
# 4.13: Absolutely Convergent Series. Power Series


I. A series $$\sum f_{m}$$ is said to be absolutely convergent on a set $$B$$ iff the series $$\sum\left|f_{m}(x)\right|$$ (briefly, $$\sum\left|f_{m}\right| )$$ of the absolute values of $$f_{m}$$ converges on $$B$$ (pointwise or uniformly). Notation:

$f=\sum\left|f_{m}\right| \text { (pointwise or uniformly } ) \text { on } B.$

In general, $$\sum f_{m}$$ may converge while $$\sum\left|f_{m}\right|$$ does not (see Problem 12$$) .$$ In this case, the convergence of $$\sum f_{m}$$ is said to be conditional. (It may be absolute for some $$x$$ and conditional for others.) As we shall see, absolute convergence ensures the commutative law for series, and it implies ordinary convergence (i.e., that of $$\sum f_{m} ),$$ if the range space of the $$f_{m}$$ is complete.

Note 1. Let

$\sigma_{m}=\sum_{k=1}^{m}\left|f_{k}\right|.$

Then

$\sigma_{m+1}=\sigma_{m}+\left|f_{m+1}\right| \geq \sigma_{m} \quad \text { on } B;$

i.e., the $$\sigma_{m}(x)$$ form a monotone sequence for each $$x \in B .$$ Hence by Theorem 3 of Chapter 3, §15,

$\lim _{m \rightarrow \infty} \sigma_{m}=\sum_{m=1}^{\infty}\left|f_{m}\right| \text { always exists in } E^{*};$

$$\sum\left|f_{m}\right|$$ converges iff $$\sum_{m=1}^{\infty}\left|f_{m}\right|<+\infty$$.

For the rest of this section we consider only complete range spaces.

Theorem $$\PageIndex{1}$$

Let the range space of the functions $$f_{m}$$ (all defined on $$A )$$ be $$E^{1}$$, $$C,$$ or $$E^{n}\left(^{*} \text { or another complete normed space). Then for } B \subseteq A, \text { we have the }\right.$$ following:

(i) If $$\sum\left|f_{m}\right|$$ converges on $$B$$ (pointwise or uniformly), so does $$\sum f_{m}$$ itself. Moreover,

$\left|\sum_{m=1}^{\infty} f_{m}\right| \leq \sum_{m=1}^{\infty}\left|f_{m}\right| \quad \text { on } B.$

(ii) (Commutative law for absolute convergence.) If $$\sum\left|f_{m}\right|$$ converges (pointwise or uniformly on $$B,$$ so does any series $$\sum\left|g_{m}\right|$$ obtained by rearranging the $$f_{m}$$ in any different order. Moreover,

$\sum_{m=1}^{\infty} f_{m}=\sum_{m=1}^{\infty} g_{m} \quad(\text { both exist on } B)/$

Note 2. More precisely, a sequence $$\left\{g_{m}\right\}$$ is called a rearrangement of $$\left\{f_{m}\right\}$$ iff there is a map $$u : N \longleftrightarrow_{onto} N$$ such that

$(\forall m \in N) \quad g_{m}=f_{u(m)}.$

Proof

(i) If $$\sum\left|f_{m}\right|$$ converges uniformly on $$B,$$ then by Theorem $$3^{\prime}$$ of §12,

$\begin{array}{l}{(\forall \varepsilon>0)(\exists k)(\forall n>m>k)(\forall x \in B)} \\ {\qquad \varepsilon>\sum_{i=m}^{n}\left|f_{i}(x)\right| \geq\left|\sum_{i=m}^{n} f_{i}(x)\right| \text { (triangle law) }}\end{array}.$

However, this shows that $$\sum f_{n}$$ satisfies Cauchy's criterion (6) of §12, so it converges uniformly on $$B$$.

Moreover, letting $$n \rightarrow \infty$$ in the inequality

$\left|\sum_{m=1}^{n} f_{m}\right| \leq \sum_{m=1}^{n}\left|f_{m}\right|,$

we get

$\left|\sum_{m=1}^{\infty} f_{m}\right| \leq \sum_{m=1}^{\infty}\left|f_{m}\right|<+\infty \quad \text { on } B, \text { as claimed. }$

By Note $$1,$$ this also proves the theorem for pointwise convergence.

