Absolutely Convergent

Frequently we want to manipulate infinite series. We may want to multiply them together and identify the product as another infinite series. For example, we could consider the product of the infinite geometric series

$$(1 - x)^{-1} \; = \; \sum_{n=0}^\infty x^n, ~|x| < 1$$

with itself, obtaining $(1-x)^{-2}$. It is natural to try to multiply out the infinite series term by term and then collect like terms, obtaining a new infinite series

$$(1 - x)^{-2} \; = \; \left(\sum_{m=0}^\infty x^m\right)\left(\sum_{n=0}^\infty x^n\right) \; = \; \sum_{m,n=0}^\infty x^{m+n} \; = \; \sum_{N=0}^\infty (N+1)x^N, ~|x| < 1.$$

The result of this manipulation is valid, but why? If we tried this trick with a different pair of series, the end result might not be true. The property of absolute convergence is what is needed to make calculations like the one above valid.

For simplicity, we shall restrict ourselves to considering real series. The results mostly hold for series of complex terms, but the proofs can be more complex.

An infinite series

$$\sum_{n=0}^\infty a_n$$

of real terms is called absolutely convergent if the series of positive terms

$$\sum_{n=0}^\infty |a_n|$$

converges. Obviously, any convergent series of positive terms is absolutely convergent, but there are plenty of series with both positive and negative terms to consider!

Proposition 1

Any absolutely convergent series is itself convergent.

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Note that

$$\left| \sum_{n=r+1}^s a_n\right| \; \le \; \sum_{n=r+1}^s |a_n|$$

for all $s > r$. Since $\sum_n |a_n|$ is convergent, it is a Cauchy series. The above inequality shows that the series $\sum_n a_n$ is therefore Cauchy itself, and hence is convergent. $_\square$

The converse is not true; there are convergent series which are not absolutely convergent. Consider

$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}.$$

This series converges to $\ln 2$ but is not absolutely convergent, since $\sum_{n=1}^\infty n^{-1}$ diverges.

If we multiply the series

$$\sum_{n=0}^\infty a_n x^n ~~\text{ and }~~ \sum_{n=0}^\infty b_n x^n$$

together term by term and collect like powers of $x$, we are led to the expression

$$\begin{array}{rcl} \displaystyle \left(\sum_{n=0}^\infty a_nx^n\right)\left(\sum_{n=0}^\infty b_nx^n\right) & = & \displaystyle a_0b_0 + \big(a_0b_1 + a_1b_0\big)x + \big(a_0b_2 + a_1b_1 + a_2b_0\big)x^2 + \cdots \\ & = & \displaystyle \sum_{n=0}^\infty \left(\sum_{m=0}^n a_mb_{n-m}\right)x^n, \end{array}$$

which leads us to ask the relationship among the series

$$\sum_{n=-0}^\infty a_n, \quad \sum_{n=0}^\infty b_n, \quad \sum_{n=0}^\infty c_n,$$

where the third series is obtained from the first two by defining

$$c_n \; = \; \sum_{m=0}^n a_m b_{n-m}, ~n \ge 0.$$

If the starting series is absolutely convergent, then we get the best possible result.

Proposition 2

If the series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$ are absolutely convergent, then so is $\sum_{n=0}^\infty c_n$, and

$$\sum_{n=0}^\infty c_n \; = \; \left(\sum_{n=0}^\infty a_n\right)\left(\sum_{n=0}^\infty b_n\right).$$

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Suppose that $\sum_{n=0}^\infty |a_n| = A$ and $\sum_{n=0}^\infty |b_n| = B$. Then

$$\begin{array}{rcl} \displaystyle \sum_{n=0}^N |c_n| & \le & \displaystyle \sum_{n=0}^N \sum_{m=0}^n |a_m|\,|b_{n-m}| \\ & = & \displaystyle \sum_{m+n \le N} |a_m|\,|b_n| \; \le \; \left(\sum_{m=0}^N |a_m|\right)\left(\sum_{n=0}^N |b_n|\right) \; \le \; AB \end{array}$$

for all $N \ge 0$, and hence $\sum_{n=0}^\infty |c_n|$ converges, which means that $\sum_{n=0}^\infty c_n$ is absolutely convergent.

