as written, not a coefficient in the formula. For the series of two thirds to the n plus 2, starting at 0, the first term is four ninths, not 1 and not two thirds.

The ratio can be negative, and then the terms alternate in sign. That is fine as long as its size is less than one. The formula a over one minus r is simply the limit of the partial-sum formula once r to the k has vanished.

The last annotation is the one that costs marks. Outside the interval from minus one to one, the formula still produces a number, and that number means nothing. Always check the ratio first.

36. Slide the ratio

Tweak it

Start at a ratio of one half and watch the staircase settle at 2. Then slide the ratio toward 1: the level the stairs approach, one over one minus r, climbs higher and higher, and the stairs take longer to get there.

Now slide the ratio to negative values. The staircase starts to zigzag, stepping above and below its limit, but for any ratio strictly between minus one and zero the zigzags still shrink and the sum is still one over one minus r. At minus one, the zigzag stops shrinking and the partial sums flip between 1 and 0.

Finally push the ratio past 1 or below minus 1. The steps grow without bound. There is no gradual transition at the boundary: for every ratio of size less than one the series converges, and at size one it already fails.

37. Rewriting a series into geometric form

Worked example

\[ \sum_{n=0}^{\infty}\left(\frac23\right)^{n+2} \]

\[ = \left(\frac23\right)^2 + \left(\frac23\right)^3 + \left(\frac23\right)^4 + \cdots \]

\[ a = \frac49, \qquad r = \frac23 \]

\[ = \sum_{n=1}^{\infty}\frac49\left(\frac23\right)^{n-1} \]

\[ |r| = \frac23 < 1 \]

\[ \frac{4/9}{1 - 2/3} = \frac{4/9}{1/3} = \frac43 \]

\[ S_{10} = \frac43\left(1 - \left(\tfrac23\right)^{10}\right) \approx 1.3102, \quad 1.3102 + 0.0231 = 1.3333 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, pp. 457-458 — rewriting a geometric series

This is the book's example of a geometric series in disguise. The exponent is n plus 2 and the index starts at 0, so reading a and r directly from the formula is risky. Writing out the first few terms removes all doubt: they are two thirds squared, cubed, to the fourth, and so on.

So the first term is four ninths and each term is two thirds of the one before. Rewriting in standard form is optional, but it makes the match with the formula explicit. The ratio is less than one, so the sum is four ninths over one third, which is four thirds.

The check uses the partial-sum formula at ten terms, 1.3102, and notices that what is missing is exactly the remaining tail, four thirds times two thirds to the tenth, about 0.0231. Together they make four thirds.

38. Example 5.9(a): a negative ratio

Worked example

\[ \sum_{n=1}^{\infty}\frac{(-3)^{n+1}}{4^{n-1}} \]

\[ = \frac{(-3)^2}{4^0} + \frac{(-3)^3}{4^1} + \frac{(-3)^4}{4^2} + \cdots \]

\[ = 9 - \frac{27}{4} + \frac{81}{16} - \cdots \]

\[ r = \frac{-27/4}{9} = -\frac34 \]

\[ a = 9, \qquad |r| = \frac34 < 1 \]

\[ \frac{9}{1 - (-3/4)} = \frac{9}{7/4} = \frac{36}{7} \]

\[ S_{20} \approx 5.1265 < \frac{36}{7} \approx 5.1429 < 5.1551 \approx S_{21} \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 458 — Example 5.9a

Here both the numerator and denominator are powers, and the signs alternate. Writing out the first three terms turns it into something you can read: 9, then minus twenty-seven quarters, then eighty-one sixteenths.

Dividing the second term by the first gives the ratio, minus three quarters. Its size is three quarters, less than one, so the series converges, and the sum is 9 over one plus three quarters, which is thirty-six sevenths, about 5.14.

The book's solution has a misprint here: it says the initial term is minus 3, then correctly divides 9 by seven quarters. The first term is 9. The check shows the zigzag in action: the twentieth partial sum is just below thirty-six sevenths and the twenty-first just above, closing in from both sides.