(ii) Again, if $$\sum f_{m} |$$ converges uniformly on $$B,$$ the inequalities $$(1)$$ hold for all $$f_{i}$$ except (possibly) for $$f_{1}, f_{2}, \ldots, f_{k}$$ . Now when $$\sum f_{m}$$ is rearranged, these $$k$$ functions will be renumbered as certain $$g_{i} .$$ Let $$q$$ be the largest of their new subscripts i. Then all of them (and possibly some more functions) are among $$g_{1}, g_{2}, \ldots, g_{q}$$ (so that $$q \geq k ) .$$ Hence if we exclude $$g_{1}, \ldots, g_{q},$$ the inequalities $$(1)$$ will certainly hold for the remaining $$g_{i}$$ $$(i>q) .$$ Thus

$(\forall \varepsilon>0)(\exists q)(\forall n>m>q)(\forall x \in B) \quad \varepsilon>\sum_{i=m}^{n}\left|g_{i}\right| \geq\left|\sum_{i=m}^{n} g_{i}\right|.$

By Cauchy's criterion, then, both $$\sum g_{i}$$ and $$\sum\left|g_{i}\right|$$ converge uniformly.

Moreover, by construction, the two partial sums

$s_{k}=\sum_{i=1}^{k} f_{i} \text { and } s_{q}^{\prime}=\sum_{i=1}^{q} g_{i}$

can differ only in those terms whose original subscripts (before the rearrangement) were $$>k . \mathrm{By}(1),$$ however, any finite sum of such terms is less than $$\varepsilon$$ in absolute value. Thus $$\left|s_{q}^{\prime}-s_{k}\right|<\varepsilon$$.

This argument holds also if $$k$$ in $$(1)$$ is replaced by a larger integer.
(Then also $$q$$ increases, since $$q \geq k$$ as noted above.) Thus we may let $$k \rightarrow+\infty(\text { hence also } q \rightarrow+\infty)$$ in the inequality $$\left|s_{q}^{\prime}-s_{k}\right|<\varepsilon,$$ with $$\varepsilon$$ fixed. Then

$s_{k} \rightarrow \sum_{m=1}^{\infty} f_{m} \text { and } s_{q}^{\prime} \rightarrow \sum_{i=1}^{\infty} g_{i},$

so

$\left|\sum_{i=1}^{\infty} g_{i}-\sum_{m=1}^{\infty} f_{m}\right| \leq \varepsilon.$

Now let $$\varepsilon \rightarrow 0$$ to get

$\sum_{i=1}^{\infty} g_{i}=\sum_{m=1}^{\infty} f_{m};$

similarly for pointwise convergence. $$\square$$

II. Next, we develop some simple tests for absolute convergence.

Theorem $$\PageIndex{2}$$

(comparison test). Suppose

$(\forall m) \quad\left|f_{m}\right| \leq\left|g_{m}\right| \text { on } B.$

Then

(i) $$\sum_{m=1}^{\infty}\left|f_{m}\right| \leq \sum_{m=1}^{\infty}\left|g_{m}\right|$$ on $$B$$;

(ii) $$\sum_{m=1}^{\infty}\left|f_{m}\right|=+\infty$$ implies $$\sum_{m=1}^{\infty}\left|g_{m}\right|=+\infty$$ on $$B ;$$ and

(iii) If $$\sum\left|g_{m}\right|$$ converges (pointwise or uniformly $$)$$ on $$B,$$ so does $$\sum\left|f_{m}\right|$$.

Proof

Conclusion (i) follows by letting $$n \rightarrow \infty$$ in

$\sum_{m=1}^{n}\left|f_{m}\right| \leq \sum_{m=1}^{n}\left|g_{m}\right|.$

In turn, (ii) is a direct consequence of $$(\mathrm{i})$$.

