Calculus 2

Alternating series and their error bound

A balance pointer swings past its mark, back, and past again, each swing shorter than the last. Does it settle, and how close to the resting place is the pointer after a given number of swings?

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Absolute and conditional convergenceCalculus 2

Where you are 6 in Convergence tests

Calculus8

Sequences and series in calculus8

Convergence tests6

Absolute and conditional convergence

Alternating series and their error bound

The direct and limit comparison tests The integral test and p series The ratio and root tests Choosing a convergence test

Why this exists

Wirrin releases the pointer of a laboratory balance at the zero mark. It swings 4 millimeters to the right, then 3 back, then 2.4 forward again. Wirrin models the nth swing as \displaystyle \frac{12}{n+2} millimeters, each in the opposite direction from the one before. This is a simplified model, not the physics of a real balance, whose friction shrinks the swings much faster; it is chosen because swings that shrink this slowly are the hard case. Measuring positions to the right of the mark as positive, the pointer’s turning points are the partial sums of

\begin{aligned} &\sum_{n=1}^{\infty} (-1)^{n+1}\frac{12}{n+2} \\ &\quad = 4 - 3 + \frac{12}{5} - 2 + \frac{12}{7} - \cdots \end{aligned}

The swing lengths shrink like \displaystyle \frac{1}{n}, as slowly as the terms of the harmonic series, whose sum is infinite (Infinite series and partial sums). Yet the turning points visibly close in on one position, because each swing takes back most of what the one before it gave. This lesson makes that cancellation exact: a test that proves such a series converges, and a bound on how far any turning point can be from the rest position. The The nth term test for divergence can only prove divergence; this is a test that proves convergence for a series with negative terms.

Definition

An alternating series is a series whose terms alternate in sign:

\sum_{n=1}^{\infty} (-1)^{n+1} b_n = b_1 - b_2 + b_3 - \cdots

with every b_n > 0.

The form \sum (-1)^n b_n = -b_1 + b_2 - \cdots is the negative of this one and converges exactly when it does, so everything below applies to both. A sign written as \cos(n\pi) is the same thing, since \cos(n\pi) = (-1)^n.

The alternating series test. Suppose that

b_n > 0 for every n,
the b_n are eventually decreasing: there is an M with b_{n+1} \le b_n for every n \ge M, and
\lim_{n \to \infty} b_n = 0.

Then \sum_{n=1}^{\infty} (-1)^{n+1} b_n converges.

The alternating series error bound. Under the same hypotheses, with S the sum and S_N the Nth partial sum, for every N \ge M - 1

\lvert S - S_N \rvert \le b_{N+1}

In plain words: the error after stopping is no bigger than the first term left out. Moreover S - S_N is either 0 or has the sign of that first omitted term, (-1)^{N+2} b_{N+1}. If the next term would have been added, S_N is an underestimate; if it would have been subtracted, S_N is an overestimate.

Intuition

Each partial sum is a turning point of the pointer. Since no swing is longer than the one before it, a swing never carries the pointer past the turning point before last. So the right-hand turning points S_1, S_3, S_5, \ldots step leftward, the left-hand ones S_2, S_4, S_6, \ldots step rightward, and every left turning point stays left of every right one. Figure 1 shows the zigzag closing in.

Figure 1Wirrin’s turning points S_N zigzag around the rest position S = 12\ln 2 - 6 \approx 2.318. The upper points step down, the lower points step up, and the gap between neighbors is the next swing.

The gap between neighbors S_N and S_{N+1} is exactly the swing b_{N+1}, and the rest position sits inside every such gap (Figure 2). So the rest position is within one swing of any turning point: that is the error bound, and which side of the gap it is on is the sign of the error.

Figure 2The even partial sums climb and the odd ones fall, and the sum S lies between every even one and every odd one.

Results

Day 1: the alternating series test

The test. If b_n > 0, the b_n are eventually decreasing and b_n \to 0, then \sum_{n=1}^{\infty} (-1)^{n+1} b_n converges.