39. Example 5.9(b): a ratio bigger than one

Worked example

\[ \sum_{n=1}^{\infty}e^{2n} \]

\[ e^{2n} = e^{2}\cdot e^{2(n-1)} = e^2\left(e^2\right)^{n-1} \]

\[ a = e^2, \qquad r = e^2 \approx 7.389 \]

\[ |r| = e^2 > 1 \;\Longrightarrow\; \sum e^{2n} \text{ diverges} \]

\[ S_k > e^{2k} \to \infty \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 458 — Example 5.9b

The series of e to the 2n does not look geometric at first, but every term is e squared times the previous one, which is the defining property. Splitting the exponent as 2 plus 2 times n minus 1 makes the standard form visible.

Both the first term and the ratio are e squared, about 7.389, which is far bigger than one. So the series diverges; you should not even think about the formula a over one minus r here.

The check needs no theory at all. Every term is positive, so any partial sum is bigger than its own last term, and the last term, e to the 2k, grows without bound. Whenever the terms themselves blow up, the running total certainly does.

40. Checkpoint 5.9: ratio minus two fifths

Prediction

\[ \sum_{n=1}^{\infty}\left(-\frac25\right)^{n-1} \]

\[ \frac{1}{1-(-2/5)} = \frac{1}{7/5} = \frac57 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 458 — Checkpoint 5.9

Commit to an answer first. This is a geometric series in standard form, so a and r can be read off safely: the first term is 1 and the ratio is minus two fifths.

The size of the ratio is two fifths, less than one, so the series converges. One minus minus two fifths is seven fifths, and 1 divided by seven fifths is five sevenths, about 0.714.

If you picked five thirds, you lost the minus sign on the ratio; if you picked minus two thirds, you probably subtracted in the wrong order. A quick sanity check is to look at the first two partial sums, 1 and three fifths: the sum must lie between them, since the partial sums zigzag, and only five sevenths does.

41. Example 5.10: a repeating decimal as a fraction

Worked example

\[ 3.\overline{26} = 3.262626\ldots \]

\[ = 3 + \frac{26}{100} + \frac{26}{10^4} + \frac{26}{10^6} + \cdots \]

\[ a = \frac{26}{100}, \qquad r = \frac{1}{100} \]

\[ \frac{26/100}{1 - 1/100} = \frac{26/100}{99/100} = \frac{26}{99} \]

\[ 3 + \frac{26}{99} = \frac{297 + 26}{99} = \frac{323}{99} \]

\[ 323 = 3\cdot 99 + 26, \qquad \frac{26}{99} = 0.262626\ldots \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 459 — Example 5.10

Every repeating decimal is a geometric series in disguise, and this example shows how to unmask one. The digits 26 repeat, and each repetition is worth one hundredth of the previous one, because it sits two decimal places further along.

Set the whole number 3 aside; it is not part of the repeating pattern. The rest is geometric with first term twenty-six hundredths and ratio one hundredth, so its sum is twenty-six ninety-ninths. Add back the 3 and you get three hundred twenty-three ninety-ninths.

The check reverses the process by division, confirming that 323 over 99 really is 3.2626 and so on. The book's first line has a small misprint, writing 26 over a thousand for the second term where it should be 26 over ten thousand; its next line has the correct powers of ten. This method is also the proof that every repeating decimal is a rational number.

42. Checkpoint 5.10: five point two seven repeating

Fill the middle

\[ 5.\overline{27} = 5 + \frac{27}{100} + \frac{27}{10^4} + \cdots = 5 + \frac{27/100}{1 - 1/100} \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 459 — Checkpoint 5.10

Fill in the blank before checking. The structure is exactly the same as the example: a whole-number part, and a two-digit block repeating every two places, so the ratio is one hundredth.

