Calculus 2

Absolute and conditional convergence

A sailboat zigzags upwind and its sideways position settles down, yet the sideways distance it covers grows without end. When does a series converge only because its terms cancel, and what changes when they do?

You need

Alternating series and their error boundCalculus 2 The direct and limit comparison testsCalculus 2

Unlocks

The ratio and root testsCalculus 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

Mihai sails a dinghy up a channel against the wind. A boat cannot sail straight into the wind, so he tacks: each leg carries him across the wind, alternately to the right and to the left, while he gains ground upwind. In a simplified model, chosen so the arithmetic comes out exactly, the sideways part of his nth leg is \displaystyle \frac{2n+1}{n(n+1)} kilometers. Counting rightward as positive, his sideways offset from the start line after N legs is the partial sum S_N of

\sum_{n=1}^{\infty} (-1)^{n+1}\frac{2n+1}{n(n+1)} = \frac{3}{2} - \frac{5}{6} + \frac{7}{12} - \cdots

while the sideways distance he has actually sailed is the partial sum T_N of the same terms with every sign made positive. Dragos, watching from the shore, sees the boat’s offset settle. Anca, who logs the distance sailed, sees her total keep growing. Both are right.

The alternating series test needs signs that alternate strictly, and the comparison and integral tests need positive terms. A series such as \displaystyle \sum \frac{\cos(3n)}{n^2+1} fits neither. Absolute convergence closes that gap: make every term positive, test that series, and convergence of the original follows. It also separates series that converge robustly from those that converge only by cancellation, and only the first kind may be added in any order.

Definition

Let \sum a_n be a series whose terms may have either sign.

\sum a_n converges absolutely if \sum \lvert a_n \rvert converges.
\sum a_n converges conditionally if \sum a_n converges but \sum \lvert a_n \rvert diverges.

Every series falls into exactly one of three classes: absolutely convergent, conditionally convergent, or divergent. For a series of nonnegative terms, \lvert a_n \rvert = a_n, so convergence and absolute convergence are the same thing; the distinction matters only when terms of both signs occur infinitely often.

A rearrangement of \sum a_n is a series \sum a_{\sigma(n)} that uses every term exactly once in some other order: \sigma is a one-to-one correspondence of the positive integers with themselves.

Intuition

Think of \lvert a_n \rvert as the length of the nth leg and a_n as that leg with its direction. If the total length \sum \lvert a_n \rvert is finite, the legs still to be sailed have a total length that shrinks to 0, so the position settles however the directions are chosen. If the total length is infinite, the position may still settle, as Mihai’s does, but only because legs in one direction keep undoing legs in the other (Figure 1).

Figure 1Mihai’s sideways offset S_N settles toward 1 kilometer, while the sideways distance sailed, T_N, keeps climbing. The series converges, but not absolutely.

That cancellation is fragile. A conditionally convergent series has infinitely much “right” and infinitely much “left” in it. Take the legs in a different order, using rightward legs faster than leftward ones, and the boat can be made to drift off without bound, or to settle anywhere you like (Figure 2).

Figure 2The same terms in two orders. In the original order the partial sums S_m settle near 0.605; taken two positive terms then one negative, the partial sums R_m grow without bound.

Results

Absolute convergence implies convergence. If \sum \lvert a_n \rvert converges, then \sum a_n converges, and

\Bigl\lvert \sum_{n=1}^{\infty} a_n \Bigr\rvert \le \sum_{n=1}^{\infty} \lvert a_n \rvert

Tails. If \sum a_n converges absolutely with sum S, then \lvert S - S_N \rvert \le \sum_{n=N+1}^{\infty} \lvert a_n \rvert.

Positive and negative parts. Let p_n = \max(a_n, 0) and q_n = \max(-a_n, 0), so a_n = p_n - q_n and \lvert a_n \rvert = p_n + q_n. If \sum a_n converges absolutely, then \sum p_n and \sum q_n both converge and \sum a_n = \sum p_n - \sum q_n. If \sum a_n converges conditionally, then \sum p_n and \sum q_n both diverge.