Nesting. If b_{n+1} \le b_n for every n, then

\begin{aligned} &S_2 \le S_4 \le S_6 \le \cdots \le S \\ &\quad \le \cdots \le S_5 \le S_3 \le S_1 \end{aligned}

Checking decrease. Compare b_{n+1} with b_n directly, use the ratio \displaystyle \frac{b_{n+1}}{b_n} \le 1, or find a differentiable f with f(n) = b_n and f'(x) < 0 for x > M (monotone sequences, Convergence of sequences).

Day 2: the error bound

The bound. For every N \ge M - 1, \lvert S - S_N \rvert \le b_{N+1}, and S - S_N is 0 or has the sign of the first omitted term. If the b_n strictly decrease from M on, then 0 < \lvert S - S_N \rvert < b_{N+1}.

The bracket. For N \ge M - 1, S lies between S_N and S_{N+1}, so the midpoint \displaystyle \frac{S_N + S_{N+1}}{2} is within \displaystyle \frac{b_{N+1}}{2} of S.

Terms for a given accuracy. To guarantee \lvert S - S_N \rvert \le \varepsilon, it is enough to take N \ge M - 1 with b_{N+1} \le \varepsilon. This is a guarantee, not the least N that happens to work.

Why it is true

The test, when b_{n+1} \le b_n for every n. Grouping in pairs,

S_{2k+2} = S_{2k} + (b_{2k+1} - b_{2k+2}) \ge S_{2k}

so the even partial sums increase. Grouping the other way,

\begin{aligned} S_{2k} &= b_1 - (b_2 - b_3) - \cdots \\ &\quad - (b_{2k-2} - b_{2k-1}) - b_{2k} \le b_1 \end{aligned}

so they are bounded above. By the monotone convergence theorem (Convergence of sequences) they converge to some number S. Since S_{2k+1} = S_{2k} + b_{2k+1} and b_{2k+1} \to 0, the odd partial sums tend to S as well, and a sequence whose even-indexed and odd-indexed terms both tend to S tends to S. The same pairing gives S_{2k+1} = S_{2k-1} - (b_{2k} - b_{2k+1}) \le S_{2k-1}, so the odd partial sums decrease toward S while the even ones increase toward it. That is the nesting.

Eventually decreasing. If the b_n decrease only from n = M on, the tail \sum_{n \ge M} (-1)^{n+1} b_n is plus or minus an alternating series of the first kind, so it converges, and the finitely many terms before it do not affect convergence (Infinite series and partial sums).

The error bound. Take N \ge M - 1, so every index from N + 1 on is at least M. Then

\begin{aligned} S - S_N &= (-1)^{N+2}\, T, \\ T &= b_{N+1} - b_{N+2} + b_{N+3} - \cdots \end{aligned}

and T is an alternating series whose terms decrease to 0. By the nesting applied to T, its sum lies between its second partial sum and its first:

0 \le b_{N+1} - b_{N+2} \le T \le b_{N+1}

So \lvert S - S_N \rvert = T \le b_{N+1}, and S - S_N has the sign of (-1)^{N+2} unless T = 0. For strictly decreasing terms the inequalities become strict (Exercise 18).

Worked examples

Example 1

Day 1: Wirrin’s balance

Show that the series of Wirrin’s turning points, \sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{12}{n+2}, converges. List the first six turning points and say what they guarantee about the rest position.

Solution

Here \displaystyle b_n = \frac{12}{n+2}, which is positive. It decreases for every n, because a larger denominator gives a smaller fraction: n + 3 > n + 2, so \displaystyle \frac{12}{n+3} < \frac{12}{n+2}. And b_n \to 0, because the denominator grows without bound. All three hypotheses hold, so the series converges: the pointer has a rest position S.

The turning points are the partial sums:

N	1	2	3	4	5	6
S_N	4	1	\displaystyle \frac{17}{5}	\displaystyle \frac{7}{5}	\displaystyle \frac{109}{35}	\displaystyle \frac{113}{70}

By the nesting, S lies between every left turning point and every right one. The tightest statement from these six is

\frac{113}{70} \le S \le \frac{109}{35}

about 1.614 \le S \le 3.114, an interval of width \displaystyle S_5 - S_6 = b_6 = \frac{3}{2} millimeters. Power series will later show that the sum of the alternating harmonic series \displaystyle 1 - \frac12 + \frac13 - \cdots is \ln 2, and from it S = 12\ln 2 - 6 \approx 2.318 millimeters, inside the interval.