The geometric part is twenty-seven hundredths divided by ninety-nine hundredths, which is twenty-seven ninety-ninths, and that reduces to three elevenths. Adding 5, which is fifty-five elevenths, gives fifty-eight elevenths.

The pattern you may have noticed is a real shortcut: a block of two repeating digits right after the decimal point becomes those digits over 99. A block of three would go over 999. The geometric series is the reason the shortcut works.

43. Building the Koch snowflake

Picture it

Start with an equilateral triangle of side 1. At every stage, remove the middle third of each side and build an outward equilateral triangle on the gap. The limiting shape is Koch's snowflake, and Example 5.11 asks for its perimeter and its area.

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, pp. 459-460 — Example 5.11 and Figure 5.11

This is the chapter opener, and it is a beautiful use of geometric series. Look across the four panels. At each stage, every side is split into thirds, and the middle third is replaced by two sides of a small outward triangle. One side becomes four, each a third as long.

The side counts under the pictures go 3, 12, 48, 192, multiplying by four each time, while the side lengths divide by three. The figures are drawn to the same scale, so you can also see that the new triangles get tiny very quickly and the overall size barely changes after the first step.

Two questions follow. What happens to the length of the boundary as the stages go on forever, and what happens to the enclosed area? The answers turn out to be opposite, and both come from geometric sequences and series.

44. Example 5.11(a): the snowflake's perimeter

Worked example

\[ N_0 = 3 \text{ sides}, \qquad l_0 = 1 \]

\[ N_1 = 4\cdot 3, \qquad l_1 = \frac13 \]

\[ N_n = 4^n\cdot 3, \qquad l_n = \left(\frac13\right)^n \]

\[ L_n = N_n\, l_n = 3\left(\frac43\right)^n \]

\[ \frac43 > 1 \;\Longrightarrow\; L = \lim_{n\to\infty}L_n = \infty \]

\[ L_1 = 12\cdot\tfrac13 = 4 = 3\left(\tfrac43\right), \qquad L_2 = 48\cdot\tfrac19 = \tfrac{16}{3} = 3\left(\tfrac43\right)^2 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, pp. 460-461 — Example 5.11a

The perimeter is the number of sides times the length of each side, so track those two separately. At every stage the number of sides is multiplied by four and the length of each side is divided by three.

After n stages there are 3 times 4 to the n sides, each of length one third to the n, and the perimeter is 3 times four thirds to the n. That is a geometric sequence with ratio four thirds, bigger than one, so it grows without bound. The snowflake's boundary is infinitely long.

The check counts sides in the pictures directly. The star has twelve sides of one third each, a perimeter of 4; the next figure has forty-eight sides of one ninth, sixteen thirds. Both agree with the formula.

45. Example 5.11(b): the snowflake's area

Worked example

\[ A_0 = T_0 = \frac{\sqrt3}{4} \]

\[ T_n = \frac19 T_{n-1} = \left(\frac19\right)^n\frac{\sqrt3}{4} \]

\[ A_n = A_{n-1} + N_{n-1}T_n \]

\[ N_{n-1}T_n = 3\cdot 4^{n-1}\left(\frac19\right)^n\frac{\sqrt3}{4} \]

\[ N_{n-1}T_n = \frac34\left(\frac49\right)^n\frac{\sqrt3}{4} \]

\[ A_n = \frac{\sqrt3}{4}\left[1 + \frac13\left(1 + \frac49 + \cdots + \left(\frac49\right)^{n-1}\right)\right] \]

\[ A_n = \frac{\sqrt3}{4}\left[1 + \frac13\cdot\frac{1 - (4/9)^n}{5/9}\right] = \frac{\sqrt3}{4}\left[\frac85 - \frac35\left(\frac49\right)^n\right] \]

\[ A = \frac{\sqrt3}{4}\cdot\frac85 = \frac{2\sqrt3}{5} \approx 0.6928 \]

\[ \frac{\sqrt3}{4}\left(1 + \tfrac39\right) = \frac{\sqrt3}{4}\cdot\frac43 = \frac{\sqrt3}{4}\left[\tfrac85 - \tfrac35\cdot\tfrac49\right] \approx 0.5774 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, pp. 461-462 — Example 5.11b

Area needs more care than perimeter. The original triangle has area root three over four. At each stage you add one new triangle on every side of the previous figure, and each new triangle has sides a third as long as the last batch, so a ninth of the area.