Rearranging an absolutely convergent series (stated without proof). Every rearrangement of an absolutely convergent series converges absolutely, to the same sum.

The Riemann rearrangement theorem (stated without proof). If \sum a_n converges conditionally and L is any real number, some rearrangement of \sum a_n converges to L. Other rearrangements diverge to \infty, and others to -\infty.

Why it is true

Absolute convergence implies convergence. Since -\lvert a_n \rvert \le a_n \le \lvert a_n \rvert, adding \lvert a_n \rvert throughout gives

0 \le a_n + \lvert a_n \rvert \le 2\lvert a_n \rvert

The series \sum 2\lvert a_n \rvert converges, so by direct comparison (The direct and limit comparison tests) the nonnegative series \sum (a_n + \lvert a_n \rvert) converges. The difference of two convergent series converges, and

a_n = (a_n + \lvert a_n \rvert) - \lvert a_n \rvert

so \sum a_n converges. For the inequality, the Triangle inequality for finite sums gives \lvert S_N \rvert \le \sum_{n=1}^{N} \lvert a_n \rvert \le \sum_{n=1}^{\infty} \lvert a_n \rvert for every N, and letting N \to \infty keeps the inequality. Applied to the tail series \sum_{n > N} a_n, whose sum is S - S_N, the same inequality gives the tail bound.

Positive and negative parts. From the definitions, 0 \le p_n \le \lvert a_n \rvert and 0 \le q_n \le \lvert a_n \rvert, so absolute convergence makes both \sum p_n and \sum q_n converge by comparison. Now suppose \sum a_n converges conditionally. If \sum p_n converged, then q_n = p_n - a_n would give a convergent \sum q_n, and then \lvert a_n \rvert = p_n + q_n would give a convergent \sum \lvert a_n \rvert, which is false. So \sum p_n diverges, and the same argument with the roles swapped shows \sum q_n diverges.

Why Riemann’s theorem is plausible. The positive terms of a conditionally convergent series add to \infty and the negative terms to -\infty, while every term tends to 0. So a running sum can be steered across any target L again and again, overshooting by less each time; Example 6 carries this out. The full proof of both rearrangement results is left to a course in real analysis.

Worked examples

Example 1

Mihai’s tacks: convergent, but not absolutely

Show that the series of Mihai’s sideways offsets, \sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{2n+1}{n(n+1)}, converges conditionally, and find its sum.

Solution

Split the term into partial fractions: \displaystyle \frac{2n+1}{n(n+1)} = \frac{1}{n} + \frac{1}{n+1}, since the right side over the common denominator is \displaystyle \frac{(n+1) + n}{n(n+1)}. So

S_N = \sum_{n=1}^{N} \frac{(-1)^{n+1}}{n} + \sum_{n=1}^{N} \frac{(-1)^{n+1}}{n+1}

In the second sum put m = n + 1: it becomes \displaystyle -\sum_{m=2}^{N+1} \frac{(-1)^{m+1}}{m}. Every term of the first sum from n = 2 to N cancels, leaving the first term of the first sum and the last of the second:

S_N = 1 + \frac{(-1)^{N+1}}{N+1}

Check: \displaystyle S_1 = \frac{3}{2} and \displaystyle S_2 = \frac{2}{3}, matching \displaystyle \frac{3}{2} and \displaystyle \frac{3}{2} - \frac{5}{6}. Since \displaystyle \frac{1}{N+1} \to 0, S_N \to 1: the series converges, and Mihai’s offset settles 1 kilometer to the right of the start line.

For absolute convergence, look at \displaystyle \lvert a_n \rvert = \frac{2n+1}{n(n+1)}. Since 2n + 1 > n + 1, we have \displaystyle \lvert a_n \rvert > \frac{1}{n}, and the harmonic series diverges, so \sum \lvert a_n \rvert diverges by direct comparison. In fact \displaystyle T_N = \sum_{n=1}^{N} \bigl(\frac1n + \frac{1}{n+1}\bigr), which is \displaystyle 2\bigl(1 + \frac12 + \cdots + \frac1N\bigr) - 1 + \frac{1}{N+1}, and it grows without bound. The series converges conditionally.

