When Signs Alternate: Absolute and Conditional Convergence

1. Absolute convergence

In the first post of this series, all terms were non-negative and the partial sums could only grow. Now the terms may be positive or negative, and the partial sums can go back and forth. The first idea is to strip off the signs and see what happens.

Definition The series \[ \sum_{n=1}^{\infty} a_n \tag{1} \] converges absolutely if the series of absolute values \[ \sum_{n=1}^{\infty} |a_n| \tag{2} \] converges.

Theorem If (2) converges, then (1) also converges. Moreover, the sum of an absolutely convergent series does not depend on the order in which its terms are added.

Why absolute convergence implies convergence: use Cauchy's criterion. By the triangle inequality,

\[ |a_{n+1} + \cdots + a_{n+p}| \le |a_{n+1}| + \cdots + |a_{n+p}|. \]

If the right side becomes small (because (2) converges), so does the left side, so (1) satisfies Cauchy's criterion. Intuitively, cancellation between positive and negative terms can only help convergence.

The practical value: to test (2) you can use every test from the first post (comparison, ratio, root, Raabe, integral), since \(|a_n| \ge 0\).

Example 1 \( \displaystyle\sum_{n=1}^\infty \frac{\sin n}{n^2} \). The signs of \(\sin n\) are irregular, but \( \left|\frac{\sin n}{n^2}\right| \le \frac{1}{n^2} \), and \( \sum\frac1{n^2} \) converges. So the series converges absolutely.

2. Conditional convergence and Riemann's theorem

Definition If the series (1) converges but the series (2) diverges, then (1) is said to converge conditionally.

The standard example is the alternating harmonic series:

\[ 1 - \frac12 + \frac13 - \frac14 + \cdots = \ln 2. \]

It converges (we prove this in Section 3), but its absolute values form the harmonic series, which diverges. It converges only "thanks to" the cancellation between terms.

Riemann's rearrangement theorem By changing the order of its terms, a conditionally convergent series can be made to converge to any prescribed value (or even to diverge).

Why this happens: in a conditionally convergent series, the positive terms alone add up to \(+\infty\) and the negative terms alone to \(-\infty\) (if either part were finite, the series would converge absolutely). Want the sum to be, say, \(10\)? Take positive terms until you pass \(10\), then negative terms until you fall below it, then positive again, and so on. You never run out of terms on either side, and since the terms tend to zero, the overshoots shrink: the partial sums settle at \(10\).

Example 2 · Same terms, half the sum Rearrange the alternating harmonic series as one positive term followed by two negative ones: \[ 1 - \frac12 - \frac14 + \frac13 - \frac16 - \frac18 + \frac15 - \frac1{10} - \frac1{12} + \cdots \] Group each positive term with the negative term right after it: \[ \Big(1 - \frac12\Big) - \frac14 + \Big(\frac13 - \frac16\Big) - \frac18 + \Big(\frac15 - \frac1{10}\Big) - \frac1{12} + \cdots = \frac12 - \frac14 + \frac16 - \frac18 + \cdots \] \[ = \frac12\Big(1 - \frac12 + \frac13 - \frac14 + \cdots\Big) = \frac12\ln 2. \] Every term of the original series appears exactly once, yet the sum is cut in half.

The lesson For finite sums, order never matters. For infinite series, it matters unless the convergence is absolute. Never rearrange, regroup in complicated ways, or split a conditionally convergent series into positive and negative parts without checking.

3. Leibniz's test for alternating series

A series whose signs strictly alternate is called alternating. For these there is a beautifully simple test.

Leibniz's test The alternating series \[ b_1 - b_2 + b_3 - b_4 + \cdots + (-1)^{n-1}b_n + \cdots \qquad (b_n \ge 0) \] converges if \[ b_n \ge b_{n+1} \quad (n = 1, 2, \dots) \qquad \text{and} \qquad \lim_{n\to\infty} b_n = 0. \] In that case the remainder \( R_n = (-1)^nb_{n+1} + (-1)^{n+1}b_{n+2} + \cdots \) satisfies \[ R_n = (-1)^n\,\theta_n\,b_{n+1} \qquad (0 \le \theta_n \le 1). \]