The amount added at stage n is the number of old sides times the area of one new triangle. Simplifying the powers turns it into three quarters of four ninths to the n, times root three over four. That is a geometric sequence with ratio four ninths.

Adding all the stages gives a geometric partial sum, which you now have a formula for. Letting n grow, four ninths to the n vanishes, and the area settles at two root three over five, about 0.693. The right-hand panel of the figure shows the areas levelling off at that value while the perimeters on the left take off. The check computes the star's area by hand and matches the formula.

46. Finite area inside an infinite boundary

Intuition

The snowflake fits inside a circle, so its area had to be finite, yet its boundary is infinitely long. Two geometric sequences, two ratios on opposite sides of one:

\[ L_n = 3\left(\frac43\right)^n \to \infty, \qquad A_n \to \frac{2\sqrt3}{5} \]

You have met this before. The region under one over x squared from 1 to infinity has infinite boundary but finite area, because the improper integral converges.

\[ \int_1^{\infty}\frac{dx}{x^2} = 1 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 462 — Example 5.11, analysis

It can feel impossible that a shape with an infinitely long boundary encloses a finite area. But the snowflake never leaves a fixed circle around the original triangle, so its area was always going to be finite; it is only the boundary that keeps wrinkling.

In terms of series, the difference is just the two ratios. Each stage multiplies the perimeter by four thirds, more than one, and multiplies the added area by four ninths, less than one. One geometric sequence diverges and the other converges.

The book points out that you have already met this in Chapter 3: the region under one over x squared from 1 to infinity has an infinitely long boundary, but its area, an improper integral, is exactly 1.

47. Trap: the formula with a ratio of two

Trap

48. A dose every six hours

Real world

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 468 — Exercises 127 and 128

Repeated doses of a medicine are a geometric series in real time. Just after a dose, your level is the new dose plus what remains of every earlier dose, each one decayed by another factor of 0.9 to the N. Long-term, that is a geometric series with ratio 0.9 to the N, whose sum is d over one minus 0.9 to the N.

The figure shows the resulting sawtooth for a dose of 0.9 every six hours. The peaks rise at first and then level off near 1.92, and the troughs level off near 1.02, both inside the band between the effective and safe limits.

The design question uses the ratio of trough to peak, which is 0.9 to the N and must be at least one half. That gives N at most about 6.58, so six hours. Then each limit gives a bound on the dose, and together they allow doses from about 0.88 to about 0.94.

49. Telescoping series

Section

Part 5

50. Partial fractions make the terms cancel

Concept

Return to Example 5.7(c), where the pattern for the partial sums was only guessed. Split each term with partial fractions:

\[ \frac{1}{n(n+1)} = \frac1n - \frac{1}{n+1} \]

\[ S_k = \left(1 - \frac12\right) + \left(\frac12 - \frac13\right) + \cdots + \left(\frac1k - \frac{1}{k+1}\right) \]

\[ S_k = 1 - \frac{1}{k+1} \;\to\; 1 \]

Every middle term appears once with a plus and once with a minus. The partial sum collapses, like a spyglass, to its first and last pieces.

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, pp. 462-463 — telescoping series

Here is the proof that the pattern from Example 5.7(c) was right. Partial fractions split each term into one over n minus one over n plus 1, and now every term shares a piece with its neighbour.

Write the kth partial sum out in full and look at it. The minus one half from the first bracket meets the plus one half from the second; the minus one third meets the plus one third; and so on all the way along. Only the very first piece, 1, and the very last piece, minus one over k plus 1, survive.