Example 2

A sign pattern with no pattern

Show that \sum_{n=1}^{\infty} \dfrac{\cos(3n)}{n^2+1} converges, and locate its sum within 0.1.

Solution

The first values of \cos(3n) are about -0.990, 0.960, -0.911, 0.844, \ldots, and the signs alternate up to n = 10; but \cos 33 \approx -0.013 and \cos 36 \approx -0.128 are both negative. The signs follow no fixed rule, so the alternating series test cannot be used, and the terms are not all positive, so neither can the comparison test.

Test the absolute values instead. Since \lvert \cos(3n) \rvert \le 1 and n^2 + 1 > n^2,

\frac{\lvert \cos(3n) \rvert}{n^2+1} \le \frac{1}{n^2}

and \displaystyle \sum \frac{1}{n^2} is a convergent p series (p = 2 > 1). By direct comparison \sum \lvert a_n \rvert converges, so the series converges absolutely, and therefore converges.

For the location, the tail bound and the integral bound from The integral test and p series give

\lvert S - S_N \rvert \le \sum_{n=N+1}^{\infty} \frac{1}{n^2} < \int_N^{\infty} \frac{dx}{x^2} = \frac{1}{N}

With N = 10, S_{10} \approx -0.3623, so -0.4623 < S < -0.2623.

Example 3

A geometric benchmark for alternating terms

Show that \sum_{n=1}^{\infty} \dfrac{(-1)^n n}{2^n} converges absolutely, and find the sum of the series and the sum of its absolute values.

Solution

The absolute values are \displaystyle \frac{n}{2^n}. A power of 2 eventually outgrows n by a geometric factor: we claim \displaystyle n \le \left(\frac32\right)^n for every n \ge 1. It holds for n = 1 and n = 2 (1 \le 1.5 and 2 \le 2.25). If it holds for some n \ge 2, then \displaystyle \left(\frac32\right)^{n+1} \ge \frac32 n = n + \frac{n}{2} \ge n + 1, so by induction it holds for all n. Dividing by 2^n,

\frac{n}{2^n} \le \left(\frac{3}{4}\right)^n

and \displaystyle \sum \left(\frac34\right)^n is a convergent geometric series, so the series converges absolutely.

For the sums, let \displaystyle x = \pm\frac12 and S_N = \sum_{n=1}^{N} n x^n. Subtracting xS_N shifts every term one place:

(1-x)S_N = x + x^2 + \cdots + x^N - Nx^{N+1}

The finite geometric sum tends to \displaystyle \frac{x}{1-x}, and \displaystyle \lvert Nx^{N+1} \rvert = \frac{N}{2^{N+1}} \le \frac12\left(\frac34\right)^N \to 0, so \displaystyle S_N \to \frac{x}{(1-x)^2}. With \displaystyle x = -\frac12 the sum is \displaystyle \frac{-1/2}{(3/2)^2} = -\frac{2}{9}; with \displaystyle x = \frac12 the absolute values sum to \displaystyle \frac{1/2}{(1/2)^2} = 2. As the theorem requires, \displaystyle \left\lvert -\frac{2}{9} \right\rvert \le 2. The ratio test, next, settles such series in one line.

Example 4

Logarithms that slow the decrease

Classify \displaystyle \sum_{n=2}^{\infty} (-1)^n \dfrac{\ln n}{\sqrt{n}} as absolutely convergent, conditionally convergent, or divergent.

Solution

Absolute values first. For n \ge 3, \ln n > 1, so \displaystyle \frac{\ln n}{\sqrt n} > \frac{1}{\sqrt n}, and \displaystyle \sum \frac{1}{\sqrt n} is a divergent p series (\displaystyle p = \frac12). So \sum \lvert a_n \rvert diverges by direct comparison, and the series does not converge absolutely.