Example 2

Day 1: terms that rise before they fall

Decide whether \sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{n}{n^2+5} converges.

Solution

Here \displaystyle b_n = \frac{n}{n^2+5} > 0. Dividing top and bottom by n^2, \displaystyle b_n = \frac{1/n}{1 + 5/n^2} \to \frac{0}{1} = 0.

The terms do not decrease from the start: \displaystyle b_1 = \frac{1}{6} and \displaystyle b_2 = \frac{2}{9}, and \displaystyle \frac{2}{9} > \frac{1}{6}. To compare neighbors in general, cross-multiply; both denominators are positive, so the direction of the inequality is kept:

\begin{aligned} &b_{n+1} \le b_n \\ &\iff (n+1)(n^2+5) \le n\bigl((n+1)^2 + 5\bigr) \end{aligned}

The left side expands to n^3 + n^2 + 5n + 5 and the right side to n^3 + 2n^2 + 6n, so the condition is 5 \le n^2 + n, which holds exactly when n \ge 2 (and strictly, since 2^2 + 2 = 6). So b_2 > b_3 > b_4 > \cdots: the terms decrease from M = 2 on. Eventually decreasing is enough, and the series converges.

The derivative tells the same story. For \displaystyle f(x) = \frac{x}{x^2+5},

\begin{aligned} f'(x) &= \frac{(x^2+5) - x \cdot 2x}{(x^2+5)^2} \\ &= \frac{5 - x^2}{(x^2+5)^2} \end{aligned}

which is negative for \displaystyle x > \sqrt{5} \approx 2.24. That gives decrease from n = 3 on, and the single comparison \displaystyle b_2 = \frac{2}{9} > \frac{3}{14} = b_3 fills the gap.

Example 3

Day 1: terms that tend to zero without decreasing

Let b_n = \dfrac{2 + (-1)^n}{n}, so that \sum_{n=1}^{\infty} (-1)^{n+1} b_n is

1 - \frac{3}{2} + \frac{1}{3} - \frac{3}{4} + \frac{1}{5} - \frac{3}{6} + \cdots

Show that the alternating series test does not apply, and that the series diverges.

Solution

The signs alternate, b_n > 0, and \displaystyle 0 < b_n \le \frac{3}{n}, so b_n \to 0 by squeezing. But \displaystyle b_{2k} = \frac{3}{2k} is larger than \displaystyle b_{2k-1} = \frac{1}{2k-1} for every k, since 3(2k - 1) > 2k means \displaystyle k > \frac{3}{4}. The terms rise at every even index, so they are not eventually decreasing and the test gives no verdict.

Pair the terms instead. The kth pair is

\begin{aligned} \frac{1}{2k-1} - \frac{3}{2k} &= \frac{2k - 3(2k-1)}{2k(2k-1)} \\ &= -\frac{4k-3}{2k(2k-1)} \end{aligned}

and \displaystyle \frac{4k-3}{2k(2k-1)} \ge \frac{1}{2k}: multiplying by 2k(2k-1) > 0, this says 4k - 3 \ge 2k - 1, that is k \ge 1. So each pair is at most \displaystyle -\frac{1}{2k}, and

S_{2k} \le -\frac{1}{2}\left(1 + \frac{1}{2} + \cdots + \frac{1}{k}\right)

The bracket is a partial sum of the harmonic series and grows without bound, so S_{2k} \to -\infty and the series diverges. The long negative steps outweigh the short positive ones every time.

Example 4

Day 2: a sum to within 0.0005, and which side

Estimate S = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n^3+2} with error at most 0.0005, and say whether the estimate is above or below S.