The figure shows the same thing geometrically. Each term is a drop from one height, one over n, to the next. Stack the drops end to end and they fill exactly the distance from the first height to the last, so all the intermediate heights disappear. As k grows the last height goes to zero, and the sum is 1. The book prints the first bracket as one plus one half; it should be one minus one half.

51. The general telescoping series

Concept

Any series whose terms are differences of consecutive members of one sequence telescopes:

\[ S_1 = b_1 - b_2, \qquad S_2 = (b_1 - b_2) + (b_2 - b_3) = b_1 - b_3 \]

\[ S_3 = (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) = b_1 - b_4 \]

\[ S_k = b_1 - b_{k+1} \]

telescoping series — A series in which most of the terms cancel in each partial sum, leaving only some of the first terms and some of the last. It converges exactly when the leftover last terms converge.

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 463 — definition of a telescoping series

The same cancellation works for any series whose terms are differences of consecutive members of one sequence, b at n minus b at n plus 1. Writing out the first three partial sums shows the pattern: the kth partial sum is always the first b minus the b just past the end.

That reduces the whole question to one sequence. If the b's converge to some number B, the series converges to the first b minus B. If they do not converge, the series diverges.

Notice what this means in practice. Evaluating a telescoping series is not a trick of cancelling infinitely many terms. It is an exact formula for every partial sum, followed by one ordinary limit.

52. Reading the telescoping formula

Notation

Step through the notes. The first is the practical skill: recognising that each term is a difference of the same sequence at two consecutive places. Sometimes it is visible, as with differences of cosines; often you have to create it with partial fractions or a logarithm rule.

The first b survives because nothing earlier cancels it. The moving piece at the end is where the convergence question lives, and the answer is only as good as your limit of that piece.

The last note is the one to take seriously. The equals sign is earned inside the finite partial sum, where rearranging and cancelling are ordinary algebra. Cancelling terms in the infinite expression directly is not justified, and the counterexample later in this part shows how it goes wrong.

53. Example 5.12: a telescoping cosine

Worked example

\[ \sum_{n=1}^{\infty}\left[\cos\left(\frac1n\right) - \cos\left(\frac1{n+1}\right)\right] \]

\[ b_n = \cos\left(\frac1n\right) \]

\[ S_2 = \left[\cos 1 - \cos\tfrac12\right] + \left[\cos\tfrac12 - \cos\tfrac13\right] = \cos 1 - \cos\tfrac13 \]

\[ S_k = \cos 1 - \cos\left(\frac{1}{k+1}\right) \]

\[ \frac{1}{k+1} \to 0 \;\Longrightarrow\; \cos\left(\frac{1}{k+1}\right) \to \cos 0 = 1 \]

\[ \sum_{n=1}^{\infty}\left[\cos\tfrac1n - \cos\tfrac1{n+1}\right] = \cos 1 - 1 \approx -0.4597 \]

\[ S_{1000} \approx -0.4596972, \qquad \cos 1 - 1 \approx -0.4596977 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, pp. 463-464 — Example 5.12

The terms are already written as differences, so the only setup is naming b sub n as the cosine of one over n. Writing the second partial sum out shows the cancellation happening: the cosine of one half appears once with each sign.

In general only the first cosine and the last one remain, so the kth partial sum is cos 1 minus the cosine of one over k plus 1. As k grows, one over k plus 1 goes to zero, and because cosine is continuous, its cosine goes to cos 0, which is 1. The sum is cos 1 minus 1, about minus 0.46.

The figure shows the partial sums dropping quickly onto that level. The check adds a thousand terms directly on a computer and matches the exact answer to six decimal places, which is a genuinely independent confirmation of the cancellation.