Now the alternating series test with \displaystyle b_n = \frac{\ln n}{\sqrt n} > 0. For \displaystyle f(x) = \frac{\ln x}{\sqrt x},

f'(x) = \frac{2 - \ln x}{2x^{3/2}}

which is negative for x > e^2 \approx 7.39, so the terms decrease from n = 8 on; and b_7 \approx 0.7355 > b_8 \approx 0.7352, so in fact they decrease from n = 7; the earlier terms do not matter. By L’Hôpital’s rule \displaystyle \frac{\ln x}{\sqrt x} has the same limit as \displaystyle \frac{1/x}{1/(2\sqrt x)} = \frac{2}{\sqrt x}, which is 0. The test applies, so the series converges, conditionally. (Numerically its sum is about 0.1933.)

Example 5

The same terms, a different order

The series \displaystyle \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{\sqrt{n}} converges conditionally. Rearrange it by taking two positive terms, then one negative term, and repeating:

1 + \frac{1}{\sqrt3} - \frac{1}{\sqrt2} + \frac{1}{\sqrt5} + \frac{1}{\sqrt7} - \frac{1}{\sqrt4} + \cdots

Show that the partial sums R_m of the rearrangement tend to \infty.

Solution

The original converges by the alternating series test, with sum about 0.6049, but not absolutely, since \displaystyle \sum \frac{1}{\sqrt n} diverges. The rearrangement uses every term exactly once, the positive terms (odd n) and the negative terms (even n) each in order, the positive ones twice as fast.

The kth block of three terms is

B_k = \frac{1}{\sqrt{4k-3}} + \frac{1}{\sqrt{4k-1}} - \frac{1}{\sqrt{2k}}

Each of the two positive terms is at least \displaystyle \frac{1}{\sqrt{4k}} = \frac{1}{2\sqrt k}, so

B_k \ge \frac{1}{\sqrt k} - \frac{1}{\sqrt 2\,\sqrt k} = \frac{1 - 1/\sqrt2}{\sqrt k}

Summing the blocks, \displaystyle R_{3k} \ge \left(1 - \frac{1}{\sqrt2}\right) \sum_{j=1}^{k} \frac{1}{\sqrt j}, a positive multiple of a partial sum of a divergent p series, so R_{3k} \to \infty. The partial sums in between, R_{3k+1} and R_{3k+2}, add positive terms to R_{3k}, so they are larger still, and every R_m \to \infty. The numbers:

m	R_m
3	0.8702
30	2.3466
300	6.4274
3000	19.1179

Example 6

Steering a rearrangement to a target

Rearrange \displaystyle \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{\sqrt{n}} so that it converges to -1, and carry out the first ten steps.

Solution

Use the rule from Riemann’s argument: while the running sum is above -1, add the next unused negative term; once it is at or below -1, add the next unused positive term. Starting from 0:

Step	Term added	Running sum
1	\displaystyle -\frac{1}{\sqrt2}	-0.707
2	\displaystyle -\frac{1}{\sqrt4}	-1.207
3	\displaystyle +\frac{1}{\sqrt1}	-0.207
4	\displaystyle -\frac{1}{\sqrt6}	-0.615
5	\displaystyle -\frac{1}{\sqrt8}	-0.969
6	\displaystyle -\frac{1}{\sqrt{10}}	-1.285
7	\displaystyle +\frac{1}{\sqrt3}	-0.708
8	\displaystyle -\frac{1}{\sqrt{12}}	-0.996
9	\displaystyle -\frac{1}{\sqrt{14}}	-1.264
10	\displaystyle +\frac{1}{\sqrt5}	-0.817

The process never stalls: the negative terms sum to -\infty and the positive terms to \infty, so the running sum always gets back across -1, and every term is eventually used. Each crossing misses -1 by at most the term just added, those terms tend to 0, and between crossings the sum moves toward -1. So the running sums converge to -1, and the same rule works for any target L.