Solution

\displaystyle b_n = \frac{1}{n^3+2} is positive, decreases for every n (its denominator increases) and tends to 0, so the bound holds for every N. We need b_{N+1} \le 0.0005, that is (N+1)^3 + 2 \ge 2000, or (N+1)^3 \ge 1998. Since 12^3 = 1728 and 13^3 = 2197, the smallest choice is N + 1 = 13, so N = 12:

\begin{aligned} S_{12} &= \frac{1}{3} - \frac{1}{10} + \frac{1}{29} - \cdots - \frac{1}{1730} \\ &\approx 0.2574466 \end{aligned}

The first omitted term is \displaystyle (-1)^{14}\frac{1}{13^3 + 2} = +\frac{1}{2199}, so S - S_{12} is positive and at most \displaystyle \frac{1}{2199} \approx 0.000455: S_{12} underestimates, and

S_{12} < S < S_{12} + \frac{1}{2199}

so 0.25744 < S < 0.25791.

The bound is a guarantee, not the least N that works. In fact S \approx 0.2577, and S_{11} \approx 0.2580 is already within 0.0005 of it, but the bound \displaystyle b_{12} = \frac{1}{1730} \approx 0.000578 could not promise that.

Example 5

Day 2: how many swings pin down the rest position

Wirrin wants the rest position of Example 1 to within 0.05 millimeter. How many turning points does the error bound require? How many if Wirrin reports the midpoint of the last two turning points instead?

Solution

Since \displaystyle b_n = \frac{12}{n+2} decreases from the start, the bound holds for every N. A single turning point S_N is within \displaystyle b_{N+1} = \frac{12}{N+3} of S, so we need \displaystyle \frac{12}{N+3} \le 0.05, that is N + 3 \ge 240: 237 turning points.

The midpoint does better. S lies between S_N and S_{N+1}, which are b_{N+1} apart, so the midpoint is within \displaystyle \frac{b_{N+1}}{2} = \frac{6}{N+3} of S. Now \displaystyle \frac{6}{N+3} \le 0.05 needs N + 3 \ge 120, so N = 117, and the midpoint uses S_{117} and S_{118}: 118 turning points, about half as many.

Even a few swings help this way. With seven turning points, the midpoint of \displaystyle S_6 = \frac{113}{70} and \displaystyle S_7 = \frac{619}{210} is \displaystyle \frac{479}{210} \approx 2.281, guaranteed within \displaystyle \frac{b_7}{2} = \frac{2}{3} millimeter, and in fact within 0.04 of S \approx 2.318. Terms that shrink like \displaystyle \frac{1}{n} make any guarantee expensive; the terms of Example 4 shrink like \displaystyle \frac{1}{n^3} and needed twelve.

Example 6

Day 2: an error bound that starts late

For S = \sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{n}{n^2+5} from Example 2, find a partial sum guaranteed to be within 0.01 of S, and say on which side of S it lies.

Solution

Example 2 showed that the b_n decrease from M = 2, so the bound holds for N \ge 1. We need b_{N+1} \le 0.01. With m = N + 1 this reads \displaystyle \frac{m}{m^2 + 5} \le \frac{1}{100}, that is m^2 - 100m + 5 \ge 0. The roots of m^2 - 100m + 5 are \displaystyle 50 \pm \sqrt{2495}, about 0.05 and 99.95, and the quadratic is positive outside them. Since m \ge 2, we need m \ge 99.95, so m = 100 and N = 99. Check: \displaystyle b_{100} = \frac{100}{10{,}005} \approx 0.009995, while \displaystyle b_{99} = \frac{99}{9806} \approx 0.010096 is too big.

So \lvert S - S_{99} \rvert \le 0.009995. The first omitted term is (-1)^{101} b_{100}, which is negative, so S < S_{99}: the partial sum overestimates. Numerically S_{99} \approx 0.06217 and S \approx 0.05715, about 0.0050 apart.

Boundaries

The terms must tend to 0. If b_n does not tend to 0, the series diverges by the The nth term test for divergence and no alternating argument applies.
Decrease is part of the test. Without it, a series with b_n \to 0 may diverge (Example 3) or converge (Exercise 16), and the test gives no verdict either way.
The error bound needs decrease from N + 1 on. It holds for N \ge M - 1. Without decrease it can fail outright: in Exercise 16, \lvert S - S_2 \rvert > b_3.
The signs must alternate term by term. A pattern such as + + - - + + \cdots is not covered as stated, and a zero term breaks the requirement b_n > 0.
Convergence only. The test says nothing about \sum b_n, which for Wirrin’s series diverges. That distinction is the next lesson.