Also, by (i),

$\sum_{m=1}^{\infty}\left|g_{m}\right|<+\infty \text { implies } \sum_{m=1}^{\infty}\left|f_{m}\right|<+\infty.$

This proves (iii) for the pointwise case (see Note 1$$) .$$ The uniform case follows exactly as in Theorem 1$$(\mathrm{i})$$ on noting that

$\sum_{k=m}^{n}\left|f_{k}\right| \leq \sum_{k=m}^{n}\left|g_{k}\right|$

and that the functions $$\left|f_{k}\right|$$ and $$\left|g_{k}\right|$$ are real (so Theorem $$3^{\prime}$$ in §12 does apply). $$\square$$

Theorem $$\PageIndex{3}$$ (Weierstrass "M-test")

If $$\sum M_{n}$$ is a convergent series of real constants $$M_{n} \geq 0$$ and if

$(\forall n) \quad\left|f_{n}\right| \leq M_{n}$

on a set $$B,$$ then $$\sum\left|f_{n}\right|$$ converges uniformly on $$B.$$ Moreover,

$\sum_{n=1}^{\infty}\left|f_{n}\right| \leq \sum_{n=1}^{\infty} M_{n} \quad \text { on } B.$

Proof

Use Theorem 2 with $$\left|g_{n}\right|=M_{n},$$ noting that $$\sum\left|g_{n}\right|$$ converges uniformly since the $$\left|g_{n}\right|$$ are constant (§12, Problem 7). $$\square$$

ExampleS

(a) Let

$f_{n}(x)=\left(\frac{1}{2} \sin x\right)^{n} \text { on } E^{1}.$

Then

$(\forall n)\left(\forall x \in E^{1}\right) \quad\left|f_{n}(x)\right| \leq 2^{-n},$

and $$\sum 2^{-n}$$ converges (geometric series with ratio $$\frac{1}{2}$$; see §12, Problem 18). Thus, setting $$M_{n}=2^{-n}$$ in Theorem 3, we infer that the series $$\sum\left|\frac{1}{2} \sin x\right|^{n}$$ converges uniformly on $$E^{1},$$ as does $$\sum\left(\frac{1}{2} \sin x\right)^{n};$$ moreover,

$\sum_{n=1}^{\infty}\left|f_{n}\right| \leq \sum_{n=1}^{\infty} 2^{-n}=1.$

Theorem $$\PageIndex{4}$$ (necessary condition of convergence)

If $$\sum f_{m}$$ or $$\sum\left|f_{m}\right|$$ converges on $$B$$ (pointwise or uniformly), then $$\left|f_{m}\right| \rightarrow 0$$ on $$B$$ (in the same sense).

Thus a series cannot converge unless its general term tends to 0 (respectively, $$\overline{0})$$.

Proof

If $$\sum f_{m}=f,$$ say, then $$s_{m} \rightarrow f$$ and also $$s_{m-1} \rightarrow f .$$ Hence

$s_{m}-s_{m-1} \rightarrow f-f=\overline{0}.$

However, $$s_{m}-s_{m-1}=f_{m} .$$ Thus $$f_{m} \rightarrow \overline{0},$$ and $$\left|f_{m}\right| \rightarrow 0,$$ as claimed.

This holds for pointwise and uniform convergence alike (see Problem 14 in §12) . $$\quad \square$$

Caution: The condition $$\left|f_{m}\right| \rightarrow 0$$ is necessary but not sufficient. Indeed, there are divergent series with general term tending to $$0,$$ as we show next.

Examples (Continued)

(b) $$\sum_{n=1}^{\infty} \frac{1}{n}=+\infty$$ (the so-called harmonic series).

Indeed, by Note 1,

$\sum_{n=1}^{\infty} \frac{1}{n} \quad \text { exists }\left(\text {in } E^{*}\right),$

so Theorem 4 of §12 applies. We group the series as follows:

\begin{aligned} \sum \frac{1}{n} &=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\cdots+\frac{1}{16}\right)+\cdots \\ & \geq \frac{1}{2}+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\left(\frac{1}{16}+\cdots+\frac{1}{16}\right)+\cdots. \end{aligned}

Each bracketed expression now equals $$\frac{1}{2}.$$ Thus

$\sum \frac{1}{n} \geq \sum g_{m}, \quad g_{m}=\frac{1}{2}.$

As $$g_{m}$$ does not tend to $$0, \sum g_{m}$$ diverges, i.e., $$\sum_{m=1}^{\infty} g_{m}$$ is infinite, by Theorem 4. A fortiori, so is $$\sum_{n=1}^{\infty} \frac{1}{n}$$.