Boundaries

The implication runs one way. Absolute convergence implies convergence; divergence of \sum \lvert a_n \rvert says nothing about \sum a_n. Mihai’s series converges and \sum (-1)^n diverges, and both have divergent absolute series.
Comparison needs the absolute values. The comparison and integral tests are for nonnegative terms. Applied to \lvert a_n \rvert they can prove absolute convergence; applied to a_n itself they prove nothing when the signs vary.
Order matters only for conditional convergence. Rearranging an absolutely convergent series never changes the sum; rearranging a conditionally convergent one can change it to any value, or destroy convergence. Grouping consecutive terms with parentheses, without reordering them, never changes the sum of a convergent series.

Where this goes next

The ratio and root tests, next, are tests for absolute convergence: they compare \lvert a_n \rvert with a geometric series, so whenever they prove convergence it is absolute. A power series converges absolutely inside its interval of convergence, and at an endpoint (Power series and the interval of convergence) it may converge absolutely, conditionally, or not at all. Absolute convergence is also what lets the terms of a Taylor and Maclaurin series be reordered freely. Choosing a convergence test collects the tests in one place.

Exercises

Classifying series

Classify each series as absolutely convergent, conditionally convergent, or divergent.

\displaystyle \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n\sqrt{n}+2}
\sum_{n=1}^{\infty} \dfrac{(-1)^n}{5n-2}
\sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{n^2}{n^2+4}
\sum_{n=1}^{\infty} \dfrac{\sin(4n)}{n^{3/2}}
\sum_{n=1}^{\infty} (-1)^n\dfrac{n}{n^2+n+3}
\sum_{n=1}^{\infty} \dfrac{(-2)^n}{3^n+1}
\sum_{n=2}^{\infty} \dfrac{(-1)^n}{(\ln n)^2}
\sum_{n=1}^{\infty} (-1)^n\sin\dfrac{1}{n^2}
\displaystyle \sum_{n=1}^{\infty} (-1)^n\left(\sqrt{n+1}-\sqrt{n}\right)
\sum_{n=1}^{\infty} (-1)^n\dfrac{2^n}{n^2}

Bounds and sums

Find the smallest N for which the tail bound together with \displaystyle \sum_{n > N} \frac{1}{n^3} < \int_N^{\infty} \frac{dx}{x^3} guarantees that S_N is within 0.001 of S = \sum_{n=1}^{\infty} \dfrac{\cos(5n)}{n^3+1}.
Show that \sum_{n=1}^{\infty} \dfrac{(-1)^n n}{3^n} converges absolutely, and find its sum and the sum of its absolute values.
Anca’s boat tacks with sideways legs of \dfrac{6}{n(n+2)} kilometers, the first to the right and then alternating. Find the total sideways distance she sails and where her sideways offset settles.

Rearrangements

Rearrange \displaystyle \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{\sqrt{n}} by taking one positive term, then four negative terms, and repeating. Show that the partial sums tend to -\infty.
The geometric series \displaystyle \sum_{n=1}^{\infty} \left(-\frac12\right)^n is rearranged as two positive terms, then one negative term, repeated: \displaystyle \frac14 + \frac{1}{16} - \frac12 + \frac{1}{64} + \frac{1}{256} - \frac18 + \cdots. What is the sum of the rearrangement? Find the sums of the positive and of the negative terms.
Dragos rearranges Mihai’s series \sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{2n+1}{n(n+1)} to converge to 0: while the running sum is at most 0 he adds the next unused positive term, and otherwise the next unused negative term. Which terms (by their index n) are used in the first six steps, and what are the running sums, to three decimals?

Reasoning

Dragos says that adding the terms of \displaystyle \sum \frac{\cos(3n)}{n^2+1} in a different order could change the sum. Is he right? Explain.
Prove that if \sum a_n converges absolutely, then \sum a_n^2 converges.
Give a series \sum a_n that diverges while \sum a_n^2 converges, and one that converges while \sum a_n^2 diverges.
Prove that if \sum a_n converges absolutely and \lvert c_n \rvert \le B for every n, then \sum a_n c_n converges absolutely.