Where this goes next

Absolute and conditional convergence separates series like Wirrin’s, which converge only because of cancellation, from those whose absolute values also converge, and shows that rearranging the terms of the first kind can change the sum. The error bound returns at the endpoints of an interval of convergence (Power series and the interval of convergence), and for an alternating Taylor series it is a quick alternative to the The Taylor remainder and the Lagrange error bound. Choosing a convergence test places this test among the others.

Exercises

Day 1: the alternating series test

Decide whether each series converges or diverges. Where the alternating series test applies, give the index from which the terms decrease.

\sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{3}{2n+7}
\sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{4n+1}{3n+2}
\displaystyle \sum_{n=1}^{\infty} \dfrac{\cos(n\pi)}{\sqrt{n+3}}
\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{\sqrt{n}}{n+6}
\sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{\ln(n+1)}{n+1}
\sum_{n=1}^{\infty} (-1)^{n+1}\sin\dfrac{2}{n}
\sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{n^2}{2^n}
For which real numbers p does \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n^p} converge? Name the test that settles each case.

Day 2: error bounds and accuracy

Find the smallest N for which the error bound guarantees \lvert S - S_N \rvert \le 0.001 for S = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n \cdot 3^n}. Compute S_N as a fraction and say whether it is above or below S.
How many terms of \displaystyle \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{\sqrt{n}+1} does the error bound need for an error of at most 0.1? Is that partial sum above or below the sum?
For \sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{2^n}{n!}, check the hypotheses, find the smallest N that the bound guarantees to within 0.001, and give the sign of S - S_N.
For the series of Exercise 1, bound \lvert S - S_{10} \rvert and say whether S_{10} overestimates or underestimates S.
For \sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{n}{4^n}, find the smallest N that the bound guarantees to within 0.0001, and say on which side of S the partial sum lies.
Yarran’s balance pointer, released at the mark, turns at 5.20, 1.10, 4.05, 1.85 and 3.60 millimeters to the right of it, in that order. Assume the swings keep shrinking to 0. Give the smallest interval these readings guarantee for the rest position, and the midpoint estimate with its error bound.

Reasoning

Suppose b_n > 0, b_n \to 0, and b_{n+1} \le b_n only for n \ge M. Prove that \lvert S - S_N \rvert \le b_{N+1} for every N \ge M - 1, without quoting the error bound.
Let b_n = \dfrac{2 + (-1)^n}{n^2}. Show that \sum_{n=1}^{\infty} (-1)^{n+1} b_n converges although the b_n are not eventually decreasing. Taking as known that \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} (The integral test and p series), show that the sum is 0, and conclude that \lvert S - S_2 \rvert > b_3.
For 0 < r < 1, the series \sum_{n=1}^{\infty} (-r)^{n-1} = 1 - r + r^2 - \cdots has sum \displaystyle \frac{1}{1+r}. Find S - S_N exactly, compare it with the bound b_{N+1}, and say for which r the bound is nearly exact and for which it overstates the error by almost a factor of 2.
Suppose the b_n strictly decrease for every n and tend to 0. Prove that 0 < \lvert S - S_N \rvert < b_{N+1} for every N, so that no partial sum equals the sum.