54. Checkpoint 5.11: a telescoping exponential

Worked example

\[ \sum_{n=1}^{\infty}\left[e^{1/n} - e^{1/(n+1)}\right] \]

\[ b_n = e^{1/n}, \qquad b_1 = e \]

\[ S_k = e - e^{1/(k+1)} \]

\[ e^{1/(k+1)} \to e^0 = 1 \]

\[ \sum_{n=1}^{\infty}\left[e^{1/n} - e^{1/(n+1)}\right] = e - 1 \approx 1.71828 \]

\[ S_{1000} = e - e^{1/1001} \approx 1.71728 = 1.71828 - 0.00100 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 464 — Checkpoint 5.11

This is the same structure as the cosine example, with the exponential in place of the cosine. The sequence being differenced is e to the one over n, whose first member is e itself.

The kth partial sum collapses to e minus e to the one over k plus 1. The exponent shrinks to zero and the exponential is continuous, so the last piece tends to 1, and the series converges to e minus 1, about 1.718.

The check is a nice bonus. The gap between the thousandth partial sum and the true sum is exactly the leftover piece, e to the one over 1001, minus 1, which is about one thousandth. For a telescoping series you know the error of every partial sum exactly.

55. Exercise 105: a telescoping series that diverges

Worked example

\[ \sum_{n=1}^{\infty}\ln\left(\frac{n}{n+1}\right) \]

\[ \ln\left(\frac{n}{n+1}\right) = \ln n - \ln(n+1) \]

\[ b_n = \ln n, \qquad b_1 = \ln 1 = 0 \]

\[ S_k = \ln 1 - \ln(k+1) = -\ln(k+1) \]

\[ -\ln(k+1) \to -\infty \;\Longrightarrow\; \text{diverges} \]

\[ \sum_{n=1}^{5}\ln\frac{n}{n+1} \approx -1.7918 = -\ln 6 \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 467 — Exercise 105

Telescoping does not mean converging, and this exercise shows it. The log of a quotient splits into a difference of logs, ln n minus ln of n plus 1, which is a perfect telescoping pattern with b sub n equal to ln n.

The first piece is ln 1, which is zero, so the kth partial sum is just minus ln of k plus 1. The cancellation worked perfectly, but what is left has no limit: the logarithm grows without bound, so the partial sums fall toward minus infinity, as the figure shows.

Notice also that the terms, ln of n over n plus 1, tend to ln 1, which is zero. So this is another series with shrinking terms that diverges, just like the harmonic series. The terms tending to zero never settles the question on their own.

56. Exercise 112: a three-factor denominator

Worked example

\[ \sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} \]

\[ \frac{1}{n(n+1)(n+2)} = \frac12\left[\frac1n - \frac2{n+1} + \frac1{n+2}\right] \]

\[ = \frac12\left[\left(\frac1n - \frac1{n+1}\right) - \left(\frac1{n+1} - \frac1{n+2}\right)\right] \]

\[ S_k = \frac12\left[\left(1 - \frac1{k+1}\right) - \left(\frac12 - \frac1{k+2}\right)\right] \]

\[ S_k \to \frac12\left[1 - \frac12\right] = \frac14 \]

\[ \frac16 + \frac1{24} + \frac1{60} = 0.225 = \frac12\left[\left(1 - \tfrac14\right) - \left(\tfrac12 - \tfrac15\right)\right] \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 467 — Exercise 112

With three factors in the denominator, partial fractions produce three pieces: one half of one over n, minus one over n plus 1, plus one half of one over n plus 2. On its own that does not look like a single telescoping difference.

The trick is to split the middle piece in two, which regroups the term as a difference of two telescoping differences. Each bracket telescopes separately, and you already know how to finish each one: the first leaves 1 minus one over k plus 1, the second one half minus one over k plus 2.

Both moving pieces vanish in the limit, leaving one half of one half, which is one quarter. The check adds the first three terms by hand, getting 0.225, and confirms the partial-sum formula gives the same value.