Answers

Absolutely convergent. \displaystyle \lvert a_n \rvert = \frac{1}{n\sqrt n + 2} < \frac{1}{n^{3/2}}, a convergent p series (\displaystyle p = \frac32).
Conditionally convergent. \displaystyle \frac{1}{5n-2} > \frac{1}{5n} and \displaystyle \sum \frac{1}{5n} diverges, so no absolute convergence; \displaystyle \frac{1}{5n-2} is positive, decreasing and tends to 0, so the alternating series test gives convergence.
Divergent. \displaystyle \frac{n^2}{n^2+4} \to 1, so the terms approach 1 and -1 in turn and do not tend to 0 (The nth term test for divergence).
Absolutely convergent. \displaystyle \frac{\lvert \sin(4n) \rvert}{n^{3/2}} \le \frac{1}{n^{3/2}}, a convergent p series.
Conditionally convergent. \displaystyle \frac{n}{n^2+n+3} \big/ \frac1n = \frac{n^2}{n^2+n+3} \to 1, so \sum \lvert a_n \rvert diverges by limit comparison with the harmonic series. For \displaystyle f(x) = \frac{x}{x^2+x+3}, \displaystyle f'(x) = \frac{3 - x^2}{(x^2+x+3)^2} < 0 for \displaystyle x > \sqrt3, and \displaystyle b_2 = \frac29 > b_3 = \frac15, so the terms decrease from n = 2 (with \displaystyle b_1 = \frac15) and tend to 0; the alternating series test applies.
Absolutely convergent. \displaystyle \lvert a_n \rvert = \frac{2^n}{3^n+1} < \left(\frac23\right)^n, a convergent geometric series.
Conditionally convergent. \displaystyle g(x) = \sqrt x - \ln x has \displaystyle g'(x) = \frac{1}{2\sqrt x} - \frac1x, zero at x = 4, so its least value is g(4) = 2 - \ln 4 > 0. Hence \displaystyle \ln n < \sqrt n, (\ln n)^2 < n and \displaystyle \frac{1}{(\ln n)^2} > \frac1n: no absolute convergence. The terms \displaystyle \frac{1}{(\ln n)^2} decrease (the denominator increases for n \ge 2) and tend to 0, so the series converges.
Absolutely convergent. \displaystyle 0 < \sin\frac{1}{n^2} \le \frac{1}{n^2}, since \sin x \le x for x \ge 0, and \displaystyle \sum \frac{1}{n^2} converges.
Conditionally convergent. \displaystyle \sqrt{n+1} - \sqrt n = \frac{1}{\sqrt{n+1} + \sqrt n}, and dividing by \displaystyle \frac{1}{\sqrt n} gives \displaystyle \frac{\sqrt n}{\sqrt{n+1}+\sqrt n} \to \frac12, so \sum \lvert a_n \rvert diverges by limit comparison with \displaystyle p = \frac12. The denominator increases, so the terms decrease, and they tend to 0; the alternating series test applies.
Divergent. \displaystyle \frac{2^n}{n^2} \to \infty (two applications of L’Hôpital’s rule to \displaystyle \frac{2^x}{x^2}), so the terms do not tend to 0.
N = 23. \displaystyle \lvert a_n \rvert \le \frac{1}{n^3+1} < \frac{1}{n^3}, so \displaystyle \lvert S - S_N \rvert < \int_N^{\infty} \frac{dx}{x^3} = \frac{1}{2N^2}. We need \displaystyle \frac{1}{2N^2} \le 0.001, that is N^2 \ge 500; 22^2 = 484 is too small and 23^2 = 529 is enough.
Sum \displaystyle -\frac{3}{16}; absolute values sum to \displaystyle \frac34. Since \displaystyle n \le \left(\frac32\right)^n, \displaystyle \frac{n}{3^n} \le \left(\frac12\right)^n, so the convergence is absolute. As in Example 3, (1-x)S_N = x + \cdots + x^N - Nx^{N+1} with \displaystyle \lvert Nx^{N+1} \rvert = \frac{N}{3^{N+1}} \to 0, so \displaystyle S_N \to \frac{x}{(1-x)^2}. With \displaystyle x = -\frac13: \displaystyle \frac{-1/3}{(4/3)^2} = -\frac{3}{16}. With \displaystyle x = \frac13: \displaystyle \frac{1/3}{(2/3)^2} = \frac34.