Answers

Converges; decreasing from n = 1. \displaystyle b_n = \frac{3}{2n+7} is positive, its denominator increases so it decreases, and b_n \to 0.
Diverges. \displaystyle b_n = \frac{4n+1}{3n+2} \to \frac{4}{3}, so the terms approach \displaystyle \frac{4}{3} and \displaystyle -\frac{4}{3} in turn and have no limit; the nth term test settles it, and the alternating series test does not apply.
Converges; decreasing from n = 1. Since \cos(n\pi) = (-1)^n, this is \sum (-1)^n b_n with \displaystyle b_n = \frac{1}{\sqrt{n+3}}, which is positive, decreasing and tends to 0.
Converges; decreasing from n = 6. For \displaystyle f(x) = \frac{\sqrt{x}}{x+6}, \displaystyle f'(x) = \frac{6 - x}{2\sqrt{x}\,(x+6)^2}, negative for x > 6, so b_6 > b_7 > \cdots; before that the terms rise (\displaystyle b_5 = \frac{\sqrt5}{11} \approx 0.2033 is less than \displaystyle b_6 = \frac{\sqrt6}{12} \approx 0.2041). And \displaystyle b_n = \frac{1/\sqrt{n}}{1 + 6/n} \to 0.
Converges; decreasing from n = 2. b_n = g(n+1) with \displaystyle g(x) = \frac{\ln x}{x} and \displaystyle g'(x) = \frac{1 - \ln x}{x^2} < 0 for x > e, so the terms decrease once n + 1 \ge 3; \displaystyle b_1 = \frac{\ln 2}{2} \approx 0.347 is less than \displaystyle b_2 = \frac{\ln 3}{3} \approx 0.366. By L’Hôpital’s rule \displaystyle \frac{\ln x}{x} \to 0.
Converges; decreasing from n = 1. For n \ge 2, \displaystyle 0 < \frac{2}{n} \le 1 < \frac{\pi}{2}, where sine is positive and increasing, so \displaystyle \sin\frac{2}{n} decreases as n grows; also b_1 = \sin 2 \approx 0.909 exceeds b_2 = \sin 1 \approx 0.841. By continuity, \displaystyle \sin\frac{2}{n} \to \sin 0 = 0.
Converges; decreasing from n = 3. The terms run \displaystyle \frac{1}{2}, 1, \frac{9}{8}, 1, \frac{25}{32}, \ldots, and \displaystyle \frac{b_{n+1}}{b_n} = \frac{(n+1)^2}{2n^2} \le 1 exactly when \displaystyle n + 1 \le \sqrt{2}\,n, that is \displaystyle n \ge \frac{1}{\sqrt2 - 1} \approx 2.41. Applying L’Hôpital’s rule twice to \displaystyle \frac{x^2}{2^x} gives b_n \to 0.
Converges for p > 0, diverges for p \le 0. For p > 0, \displaystyle b_n = \frac{1}{n^p} decreases and tends to 0, so the alternating series test applies. For p = 0 the terms are \pm 1, and for p < 0 their size n^{-p} tends to \infty; either way the terms do not tend to 0 and the nth term test gives divergence.
N = 4, \displaystyle S_4 = \frac{31}{108} \approx 0.28704, below S. Here \displaystyle b_{N+1} = \frac{1}{(N+1)3^{N+1}}: \displaystyle b_4 = \frac{1}{324} is too big and \displaystyle b_5 = \frac{1}{1215} \approx 0.00082 is not. \displaystyle S_4 = \frac{1}{3} - \frac{1}{18} + \frac{1}{81} - \frac{1}{324}, which is \displaystyle \frac{108 - 18 + 4 - 1}{324} = \frac{31}{108}. The first omitted term \displaystyle +\frac{1}{1215} is positive, so S > S_4. (The sum is \displaystyle \ln\frac{4}{3} \approx 0.28768.)
80 terms, and S_{80} is below the sum. The terms decrease to 0, and \displaystyle b_{N+1} = \frac{1}{\sqrt{N+1} + 1} \le 0.1 exactly when \displaystyle \sqrt{N+1} \ge 9, that is N \ge 80. The first omitted term, n = 81, has sign (-1)^{82} = 1, so S > S_{80}.
