Absolute and Conditional Convergence

So far most of our work has been with series all of whose terms are positive. There is a good reason for this; there is very little we can say about series with mixed signs. Indeed there is just one useful result at this level, which is the topic of this section.

The easiest case occurs when the series really can be thought of as a series of positive terms.

Definition 6.23   The series $\sum_{{n=1}}^{\infty}$a_n is absolutely convergent iff the series $\sum_{{n=1}}^{\infty}$| a_n| is convergent.

Definition 6.24   The series $\sum_{{n=1}}^{\infty}$a_n is conditionally convergent if and only if the series $\sum_{{n=1}}^{\infty}$a_n is convergent but not absolutely convergent.

Example 6.25   Show that the series $$\sum_{{n=1}}^{\infty}$$$${\frac{{(-1)^{n+1}}}{{n^2}}}$$ is absolutely convergent.

Solution. We have $$\sum_{{n=1}}^{\infty}$$$$\left\vert\vphantom{\frac{(-1)^{n+1}}{n^2}}\right.$$$${\frac{{(-1)^{n+1}}}{{n^2}}}$$$$\left.\vphantom{\frac{(-1)^{n+1}}{n^2}}\right\vert$$ = $$\sum_{{n=1}}^{\infty}$$$${\frac{{1}}{{n^2}}}$$ and this is convergent by 6.19. So $$\sum_{{n=1}}^{\infty}$$$${\frac{{(-1)^{n+1}}}{{n^2}}}$$ is absolutely convergent.

Note:We choose to work with the sign (- 1)ⁿ⁺¹ rather than (- 1)ⁿ simply for tidiness; it is usual to start a series with a positive term, so the coefficient of a₁ is chosen to be +. Thus that of a₂ must be - etc if the series is to alternate.

Exercise 6.26   Show that the series $$\sum_{{n=2}}^{\infty}$$$${\frac{{(-1)^n}}{{n^2\log n}}}$$ is absolutely convergent.

Exercise 6.27   Show that the series $$\sum_{{n=1}}^{\infty}$$$${\frac{{\sin n}}{{n^2}}}$$ is absolutely convergent. [Hint: note that | sin n|$\le$1, and use the comparison test.]

Our interest in absolutely convergent series starts by observing that they are in fact all convergent. Indeed this is the easiest way to show a series is convergent if the terms are not all positive.

Proposition 6.28   An absolutely convergent series is convergent.

Proof. Assume that $\sum_{{n=1}}^{\infty}$a_n is absolutely convergent, and define

a_n⁺ = $$\left\{\vphantom{\begin{array}{ll} a_n&\mbox{if $a_n > 0$}\\ 0 &\mbox{if $a_n \le 0$} \end{array} }\right.$$$$\begin{array}{ll} a_n&\mbox{if $a_n > 0$}\\ 0 &\mbox{if $a_n \le 0$} \end{array}$$    and    a_n^- = $$\left\{\vphantom{\begin{array}{ll} \vert a_n\vert&\mbox{if $a_n < 0$}\\ 0 &\mbox{if $a_n \ge 0$} \end{array} }\right.$$$$\begin{array}{ll} \vert a_n\vert&\mbox{if $a_n < 0$}\\ 0 &\mbox{if $a_n \ge 0$} \end{array}$$

The point of this definition is that

$$\le \le \le \le$$ (6.1)

so we have two new series of positive terms, while

| a_n| = a_n⁺ + a_n^-    and    a_n = a_n⁺ - a_n^-.

Using equation 6.1 to compare with the convergent series $\sum_{{n=1}}^{\infty}$| a_n|, we see that each of

$$\sum_{{n=1}}^{\infty}$$a_n⁺    and    $$\sum_{{n=1}}^{\infty}$$a_n^-

is a convergent series of positive terms. Thus

$$\sum_{{n=1}}^{\infty}$$a_n = $$\sum_{{n=1}}^{\infty}$$a_n⁺ - a_n^- = $$\sum_{{n=1}}^{\infty}$$a_n⁺ + $$\sum_{{n=1}}^{\infty}$$a_n^-

is also convergent using 6.11. $\qedsymbol$

This gives one way of proving that a series is convergent even if the terms are not all positive, and so we can't use the comparison test directly. There is essentially only one other way, which is a very special, but useful case known as Leibniz theorem, or the theorem on alternating signs, or the alternating series test. We give the proof because the argument is so like the proof of the convergence of the ratio of adjacent terms in the Fibonacci series 3.1.