Distance \displaystyle \frac92 kilometers; offset \displaystyle \frac32 kilometers to the right. \displaystyle \frac{6}{n(n+2)} = 3\left(\frac1n - \frac{1}{n+2}\right), which telescopes: \displaystyle \sum = 3\left(1 + \frac12\right) = \frac92. So the series converges absolutely. Signed, with m = n + 2 in the second sum, \displaystyle S_N = 3\left[\sum_{n=1}^{N} \frac{(-1)^{n+1}}{n} - \sum_{m=3}^{N+2} \frac{(-1)^{m+1}}{m}\right]. All terms cancel except \displaystyle 1 - \frac12 and the two terms with \displaystyle \frac{1}{N+1} and \displaystyle \frac{1}{N+2}, which tend to 0, so the offset is \displaystyle 3 \cdot \frac12 = \frac32.
The kth block is \displaystyle \frac{1}{\sqrt{2k-1}} minus the four terms \displaystyle \frac{1}{\sqrt{8k-6}}, \frac{1}{\sqrt{8k-4}}, \frac{1}{\sqrt{8k-2}}, \frac{1}{\sqrt{8k}}. Each of those four is at least \displaystyle \frac{1}{\sqrt{8k}}, so together they are at least \displaystyle \frac{4}{2\sqrt2\,\sqrt k} = \frac{\sqrt2}{\sqrt k}; and \displaystyle \frac{1}{\sqrt{2k-1}} \le \frac{1}{\sqrt k} since 2k - 1 \ge k. So the block is at most \displaystyle \frac{1 - \sqrt2}{\sqrt k}, and the partial sum after k blocks is at most \displaystyle (1-\sqrt2)\sum_{j=1}^{k} \frac{1}{\sqrt j} \to -\infty. Within a block the partial sums only fall after the first term, which is at most 1, so every partial sum tends to -\infty.
\displaystyle -\frac13. The series converges absolutely (\displaystyle \sum \left(\frac12\right)^n = 1), so every rearrangement has the original sum \displaystyle \frac{-1/2}{1 + 1/2} = -\frac13. The positive terms \displaystyle \frac14 + \frac{1}{16} + \cdots sum to \displaystyle \frac{1/4}{1 - 1/4} = \frac13, the negative terms \displaystyle -\frac12 - \frac18 - \cdots to \displaystyle -\frac{1/2}{1 - 1/4} = -\frac23, and \displaystyle \frac13 - \frac23 = -\frac13.
Indices 1, 2, 4, 6, 3, 8; running sums 1.500, 0.667, 0.217, -0.093, 0.490, 0.254. From 0: add \displaystyle \frac32; then subtract \displaystyle \frac56, \displaystyle \frac{9}{20} and \displaystyle \frac{13}{42} (indices 2, 4, 6) until the sum is at most 0; then add \displaystyle \frac{7}{12} (index 3) and subtract \displaystyle \frac{17}{72} (index 8).
No. Example 2 shows the series converges absolutely, and every rearrangement of an absolutely convergent series has the same sum.
Since \sum \lvert a_n \rvert converges, \lvert a_n \rvert \to 0, so \lvert a_n \rvert \le 1 for all n from some M on. For those n, a_n^2 = \lvert a_n \rvert \cdot \lvert a_n \rvert \le \lvert a_n \rvert, and \sum a_n^2 converges by direct comparison (the first M terms do not affect convergence).
\displaystyle a_n = \frac1n: \displaystyle \sum \frac1n diverges and \displaystyle \sum \frac{1}{n^2} converges. \displaystyle a_n = \frac{(-1)^n}{\sqrt n}: the series converges by the alternating series test, and \displaystyle \sum a_n^2 = \sum \frac1n diverges.
\lvert a_n c_n \rvert \le B\lvert a_n \rvert, and \sum B \lvert a_n \rvert converges, so \sum \lvert a_n c_n \rvert converges by direct comparison.