Converges; N = 9; S - S_9 < 0. \displaystyle b_n = \frac{2^n}{n!} > 0 and \displaystyle \frac{b_{n+1}}{b_n} = \frac{2}{n+1} \le 1 for every n \ge 1 (with b_1 = b_2 = 2), so the terms never increase; for n \ge 3 the ratio is at most \displaystyle \frac{1}{2}, so \displaystyle b_n \le b_3 \left(\frac{1}{2}\right)^{n-3} \to 0. Then \displaystyle b_9 = \frac{512}{362{,}880} \approx 0.00141 and \displaystyle b_{10} = \frac{1024}{3{,}628{,}800} \approx 0.000282, so N + 1 = 10. The first omitted term has sign (-1)^{11} = -1, so S < S_9 \approx 0.86490: an overestimate. (The sum is 1 - e^{-2} \approx 0.86466.)
\displaystyle \lvert S - S_{10} \rvert \le \frac{3}{29} \approx 0.103, and S_{10} underestimates. The first omitted term is \displaystyle b_{11} = \frac{3}{29} with sign (-1)^{12} = 1. (Numerically S_{10} \approx 0.1313 and S \approx 0.1848.)
N = 8, and S_8 is below S. \displaystyle \frac{b_{n+1}}{b_n} = \frac{n+1}{4n} < 1 for n \ge 1 and b_n \to 0. \displaystyle b_8 = \frac{8}{65{,}536} \approx 0.000122 is too big and \displaystyle b_9 = \frac{9}{262{,}144} \approx 0.0000343 is not. The first omitted term, n = 9, is positive, so S > S_8 \approx 0.159973. (The sum is \displaystyle \frac{4}{25} = 0.16.)
1.85 \le S \le 3.60; midpoint 2.725 millimeters, within 0.875 millimeter. The swings 5.20, 4.10, 2.95, 2.20, 1.75 shrink, so the turning points nest: the right ones 5.20, 4.05, 3.60 step left, the left ones 1.10, 1.85 step right, and the rest position lies between the last two. The midpoint is \displaystyle \frac{1.85 + 3.60}{2} = 2.725, with error at most \displaystyle \frac{3.60 - 1.85}{2} = 0.875.
For N \ge M - 1, every index from N + 1 on is at least M. Write S - S_N = (-1)^{N+2} T with T = b_{N+1} - b_{N+2} + \cdots, an alternating series whose terms decrease to 0. The pairing proof of the test shows its even partial sums increase and its odd ones decrease to T, so T_2 \le T \le T_1, that is 0 \le b_{N+1} - b_{N+2} \le T \le b_{N+1}. Hence \lvert S - S_N \rvert = T \le b_{N+1}.
Not eventually decreasing: \displaystyle b_{2k} = \frac{3}{4k^2} > b_{2k-1} = \frac{1}{(2k-1)^2}, since 3(2k-1)^2 > 4k^2 for every k \ge 2 (for example \displaystyle b_4 = \frac{3}{16} > b_3 = \frac{1}{9}). Converges: the terms are \displaystyle (-1)^{n+1}b_n = \frac{2(-1)^{n+1}}{n^2} - \frac{1}{n^2}, a convergent alternating series (test) minus a convergent p series, and a difference of convergent series converges. Sum: \displaystyle \sum \frac{(-1)^{n+1}}{n^2} = \sum \frac{1}{n^2} - 2\sum \frac{1}{(2k)^2}, which is \displaystyle \frac{\pi^2}{6} - \frac{1}{2}\cdot\frac{\pi^2}{6} = \frac{\pi^2}{12}, so \displaystyle S = 2 \cdot \frac{\pi^2}{12} - \frac{\pi^2}{6} = 0. Then \displaystyle S_2 = 1 - \frac{3}{4} = \frac{1}{4}, and \displaystyle \lvert S - S_2 \rvert = \frac{1}{4} > \frac{1}{9} = b_3.
\displaystyle S - S_N = \frac{(-r)^N}{1+r}, so \displaystyle \lvert S - S_N \rvert = \frac{r^N}{1+r}, while b_{N+1} = r^N. From the finite geometric sum, \displaystyle S_N = \frac{1 - (-r)^N}{1+r}, and subtracting from \displaystyle \frac{1}{1+r} gives the error, whose sign (-1)^N is that of the first omitted term (-r)^N. The true error is the bound times \displaystyle \frac{1}{1+r}: nearly exact when r is near 0, and nearly half the bound when r is near 1.
Write S - S_N = (-1)^{N+2} T with T = b_{N+1} - b_{N+2} + b_{N+3} - \cdots. The partial sums of T nest, so T_2 \le T \le T_3. Strict decrease gives T_2 = b_{N+1} - b_{N+2} > 0 and T_3 = b_{N+1} - (b_{N+2} - b_{N+3}) < b_{N+1}. So 0 < T < b_{N+1}, which is 0 < \lvert S - S_N \rvert < b_{N+1}; in particular S \ne S_N.