57. Telescoping pins down Euler's constant

Concept

The harmonic partial sums diverge, but their difference from a logarithm does not. The section's project studies that difference.

\[ T_k = \sum_{n=1}^{k}\frac1n - \ln k \]

The sequence decreases, since each step subtracts ln of one plus one over k and adds only one over k plus 1, and it stays above zero, so it converges by the Monotone Convergence Theorem. A telescoping sum of those steps shows the error after k terms is at most one over k.

\[ T_k \to \gamma \approx 0.5772, \qquad 0 < T_k - \gamma \le \frac1k \]

OpenStax Calculus Volume 2, §5.2 Infinite Series §5.2, p. 465 — student project: Euler's constant

This slide summarises the section's student project, which ties the harmonic series and telescoping together. The harmonic partial sums diverge, but they diverge in step with the natural logarithm, so their difference from ln k behaves well.

The table shows that difference falling: 1, then 0.626, then 0.582, and settling near 0.5772. Each step is a decrease, because the log grows by more than the new harmonic term adds, and the difference never drops below zero, so the Monotone Convergence Theorem guarantees a limit. That limit is Euler's constant, usually written gamma.

The project's final step uses a telescoping sum of the one-step differences to show the error after k terms is at most one over k. So a thousand terms pin gamma down to about three decimal places, which is exactly what the table shows.

58. Break the regrouping: one minus one forever

Counterexample

\[ 1 - 1 + 1 - 1 + \cdots \]

\[ (1-1) + (1-1) + \cdots = 0 \qquad\text{vs}\qquad 1 + (-1+1) + (-1+1) + \cdots = 1 \]

Commit to an answer before revealing. Both groupings look reasonable, and they give different answers, 0 and 1. That contradiction is the signal that something illegal has happened, and the illegal step is treating the infinite expression as if its terms could be regrouped freely.

Putting brackets in an infinite sum quietly changes which partial sums you are looking at: the grouping into pairs only sees the partial sums after an even number of terms. For this series the full sequence of partial sums is 1, 0, 1, 0, which has no limit, so the series diverges and there is no value for any grouping to find.

This is why every telescoping evaluation in this lesson cancels inside the finite partial sum first. Finite sums can be rearranged safely; the limit is taken only afterwards.

59. Put a telescoping evaluation in order

Ranking

Drag the steps into order before checking. The series used as the example is Exercise 81 from the book, with terms one over n plus 1 times n plus 2.

Partial fractions come first, because they create the differences. Then write the partial sum and cancel, which is safe because it is a finite sum. Only then take the limit of the leftover last piece. For this series b sub n is one over n plus 1, so the first piece is one half, the last piece tends to zero, and the sum is one half.

60. Putting it together

Section

Part 6

61. Pattern: finding the sum of a series

Pattern

1. Write out a few terms. The first terms reveal the ratio, the cancellation, or neither.
2. Constant ratio? Find a (the first term) and r. If r is less than one in size, the sum is a over one minus r; otherwise it diverges.
3. Differences of one sequence? Cancel inside S_k, then take the limit of the leftover last piece.
4. Known pieces? Split with the sum and constant multiple rules, but only into convergent series.
5. Otherwise, the partial sums decide. Never report the limit of the terms as the sum.

This is the order to work in when a question asks you to find the sum of a series. It always starts with writing out a few terms, because that is how you recognise which family, if any, you are dealing with.

If each term is a fixed multiple of the last, it is geometric: find the first term and the ratio, check the ratio, and use the formula or declare divergence. If each term is a difference of one sequence at neighbouring places, possibly after partial fractions or a log rule, it telescopes: cancel inside the partial sum and take one limit.

If the series is built from pieces you know, the algebraic rules let you split it, but only into convergent pieces. And if it is none of these, you are in the territory of Section 5.3 onward, where tests decide convergence even when no formula for the sum exists.

62. Match each series to its sum

Matching

Match each series to its sum before checking, and name the family each one belongs to as you go.