Why it works: picture the partial sums on a number line. Starting at \(S_1 = b_1\), each step goes back by \(b_2\), forward by \(b_3\), back by \(b_4\), and so on. Since the steps keep shrinking, each step overshoots less than the one before. The even partial sums \(S_2, S_4, S_6, \dots\) climb upward, the odd ones \(S_1, S_3, S_5, \dots\) come down, and they are always trapped between each other:

\[ S_2 \le S_4 \le S_6 \le \cdots \le S \le \cdots \le S_5 \le S_3 \le S_1. \]

Since the gap \( S_{2k+1} - S_{2k} = b_{2k+1} \) tends to zero, both sequences converge to the same limit \(S\).

The remainder estimate

The formula for \(R_n\) says two practical things:

Size: \( |R_n| \le b_{n+1} \). The error made by stopping after \(n\) terms is at most the first omitted term.
Sign: the error has the sign of the first omitted term, \((-1)^n\).

This makes alternating series ideal for numerical computation: you know in advance how many terms you need.

Example 3 · The alternating harmonic series \( b_n = \frac1n \) decreases and tends to \(0\), so \( \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \) converges. Since \(\sum\frac1n\) diverges, the convergence is conditional.

How good is \(S_{100}\) as an approximation of \(\ln 2\)? The error is at most \(b_{101} = \frac{1}{101} < 0.01\). Only about two correct decimals after 100 terms: conditionally convergent series often converge slowly.

Example 4 · How many terms do we need? Compute \( \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3} \) with an error less than \(0.001\).

We need \( b_{n+1} = \frac{1}{(n+1)^3} < 0.001 \), i.e. \( (n+1)^3 > 1000 \), so \( n + 1 \ge 11 \). Taking \(n = 10\) terms is enough, since \( b_{11} = \frac{1}{1331} < 0.001 \).

Both conditions are needed

\(b_n \to 0\): without it, the series diverges by the n-th term test. Example: \( \sum (-1)^{n-1}\frac{n}{n+1} \).

Monotonicity: take \( b_{2k-1} = \frac1k \), \( b_{2k} = \frac{1}{k^2} \). Then \(b_n \to 0\), but the terms do not decrease, and

\[ 1 - 1 + \frac12 - \frac14 + \frac13 - \frac19 + \cdots = \sum_{k=1}^\infty\Big(\frac1k - \frac1{k^2}\Big) = \infty. \]

The positive terms are much bigger than the negative ones, and cancellation fails.

4. The Abel and Dirichlet tests

Leibniz's test handles strict alternation \(+, -, +, -\). Many series have irregular sign patterns, like \( \sum\frac{\sin n}{n} \), where \(|\sin n|/n\) doesn't even give an absolutely convergent series. The two tests below handle series of the form \( \sum a_nb_n \), where one factor oscillates and the other is "tame".

Abel's test The series \( \displaystyle\sum_{n=1}^\infty a_nb_n \) converges if the series \( \displaystyle\sum_{n=1}^\infty a_n \) converges and the numbers \(b_n\) form a monotone, bounded sequence.

Dirichlet's test The series \( \displaystyle\sum_{n=1}^\infty a_nb_n \) converges if the partial sums \( A_n = \displaystyle\sum_{k=1}^n a_k \) are bounded and \(b_n\) tends monotonically to zero as \(n \to \infty\).

How to tell them apart: the two tests trade conditions. Dirichlet asks less of \(\sum a_n\) (bounded partial sums, not necessarily convergent) but more of \(b_n\) (must go to zero). Abel asks more of \(\sum a_n\) (convergence) but less of \(b_n\) (only monotone and bounded).

The idea behind both is summation by parts, the discrete version of integration by parts:

\[ \sum_{k=1}^{n} a_kb_k = A_nb_n + \sum_{k=1}^{n-1} A_k\,(b_k - b_{k+1}). \]

If \(A_k\) is bounded and \(b_k\) decreases monotonically to zero, the first term tends to zero and the sum on the right converges absolutely, because \( \sum (b_k - b_{k+1}) \) telescopes.

Leibniz's test is a special case of Dirichlet's: take \( a_n = (-1)^{n-1} \), whose partial sums are \(1, 0, 1, 0, \dots\) (bounded), and \(b_n\) decreasing to zero.