Warning:Note how we usually talk about the ``Fibonacci series'', even though it is a sequence rather than a series. Try not to be confused by this popular but inaccurate usage.

Theorem 6.29   Leibniz Theorem Let {a_n} be a decreasing sequence of positive terms such that a_n$\to$ 0 as n$\to$$\infty$. Then the series

$$\sum_{{n=1}}^{\infty}$$(- 1)ⁿ⁺¹a_n    is convergent.    

Proof. Write S_n for the n^th partial sum of the series $$\sum_{{n=1}}^{\infty}$$(- 1)ⁿ⁺¹a_n. We show this sequence has the same type of oscillating behaviour that the corresponding sequence of partial sums in the Fibonacci example. By definition, we have

s_2n+1 = a₁ - a₂ + a₃ -...+ a_2n-1 - a_2n + a_2n+1

s_2n-1 = a₁ - a₂ + a₃ -...+ a_2n-1
and so, subtracting, we have

s_2n+1 = s_2n-1 - a_2n + a_2n+1.

Since {a_n} is a decreasing sequence, a_2n > a_2n+1 and so s_2n+1 < s_2n-1. Thus we have a decreasing sequence

s₁ > s₃ > s₅ >...> s_2n-1 > s_2n+1 >....

Similarly s_2n > s_2n-2 and we have an increasing sequence

s₂ < s₄ < s₆...< s_2n-2 < s_2n <....

Also

s_2n+1 = s_2n + a_2n+1 > s_2n

Thus

s₂ < s₄ < s₆...< s_2n-2 < s_2n < s_2n+1 < s_2n-1 <...s₅ < s₃ < s₁

and the sequence s₁, s₃, s₅, ... is a decreasing sequence which is bounded below (by s₂), and so by 3.9 is convergent to $\alpha$ (say). Similarly s₂, s₄, s₆, ... is an increasing sequence which is bounded above (by s₁), and so by 3.9 is convergent to $\beta$ (say) also

s_2n+1 - s_2n = a_2n+1

$$\to \infty$$

$$\alpha \beta$$ = 0

So $\alpha$ = $\beta$, and all the partial sums are tending to $\alpha$, so the series converges. $\qedsymbol$

Example 6.30   Show that the series $$\sum_{{n=1}}^{\infty}$$$${\frac{{(-1)^{n+1}}}{{n}}}$$ is conditionally convergent.

Solution. We have $$\sum_{{n=1}}^{\infty}$$$$\left\vert\vphantom{\frac{(-1)^{n+1}}{n}}\right.$$$${\frac{{(-1)^{n+1}}}{{n}}}$$$$\left.\vphantom{\frac{(-1)^{n+1}}{n}}\right\vert$$ = $$\sum_{{n=1}}^{\infty}$$$${\frac{{1}}{{n}}}$$ and this is divergent by 6.19; thus the series is not absolutely convergent. We show using 6.29 that this series is still convergent, and so is conditionally convergent.

Write a_n = 1/n, so a_n > 0, a_n+1 < a_n and a_n$\to$ 0 as n$\to$$\infty$. Thus all the conditions of Leibniz's theorem are satisfied, and so the series $$\sum_{{n=1}}^{\infty}$$$${\frac{{(-1)^{n+1}}}{{n}}}$$ is convergent.

Proposition 6.31 (Re-arranging an Absolutely convergent Series)   Let $\sum_{{n=1}}^{\infty}$a_n be an absolutely convergent series and suppose that {b_n} is a re-arrangement of {a_n}. Then $\sum_{{n=1}}^{\infty}$b_n is convergent, and

$$\sum_{{n=1}}^{\infty}$$b_n = $$\sum_{{n=1}}^{\infty}$$a_n.

Proof. See next year, or (Spivak, 1967); the point here is that we need absolute convergence before series behave in a reasonable way. $\qedsymbol$

Warning:It is not useful to re-arrange conditionally convergent series (remember the rearrangement I did in section 1.1). There is a result which is an extreme form of this:

Pick x $\in$ $\mathbb {R}$, and let $\sum_{{n=1}}^{\infty}$a_n be a conditionally convergent series. then there is a re-arrangement {b_n} of {a_n} such that $\sum_{{n=1}}^{\infty}$b_n = x!

In other words, we can re-arrange to get any answer we want!