Suppose now that $\sum_{n=0}^\infty a_n = \alpha$ and $\sum_{n=0}^\infty b_n = \beta$. We note that

$$\sum_{n=0}^{2N} c_n - \alpha\beta \; = \; \left(\sum_{m=0}^Na_m\right)\left(\sum_{n=0}^Nb_n\right) - \alpha\beta + \sum_{n=0}^{N-1} \sum_{m=N+1}^{2N-n}\big[a_mb_n + a_nb_m\big]$$

and hence

$$\left|\sum_{n=0}^{2N} c_n - \alpha\beta \right| \; \le \; \left| \left(\sum_{m=0}^Na_m\right)\left(\sum_{n=0}^Nb_n\right) - \alpha\beta \right| + A\sum_{m=N+1}^\infty |b_m| + B\sum_{m=N+1}^\infty|a_m|,$$

which implies, on letting $N \to \infty$, that $\sum_{n=0}^\infty c_n = \alpha \beta$, as required. $_\square$

It is important to note that this result does not automatically hold if the series involved are not absolutely convergent. A convergent series which is not absolutely convergent is called conditionally convergent.

Suppose that $a_n = b_n = \frac{(-1)^n}{\sqrt{n+1}}$ for all $n\ge 0$. Then the identical series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$ are both conditionally convergent. If we consider the series $\sum_{n=0}^\infty c_n$, then

$$c_n \; =\; \sum_{m=0}^n a_mb_{n-m} \; = \; (-1)^n \sum_{m=0}^n \frac{1}{\sqrt{(m+1)(n+1-m)}} ~~(n \ge 0).$$

Using the AM-GM Inequality, we see that

$$|c_n| \; \ge \; \sum_{m=0}^n \frac{2}{n+2} \; = \; \frac{2(n+1)}{n+2} \; \ge \; 1$$

for all $n \ge 0$. Thus the series $\sum_{n=0}^\infty c_n$ does not even converge (let alone not converge to a desired limit).

Since multiplying series and collecting terms in this manner is a very common tool in problem-solving, it is reassuring to realize that many series are absolutely convergent. In particular, Maclaurin and Taylor series for functions are (basically) absolutely convergent wherever they make sense. A power series is an infinite series of the form

$$A(x) \; = \; \sum_{n=0}^\infty a_n x^n$$

for some constants $a_0,a_1,a_2,\ldots$. Any Maclaurin series is an expansion of this type. A key theorem states that any power series possesses a radius of convergence. In other words, there exists a number $R$, which could be anything between $0$ and $\infty$, such that $A(x) = \sum_{n=0}^\infty a_nx^n$ converges absolutely for $|x| < R$, and diverges for $|x| > R$.

Note that this tells us that $A(x)$ diverges for all nonzero $x$ in the case that $R=0$, and that $A(x)$ is absolutely convergent for all real $x$ in the case that $R = \infty$. For example,

• $\sum_{n=0}^\infty \frac{x^n}{n!}$ has radius of convergence $\infty$, and converges absolutely for all real $x$;
• $\sum_{n=0}^\infty x^n$ has radius of convergence $1$, and converges absolutely for all $|x| < 1$, while diverging for all $|x| \geq 1$;
• $\sum_{n=0}^\infty n! x^n$ has radius of convergence $0$, and diverges for all nonzero $x$.

All the formal manipulation of power series that are often performed are justified provided that they are performed for values of $x$ which are less than the radii of convergence of all power series being analyzed.

However, while every infinitely differentiable function $f$ has a Maclaurin series, and every Maclaurin series has a radius of convergence, these two facts do not imply that the Maclaurin series of $f$ converges to $f$, even inside the interval of convergence! For example, the function

$$f(x) \; = \; \left\{ \begin{array}{lll} \displaystyle e^{-1/{x^2}} & \qquad & x \neq 0 \\ 0 & & x = 0 \end{array} \right.$$

is infinitely differentiable, with $f^{(n)}(0) = 0$ for all integers $n \ge 0$. Thus the Maclaurin series of $f$ is identically zero and therefore has an infinite radius of convergence. However, the Maclaurin series of $f$ only agrees with $f$ at the point $x=0$. Determining which functions can be expressed by their Maclaurin series is a nontrivial matter, most easily resolved using complex analysis; it is a matter of great relief that all the "standard" functions (powers, exponentials, trigonometric functions, etc.) do agree with their Maclaurin series, and we have to find examples as pathological as the one above to find a counterexample.