Theorem $$\PageIndex{5}$$ (root and ratio tests)

A series of constants $$\sum a_{n}\left(\left|a_{n}\right| \neq 0\right)$$ converges absolutely if

$\overline{\lim } \sqrt[n]{\left|a_{n}\right|}<1 \text { or } \overline{\lim }\left(\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}\right)<1.$

It diverges if

$\overline{\lim } \sqrt[n]{\left|a_{n}\right|}>1 \text { or } \underline{\lim } \left(\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}\right)>1.$

It may converge or diverge if

$\overline{\lim } \sqrt[n]{\left|a_{n}\right|}=1$

or if

$\underline{\lim } \left(\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}\right) \leq 1 \leq \overline{\lim }\left(\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}\right).$

(The $$a_{n}$$ may be scalars or vectors.)

Proof

If $$\overline{\lim } \sqrt[n]{\left|a_{n}\right|}<1,$$ choose $$r>0$$ such that

$\overline{\lim } \sqrt[n]{\left|a_{n}\right|}<r<1.$

Then by Corollary 2 of Chapter 2, §13, $$\sqrt[n]{\left|a_{n}\right|}<r$$ for all but finitely many $$n .$$ Thus, dropping a finite number of terms (§12, Problem 17), we may assume that

$\left|a_{n}\right|<r^{n} \text { for all } n.$

As $$0<r<1,$$ the geometric series $$\sum r^{n}$$ converges. Hence so does $$\sum\left|a_{n}\right|$$ by Theorem 2.

In the case

$\overline{\lim }\left(\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}\right)<1,$

we similarly obtain $$(\exists m)(\forall n \geq m)\left|a_{n+1}\right|<\left|a_{n}\right| r;$$ hence by induction,

$(\forall n \geq m) \quad\left|a_{n}\right| \leq\left|a_{m}\right| r^{n-m}. \quad \text { (Verify!) }$

Thus $$\sum\left|a_{n}\right|$$ converges, as before.

If $$\overline{\lim } \sqrt[n]{\left|a_{n}\right|}>1,$$ then by Corollary 2 of Chapter 2, §13,\left|a_{n}\right|>1\) for infinitely many $$n .$$ Hence $$\left|a_{n}\right|$$ cannot tend to $$0,$$ and so $$\sum a_{n}$$ diverges by Theorem 4.

Similarly, if

$\underline{\lim } \left(\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}\right)>1,$

then $$\left|a_{n+1}\right|>\left|a_{n}\right|$$ for all but finitely many $$n,$$ so $$\left|a_{n}\right|$$ cannot tend to 0 again. $$\quad \square$$

Note 3. We have

$\underline{\lim } \left(\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}\right) \leq \underline{\lim } \sqrt[n]{\left|a_{n}\right|} \leq \overline{\lim } \sqrt[n]{\left|a_{n}\right|} \leq \overline{\lim }\left(\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}\right).$

Thus

$\begin{array}{l}{\quad \overline{\lim }\left(\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}\right)<1 \text { implies } \overline{\lim } \sqrt[n]{\left|a_{n}\right|}<1 ; \text { and }} \\ {\qquad \underline{\lim } \left(\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}\right)>1 \text { implies } \overline{\lim } \sqrt[n]{\left|a_{n}\right|}>1.}\end{array}$

Hence whenever the ratio test indicates convergence or divergence, so certainly does the root test. On the other hand, there are cases where the root test yields a result while the ratio test does not. Thus the root test is stronger (but the ratio test is often easier to apply).

Examples (continued)

(c) Let $$a_{n}=2^{-k}$$ if $$n=2 k-1$$ (odd) and $$a_{n}=3^{-k}$$ if $$n=2 k$$ (even). Thus

$\sum a_{n}=\frac{1}{2^{1}}+\frac{1}{3^{1}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{2^{3}}+\frac{1}{3^{3}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\cdots.$

Here

$\underline{\lim } \left(\frac{a_{n+1}}{a_{n}}\right)=\lim _{k \rightarrow \infty} \frac{3^{-k}}{2^{-k}}=0 \text { and } \overline{\lim }\left(\frac{a_{n+1}}{a_{n}}\right)=\lim _{k \rightarrow \infty} \frac{2^{-k-1}}{3^{-k}}=+\infty,$

while

$\overline{\lim } \sqrt[n]{a_{n}}=\lim \sqrt[2n-1]{2^{-n}}=\frac{1}{\sqrt{2}}<1.\quad(\text {Verify!})$

Thus the ratio test fails, but the root test proves convergence.