Example 5 · Dirichlet: \( \sum\frac{\sin nx}{n} \) For \(x \ne 2k\pi\), a trigonometric identity gives \[ \left|\sum_{k=1}^n \sin kx\right| = \left|\frac{\sin\frac{nx}{2}\,\sin\frac{(n+1)x}{2}}{\sin\frac x2}\right| \le \frac{1}{\left|\sin\frac x2\right|}, \] so the partial sums are bounded. Since \(b_n = \frac1n\) decreases to zero, Dirichlet's test shows that \( \displaystyle\sum_{n=1}^\infty \frac{\sin nx}{n} \) converges for every \(x\). (For \(x = 2k\pi\) all terms are zero.) The same works for \( \sum\frac{\cos nx}{n} \) when \(x \ne 2k\pi\).

Example 6 · Abel \( \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\Big(1 + \frac1n\Big)^n \). Here \( \sum\frac{(-1)^{n-1}}{n} \) converges (Example 3), and \( b_n = \big(1 + \frac1n\big)^n \) is increasing and bounded (it tends to \(e\)). By Abel's test, the series converges.

5. Operations with convergent series

5.1 Sum and difference

Rule 1 If \( \sum a_n \) and \( \sum b_n \) converge, then \[ \sum_{n=1}^\infty a_n \pm \sum_{n=1}^\infty b_n = \sum_{n=1}^\infty (a_n \pm b_n). \]

This always works, for any convergent series: it follows directly from adding the partial sums and passing to the limit. Similarly, \( \sum c\,a_n = c\sum a_n \) for any constant \(c\).

5.2 Product: the Cauchy product

How should we multiply two infinite series? Expanding \((a_1 + a_2 + \cdots)(b_1 + b_2 + \cdots)\) gives all products \(a_ib_k\), and these must be arranged into one series. The standard choice collects the products along "diagonals":

Rule 2 · Cauchy product If \( \sum a_n \) and \( \sum b_n \) converge and at least one of them converges absolutely, then \[ \Big(\sum_{n=1}^\infty a_n\Big)\Big(\sum_{n=1}^\infty b_n\Big) = \sum_{n=1}^\infty c_n, \qquad \text{where} \quad c_n = a_1b_n + a_2b_{n-1} + \cdots + a_nb_1. \]

Why "diagonals"? Write the products \(a_ib_k\) in a table. The term \(c_n\) collects all products with \(i + k = n + 1\), one antidiagonal of the table. It's exactly how you collect equal powers when multiplying polynomials (or power series).

Example 7 · Squaring the geometric series For \(|x| < 1\), multiply \( \sum_{n=0}^\infty x^n = \frac{1}{1 - x} \) by itself (it converges absolutely). With indices starting at \(0\), the \(n\)-th diagonal contains \( x^0x^n, x^1x^{n-1}, \dots, x^nx^0 \), which is \(n + 1\) copies of \(x^n\): \[ \frac{1}{(1 - x)^2} = \sum_{n=0}^\infty (n + 1)\,x^n = 1 + 2x + 3x^2 + 4x^3 + \cdots \]

Example 8 · Why \( e^xe^y = e^{x+y} \) The series \( e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \) converges absolutely for all \(x\). Its Cauchy product with \( e^y \) has terms \[ c_n = \sum_{k=0}^{n}\frac{x^k}{k!}\cdot\frac{y^{n-k}}{(n-k)!} = \frac{1}{n!}\sum_{k=0}^n\binom nk x^ky^{n-k} = \frac{(x + y)^n}{n!}, \] by the binomial theorem. So \( e^x\cdot e^y = \sum\frac{(x+y)^n}{n!} = e^{x+y} \): the law of exponents follows from the Cauchy product.

Absolute convergence is not a formality Take \( a_n = b_n = \frac{(-1)^{n-1}}{\sqrt n} \). Both series converge (Leibniz), but only conditionally. In their Cauchy product, all terms of \(c_n\) have the same sign \((-1)^{n-1}\), and since \( \sqrt{k}\sqrt{n+1-k} \le \frac{n+1}{2} \), \[ |c_n| = \sum_{k=1}^n \frac{1}{\sqrt k\,\sqrt{n + 1 - k}} \ge n\cdot\frac{2}{n + 1} \longrightarrow 2. \] The terms \(c_n\) don't tend to zero, so the product series diverges. Two convergent series, a divergent product.