What happens if we take the terms of an infinite series, and add them in a different order? If the series is absolutely convergent, the answer is "nothing"! A rearrangement of a series $\sum_n a_n$ is obtained by a permutation $\pi$, a bijection $\pi\,:\, \mathbb{N} \to \mathbb{N}$, which can be used to define the rearranged series $\sum_n a_{\pi n}$.

Proposition 3

If $\sum_n a_n$ is an absolutely convergent series, and if $\pi \,:\,\mathbb{N} \to \mathbb{N}$ is a rearrangement of its terms, then $\sum_n a_{\pi n}$ is absolutely convergent, and $\sum_n a_{\pi n} \,=\, \sum_n a_n$.

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For any $N \in \mathbb{N}$, let $M(N) \,=\, \mathrm{max}\big\{\pi1,\pi2,\ldots,\pi N\big\}$. Then

$$\sum_{n=1}^N \big|a_{\pi n}\big| \; \le \; \sum_{n=1}^{M(N)} |a_n| \; \le \; \sum_{n=1}^\infty |a_n|,$$

and hence $\sum_n \big|a_{\pi n}\big| \,<\, \infty$, and the rearranged series is absolutely convergent.

Suppose that $\sum_n a_n = \alpha$. For any $N \in \mathbb{N}$, we can find $L(N) \in \mathbb{N}$ such that $\{1,2,\ldots,N\} \,\subseteq\, \{\pi1,\pi2, \ldots, \pi L(N)\}$, so that $\sum_{n=1}^m a_{\pi n}$ is equal to $\sum_{n=1}^N a_n$ plus a sum over some additional terms for any $m \ge L(N)$. But this means that

$$\left| \sum_{n=1}^m a_{\pi n} - \alpha\right| \; \le \; \left| \sum_{n=1}^N a_n - \alpha\right| + \sum_{n=N+1}^\infty |a_n|$$

for all $m \ge L(N)$; from this we deduce that $\sum_n a_{\pi n} \,=\, \alpha$, as required. $_\square$

What is strange is that, if the series we try to rearrange is convergent, but not absolutely convergent, then the above result is not true. A conditionally convergent series can be rearranged to converge to any limit desired.

Consider the series

$$X \; = \;\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \; = \; 1 - \tfrac12 + \tfrac13 - \tfrac14 + \tfrac15 - \cdots.$$

As it stands,

$$\sum_{n=1}^{2N} \frac{(-1)^{n-1}}{n} \; = \; \sum_{n=1}^{2N} \frac{1}{n} - 2\sum_{n=1}^N \frac{1}{2n} \; = \; \ln 2 + S_{2N} - S_N,$$

where

$$S_M \; = \; \sum_{n=1}^M n^{-1} - \ln M.$$ Since $S_M \to \gamma$, the Euler constant, as $M \to \infty$, we deduce that

$$X \; = \; 1 - \tfrac12 + \tfrac13 - \tfrac14 + \cdots \; = \; \ln 2.$$

Consider now the series where we are adding the same terms together in a different order:

$$X_1 \; = \; 1 + \tfrac13 - \tfrac12 + \tfrac15 + \tfrac17 - \tfrac14 + \tfrac19 + \tfrac{1}{11} - \tfrac16 + \cdots,$$

so that we add two positive reciprocals, then a negative one, then two positive ones, then a negative one, and so on. We are adding the same numbers, so should get the same answer of $\ln2$, you think? I am afraid not. If we add up the first $3N$ terms $($the first $N$ blocks of three$),$ we get

$$\sum_{n=1}^{2N} \frac{1}{2n-1} - \sum_{n=1}^N \frac{1}{2n} \; = \; S_{4N} - \tfrac12S_{2N} - \tfrac12S_N + \tfrac32\ln2;$$

letting $N \to \infty$ shows that $X_1 = \tfrac32\ln2$. It is possible to prove, for any real number $u$, a rearrangement of the terms of the series can be found that would make the sum converge to $u$. Equally well, rearrangements can be found which make the series diverge, or even oscillate between $-1$ and $1$ if that is what we want.

How to get a rearrangement that converges to $u$:
Add positive reciprocals until the sum exceeds $u$, and then add negative reciprocals until the sum is less than $u$. Then add positive reciprocals until the sum is greater than $x$, and then add negative reciprocals until the sum is less than $u$. We can repeat this recipe indefinitely. Since we use at least one positive and one negative reciprocal in each cycle, the new series will contain each reciprocal exactly once—we have a rearrangement of the terms. It is also clear that the rearranged series tends to $u$.