Note 4. The assumption $$\left|a_{n}\right| \neq 0$$ is needed for the ratio test only.

III. Power Series. As an application, we now consider so-called power series,

$\sum a_{n}(x-p)^{n},$

where $$x, p, a_{n} \in E^{1}(C);$$ the $$a_{n}$$ may also be vectors.

Theorem $$\PageIndex{6}$$

For any power series $$\sum a_{n}(x-p)^{n},$$ there is a unique $$r \in E^{*}$$ $$(0 \leq r \leq+\infty),$$ called its convergence radius, such that the series converges absolutely for each $$x$$ with $$|x-p|<r$$ and does not converge (even conditionally) if $$|x-p|>r.$$

Specifically,

$r=\frac{1}{d}, \text { where } d=\overline{\lim } \sqrt[n]{\left|a_{n}\right|} \quad \text { (with } r=+\infty \text { if } d=0).$

Proof

Fix any $$x=x_{0}.$$ By Theorem 5, the series $$\sum a_{n}\left(x_{0}-p\right)^{n}$$ converges absolutely if $$\overline{\lim } \sqrt[n]{\left|a_{n}\right|}\left|x_{0}-p\right|<1,$$ i.e., if

$\left|x_{0}-p\right|<r \quad\left(r=\frac{1}{\lim \sqrt[n]{\left|a_{n}\right|}}=\frac{1}{d}\right),$

and diverges if $$\left|x_{0}-p\right|>r . \quad$$ (Here we assumed $$d \neq 0;$$ but if $$d=0,$$ the condition $$d\left|x_{0}-p\right|<1$$ is trivial for any$$x_{0},$$ so $$r=+\infty$$ in this case.) Thus $$r$$ is the required radius, and clearly there can be only one such $$r.$$ (Why?) $$\square$$

Note 5. If $$\lim _{n \rightarrow \infty} \frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}$$ exists, it equals $$\lim _{n \rightarrow \infty} \sqrt[n]{\left|a_{n}\right|},$$ by Note 3 (for $$\overline{lim}$$ and $$\underline{lim}$$ coincide here). In this case, one can use the ratio test to find

$d=\lim _{n \rightarrow \infty} \frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}$

and hence (if $$d \neq 0 )$$

$r=\frac{1}{d}=\lim _{n \rightarrow \infty} \frac{\left|a_{n}\right|}{\left|a_{n+1}\right|}.$

Theorem $$\PageIndex{7}$$

If a power series $$\sum a_{n}(x-p)^{n}$$ converges absolutely for some $$x=x_{0} \neq p,$$ then $$\sum\left|a_{n}(x-p)^{n}\right|$$ converges uniformly on the closed globe $$\overline{G}_{p}(\delta)$$ $$\delta=\left|x_{0}-p\right|.$$ So does $$\sum a_{n}(x-p)^{n}$$ if the range space is complete (Theorem 1).

Proof

Suppose $$\sum\left|a_{n}\left(x_{0}-p\right)^{n}\right|$$ converges. Let

$\delta=\left|x_{0}-p\right| \text { and } M_{n}=\left|a_{n}\right| \delta^{n};$

thus $$\sum M_{n}$$ converges.

Now if $$x \in \overline{G}_{p}(\delta),$$ then $$|x-p| \leq \delta,$$ so

$\left|a_{n}(x-p)^{n}\right| \leq\left|a_{n}\right| \delta^{n}=M_{n}.$

Hence by Theorem 3, $$\sum\left|a_{n}(x-p)^{n}\right|$$ converges uniformly on $$\overline{G}_{p}(\delta). \square$$

Examples (Continued)

(d) Consider $$\sum \frac{x^{n}}{n !}$$ Here

$p=0 \text { and } a_{n}=\frac{1}{n !}, \text { so } \frac{\left|a_{n}\right|}{\left|a_{n+1}\right|}=n+1 \rightarrow+\infty.$

By Note 5, then, $$r=+\infty ;$$ i.e., the series converges absolutely on all of $$E^{1} .$$ Hence by Theorem 7, it converges uniformly on any $$\overline{G}_{0}(\delta),$$ hence on any finite interval in $$E^{1}$$. (The pointwise convergence is on all of $$E^{1}$$.)