The first is geometric, but starting at n equal to 1 means its first term is one third, not 1: one third over two thirds is one half. The second is the telescoping series from this lesson, with sum 1. The third starts at n equal to 0, so its first term is 1 and its ratio minus one half: 1 over three halves is two thirds.

The last one telescopes, but to minus ln of k plus 1, which has no limit, so it diverges. Getting the first and third right depends entirely on reading the first term correctly, which is the most common slip with geometric series.

63. Check: summing a geometric series

Check

Identify the first term and the ratio before doing any arithmetic. The first term is 3 and each term is a quarter of the previous one, so the ratio is one quarter.

The ratio is less than one, so the formula applies: 3 over three quarters is 4. The distractors each come from a specific slip: reporting a single term, using the ratio in place of one minus the ratio, or using the wrong first term.

64. Check: what shrinking terms tell you

Check

This question tests the lesson's central warning. Terms that tend to zero are a necessary feature of every convergent series, as Section 5.3 will prove, but they are not enough.

You have seen both fates in this lesson: the harmonic series and the series of ln of n over n plus 1 both have terms tending to zero and diverge, while the series of one over n times n plus 1 has terms tending to zero and converges. So the only safe conclusion is that you have not concluded anything yet.

65. Check: a telescoping sum with a gap of two

Check

Write out the first several terms before choosing. With a gap of two, each negative piece cancels a positive piece two places later rather than one, so the cancellation leaves two pieces at the start and two at the end.

The surviving first pieces are 1 and one half, and the last two, one over k plus 1 and one over k plus 2, both vanish in the limit. So the sum is three halves. If you chose 1, you applied the pattern from the gap-of-one series without checking how many pieces survive.

66. Explain where the condition enters

Explain it to yourself

Write your two sentences before revealing. Being able to point to the exact line where a hypothesis is used is the best evidence that you understand a derivation rather than having memorised its result.

Look back at the derivation. Multiplying by the ratio, subtracting and dividing were all valid for any ratio except 1, so the closed form for the partial sum has no size condition at all. The size condition appears only when you take the limit, because the partial sum approaches a over one minus r exactly when r to the k approaches zero.

That is also why the formula gives nonsense outside the condition: it is the limit of an expression whose limit does not exist there.

67. Exit ticket

Exit ticket

\[ \sum_{n=1}^{\infty}\left[\left(\frac25\right)^n + \frac{1}{(n+1)(n+2)}\right] \]

This problem uses the whole lesson at once. Identify each piece, sum it with its own method, and justify combining them.

The first piece is geometric. It starts at n equal to 1, so its first term is two fifths, and its ratio is two fifths; the sum is two fifths over three fifths, which is two thirds. The second piece telescopes after partial fractions into one over n plus 1 minus one over n plus 2; the first piece is one half and the last vanishes, so it sums to one half.

Both converge, so the sum rule lets you add them: two thirds plus one half is seven sixths. Writing the phrase both converge is what makes the split legitimate.

68. Recap

Recap

\[ \sum_{n=1}^{\infty}a_n = \lim_{k\to\infty}S_k \]

Next, Section 5.3 decides convergence when no formula for the partial sums exists, using the divergence and integral tests.

Stewart, Calculus: Early Transcendentals 8e, §11.2 Series §11.2, pp. 707-718 — the same material in Stewart

One definition carries everything: the sum of a series is the limit of its partial sums. The harmonic series showed that terms shrinking to zero is not enough for that limit to exist, and its slow logarithmic growth showed why computation alone can never decide.

Two families have sums you can write down. A geometric series converges exactly when its ratio is less than one in size, to the first term over one minus the ratio. A telescoping series collapses inside every partial sum to its first piece minus a last piece, and converges when that last piece has a limit. The algebraic rules let you combine them, provided every piece converges.

Most series are neither geometric nor telescoping, so Section 5.3 starts building tests that decide convergence without a formula for the partial sums, beginning with the divergence test and the integral test.