If
$\sum {a}_{n}$ is an absolutely convergent infinite series of complex numbers,
then it is a convergent infinite series.(Absolute convergence implies convergence.)
If
$\left\{{S}_{N}\right\}$ denotes the sequence of partial sums for
$\sum {a}_{n},$ and if
$\left\{{T}_{N}\right\}$ denotes the sequence of partial sums
for
$\sum |{a}_{n}|,$ then
$$|{S}_{M}-{S}_{N}|=|\sum _{n=N+1}^{M}{a}_{n}|\le \sum _{n=N+1}^{M}|{a}_{n}|=|{T}_{M}-{T}_{N}|$$
for all
$N$ and
$M.$ We are given that
$\left\{{T}_{N}\right\}$ is convergent and hence it is a Cauchy sequence.
So, by the inequality above,
$\left\{{S}_{N}\right\}$ must also be a Cauchy sequence.
(If
$|{T}_{N}-{T}_{M}|<\u03f5,$ then
$|{S}_{N}-{S}_{M}|<\u03f5$ as well.)
This implies that
$\sum {a}_{n}$ is convergent.
Let
$z$ be a complex number, and define a sequence
$\left\{{a}_{n}\right\}$ by
${a}_{n}={z}^{n}.$ Consider the infinite series
$\sum {a}_{n}.$ Show that
${\sum}_{n=0}^{\infty}{a}_{n}$ converges to a number
$S$ if and only if
$\left|z\right|<1.$ Show in fact that
$S=1/(1-z),$ when
$\left|z\right|<1.$
HINT: Evaluate explicitly the partial sums
${S}_{N},$ and then take their limit.
Show that
${S}_{N}=\frac{1-{z}^{N+1}}{1-z}.$
- Show that
${\sum}_{n=1}^{\infty}\frac{1}{n(n+1)}$ converges to
$1,$ by computing explicit formulas for the partial sums.
HINT: Use a partial fraction decomposition for the
${a}_{n}$ 's.
- (The Harmonic Series.) Show that
${\sum}_{n=1}^{\infty}1/n$ diverges by verifying that
${S}_{{2}^{k}}>k/2.$ HINT: Group the terms in the sum as follows,
$$1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{16})+...,$$
and then estimate the sum of each group.
Remember this example as an infinite series that diverges, despite thefact that is terms tend to 0.
The next theorem is the most important one we have concerning infinite series of numbers.
Comparison test
Suppose
$\left\{{a}_{n}\right\}$ and
$\left\{{b}_{n}\right\}$ are two sequences
of nonnegative real numbers for which there exists a positive integer
$M$ and a constant
$C$ such that
${b}_{n}\le C{a}_{n}$ for all
$n\ge M.$ If the infinite series
$\sum {a}_{n}$ converges,
so must the infinite series
$\sum {b}_{n}.$
We will show that the sequence
$\left\{{T}_{N}\right\}$ of partial sums
of the infinite series
$\sum {b}_{n}$ is a bounded sequence.
Then, by
[link] , the infinite series
$\sum {b}_{n}$ must be summable.
Write
${S}_{N}$ for the
$N$ th partial sum of the
convergent infinite series
$\sum {a}_{n}.$ Because this series is summable, its sequence of partial sums is a bounded sequence.
Let B be a number such that
${S}_{N}\le B$ for all
$N.$ We have for all
$N>M$ that
$$\begin{array}{ccc}\hfill rcl{T}_{N}& =& \sum _{n=1}^{N}{b}_{n}\hfill \\ & =& \sum _{n=1}^{M}{b}_{n}+\sum _{n=M+1}^{N}{b}_{n}\hfill \\ & \le & \sum _{n=1}^{M}{b}_{n}+\sum _{n=M+1}^{N}C{a}_{n}\hfill \\ & =& \sum _{n=1}^{M}{b}_{n}+C\sum _{n=M+1}^{N}{a}_{n}\hfill \\ & \le & \sum _{n=1}^{M}{b}_{n}+C\sum _{n=1}^{N}{a}_{n}\hfill \\ & \le & \sum _{n=1}^{M}{b}_{n}+C{S}_{N}\hfill \\ & \le & \sum _{n=1}^{M}{b}_{n}+CB,\hfill \end{array}$$
which completes the proof, since this final quantity is a fixed constant.
- Let
$\left\{{a}_{n}\right\}$ and
$\left\{{b}_{n}\right\}$ be as in the preceding theorem.
Show that if
$\sum {b}_{n}$ diverges, then
$\sum {a}_{n}$ also must diverge.
- Show by example that the hypothesis that the
${a}_{n}$ 's and
${b}_{n}$ 's of the Comparison Test are nonnegative
can not be dropped.
Let
$\left\{{a}_{n}\right\}$ be a sequence of positive numbers.
- If
$lim\; sup{a}_{n+1}/{a}_{n}<1,$ show that
$\sum {a}_{n}$ converges.
HINT: If
$lim\; sup{a}_{n+1}/{a}_{n}=\alpha <1,$ let
$\beta $ be a number for which
$\alpha <\beta <1.$ Using part (a) of
[link] , show that there exists an
$N$ such that for all
$n>N$ we must have
${a}_{n+1}/{a}_{n}<\beta ,$ or equivalently
${a}_{n+1}<\beta {a}_{n},$ and therefore
${a}_{N+k}<{\beta}^{k}{a}_{N}.$ Now use the comparison test with the geometric series
$\sum {\beta}^{k}.$
- If
$lim\; inf{a}_{n+1}/{a}_{n}>1,$ show that
$\sum {a}_{n}$ diverges.
- As special cases of parts (a) and (b), show that
$\left\{{a}_{n}\right\}$ converges if
${lim}_{n}{a}_{n+1}/{a}_{n}<1,$ and diverges if
${lim}_{n}{a}_{n+1}/{a}_{n}>1.$
- Find two examples of infinite series'
$\sum {a}_{n}$ of positive numbers, such that
$lim{a}_{n+1}/{a}_{n}=1$ for both examples, and such that
one infinite series converges and the other diverges.
- Derive the Root Test:
If
$\left\{{a}_{n}\right\}$ is a sequence of positive numbers for which
$lim\; sup{a}_{n}^{1/n}<1,$ then
$\sum {a}_{n}$ converges.
And, if
$lim\; inf{a}_{n}^{1/n}>1,$ then
$\sum {a}_{n}$ diverges.
- Let
$r$ be a positive integer.
Show that
$\sum 1/{n}^{r}$ converges if and only if
$r\ge 2.$ HINT: Use
[link] and the Comparison Test for
$r=2.$
- Show that the following infinite series are summable.
$$\sum 1/({n}^{2}+1),\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\sum n/{2}^{n},\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\sum {a}^{n}/n!,$$
for
$a$ any complex number.