6. Strategy: how to classify a series

Check \(a_n \to 0\). If not, the series diverges.
Test \( \sum|a_n| \) with the tests for positive series (comparison, ratio, root, integral). If it converges, the series converges absolutely, and you're done.
If \( \sum|a_n| \) diverges, the series may still converge conditionally:
- signs alternate regularly → Leibniz;
- a factor like \(\sin nx\), \(\cos nx\) with bounded partial sums, times a factor decreasing to zero → Dirichlet;
- a known convergent series times a monotone bounded factor → Abel.
If one of these applies, the series converges conditionally. Remember: then you may not rearrange it, and it can't be safely multiplied by another conditionally convergent series.

Type	\(\sum a_n\)	\(\sum \|a_n\|\)	Rearrangement	Example
Absolute convergence	converges	converges	sum unchanged	\( \sum\frac{(-1)^{n-1}}{n^2} \)
Conditional convergence	converges	diverges	any sum possible	\( \sum\frac{(-1)^{n-1}}{n} \)
Divergence	diverges	diverges	—	\( \sum(-1)^{n-1} \)

7. Practice problems

Try each one before opening the solution. More solved problems from this topic are in Series.

Problem 1: Classify \( \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}\,n}{n^2 + 1} \).

\( |a_n| = \frac{n}{n^2+1} \sim \frac1n \): the absolute series diverges (limit comparison). But \( f(x) = \frac{x}{x^2+1} \) decreases for \(x \ge 1\) and tends to \(0\), so Leibniz gives convergence. Conditionally convergent.

Problem 2: Classify \( \displaystyle\sum_{n=1}^\infty \frac{\cos n}{n^2} \).

\( \left|\frac{\cos n}{n^2}\right| \le \frac1{n^2} \). Absolutely convergent.

Problem 3: Classify \( \displaystyle\sum_{n=1}^\infty (-1)^n\frac{n}{2n + 1} \).

\( |a_n| \to \frac12 \ne 0 \). Divergent by the n-th term test.

Problem 4: Classify \( \displaystyle\sum_{n=2}^\infty \frac{(-1)^n}{n\ln n} \).

The absolute series \( \sum\frac{1}{n\ln n} \) diverges (integral test, see the first post). But \( \frac{1}{n\ln n} \) decreases to zero, so Leibniz gives convergence. Conditionally convergent.

Problem 5: How many terms of \( \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} \) are needed to get the sum with an error below \(0.01\)?

We need \( \frac{1}{(n+1)^2} < 0.01 \), i.e. \( n + 1 > 10 \), so \(n + 1 \ge 11\) and \( n = 10 \) terms suffice (\( b_{11} = \frac1{121} < 0.01 \)).

Problem 6: Show that \( \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\operatorname{arctg}n \) converges.

Abel's test: \( \sum\frac{(-1)^{n-1}}{n} \) converges, and \( \operatorname{arctg}n \) is increasing and bounded by \(\frac\pi2\).

Problem 7: For which \(x\) does \( \displaystyle\sum_{n=1}^\infty \frac{\cos nx}{\sqrt n} \) converge?

For \(x \ne 2k\pi\), the partial sums of \(\cos nx\) are bounded (by \( \frac{1}{|\sin(x/2)|} \)), and \( \frac1{\sqrt n} \) decreases to zero: converges by Dirichlet. For \(x = 2k\pi\), the series is \( \sum\frac1{\sqrt n} \), which diverges. So it converges exactly for \( x \ne 2k\pi \).

Final thoughts

Series with mixed signs live in two very different worlds. Absolutely convergent series behave like finite sums: you can reorder, regroup and multiply them freely. Conditionally convergent series converge only through a delicate balance of cancellations, and that balance is destroyed by rearranging. Knowing which world you are in is the first question to ask about any series with negative terms.

Next in the series: iterated and double series, then series of functions and uniform convergence, and power series.