11.1 Infinite Series

11.1   Definition (Series operator.) If $f$ is a complex sequence, we define a new sequence $\sum f$ by

$$(\sum f)(n)=\sum_{j=0}^nf(j)\mbox{ for all } n\in\mbox{{\bf N}}$$

or

$$\sum\{f(n)\}=\{\sum_{j=0}^nf(j)\}\mbox{ for all } n\in\mbox{{\bf N}}.$$

We use variations, such as

$$\sum\{f(n)\}_{n\geq 1}=\{\sum_{j=1}^nf(j)\}_{n\geq 1}.$$

$\sum$ is actually a function that maps complex sequences to complex sequences. We call $\sum f$ the series corresponding to $f$.

11.2   Remark. If $f,g$ are complex sequences and $c\in\mbox{{\bf C}}$, then

$$\sum(f+g)=\sum f+\sum g$$

and

$$\sum(cf)=c(\sum f),$$

since for all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} \left(\sum(f+g)\right)(n)&=&\sum_{j=0}^n(f+g)(j)=\sum_{j=0}^nf... ...um_{j=0}^ng(j)=(\sum f)(n)+(\sum g)(n) \\ &=&(\sum f+\sum g)(n) \end{eqnarray*}$$

and

$$\begin{eqnarray*} \left(\sum(cf)\right)(n)&=&\sum_{j=0}^n(cf)(j)=\sum_{j=0}^nc\c... ...0}^nf(j)\\ &=&c\cdot\left((\sum f)(n)\right)=(c\cdot\sum f)(n). \end{eqnarray*}$$

11.3   Examples. If $\{r^n\}$ is a geometric sequence, then $\sum\{r^n\}=\{\sum_{j=0}^nr^j\}$ is a sequence we have been calling a geometric series. If $${\{c_n(t)\}=\left\{{{t^{2n}(-1)^n}\over {(2n)!}}\right\}}$$, then $\sum\{c_n(t)\}=\{C_n(t)\}$ is the sequence for $\cos(t)$ that we studied in the last chapter.

11.4   Definition (Summable sequence.) A complex sequence $\{a_n\}$ is summable if and only if the series $\sum\{a_n\}$ is convergent. If $\{a_n\}$ is summable, we denote $\lim(\sum\{a_n\})$ by $${\sum_{n=0}^\infty a_n}$$. We call $${\sum_{n=0}^\infty a_n}$$ the sum of the series $\sum\{a_n\}$.

11.5   Example. If $r\in\mbox{{\bf C}}$ and $\vert r\vert<1$, then $${\sum_{n=0}^\infty r^n=\lim\{\sum_{j=0}^n r^j\}={1\over {1-r}}}$$.

11.6   Example (Harmonic series.) The series

$$\sum\left\{{1\over n}\right\}_{n\geq 1}=\left\{\sum_{j=1}^n{1\over j}\right\}_{n\geq 1}$$

is called the harmonic series, and is denoted by $\{H_n\}_{n\geq 1}$. Thus

$$H_n=\sum_{j=1}^n{1\over j}.$$

We will show that $\{H_n\}_{n\geq 1}$ diverges; i.e., the sequence $${ \left\{ {1\over n}\right\}_{n\geq 1}}$$ is not summable. For all $n\geq 1$, we have

$$\begin{eqnarray*} H_{2n}&=&\phantom{((+}1 \phantom{ {1\over 2}))}+{1\over 2} +{1... ...eft(1+{1\over 2}+\cdots+{1\over n}\right) \\ &=&{1\over 2}+H_n. \end{eqnarray*}$$

From the relation $${H_{2n}\geq{1\over 2}+H_n}$$, we have

$$\begin{eqnarray*} H_2&\geq&{1\over 2}+H_1={1\over 2}+1\\ H_4&\geq&{1\over 2}+H_2\geq{2\over 2}+1\\ H_8&\geq&{1\over 2}+H_4\geq{3\over 2}+1 \end{eqnarray*}$$

and (by induction),

$$H_{2^n}\geq{n\over 2}+1\mbox{ for all }n\in\mbox{{\bf Z}}_{\geq 1}.$$

Hence, $\{H_n\}_{n\geq 1}$ is not bounded, and thus $\{H_n\}$ diverges; i.e., $${ \left\{ {1\over n}\right\}_{n\geq 1}}$$ is not summable.

11.7   Theorem (Sum theorem for series.) Let $f,g$ be summable sequences and let $c\in\mbox{{\bf C}}$. Then $f+g$ and $cf$ are summable, and

$$\begin{eqnarray*} \sum_{n=0}^\infty(f+g)(n)&=&\sum_{n=0}^\infty f(n)+\sum_{n=0}^\infty g(n)\\ \sum_{n=0}^\infty cf(n)&=&c\sum_{n=0}^\infty f(n). \end{eqnarray*}$$

If $f$ is not summable, and $c\neq 0$, then $cf$ is not summable.

Proof: The proof is left to you.

11.8   Exercise. Let $f,g$ be summable sequences. Show that $f+g$ is summable and that

$$\sum_{j=0}^\infty(f+g)(j)=\sum_{j=0}^\infty f(j)+\sum_{j=0}^\infty g(j).$$

11.9   Example. The product of two summable sequences is not necessarily summable. If

$$f=\left\{1,-1,\sqrt{{1\over 2}},-\sqrt{{1\over 2}}, \sqrt{{1\... ...},\sqrt{{1\over 4}},-\sqrt{{1\over 4}},\cdots\right\}_{n\geq 1}$$

then

$$\sum f=\left\{1,0,\sqrt{{1\over 2}},0,\sqrt{{1\over 3}},0,\sqrt{{1\over 4}}, 0,\cdots\right\}_{n\geq 1}.$$

This is a null sequence, so $f$ is summable and $${\sum_{n=1}^\infty f(n)=0}$$. However,

$$\displaystyle {f^2=\left\{1,1,{1\over 2},{1\over 2},{1\over 3},{1\over 3},{1\over 4},{1\over 4},\cdots\right\}}_{n\geq 1},$$

so $${\left(\sum(f^2)\right)(2n)=2\sum_{j=1}^n{1\over j}=2H_n}$$. Thus $\sum(f^2)$ is unbounded and hence $f^2$ is not summable.

11.10   Theorem. Every summable sequence is a null sequence. [The converse is not true. The harmonic series provides a counterexample.]

Proof: Let $f$ be a summable sequence. Then $${\{\sum_{j=0}^nf(j)\}}$$ converges to a limit $L$, and by the translation theorem $${\{\sum_{j=0}^{n+1}f(j)\}\to L}$$ also. Hence

$$\{\sum_{j=0}^{n+1}f(j)\}-\{\sum_{j=0}^nf(j)\}\to L-L=0;$$

i.e.,

$$\{f(n+1)\}\to 0$$

and it follows that $f$ is a null sequence. $\mid\!\mid\!\mid$