6.6 Geometric Series

6.73   Theorem (Geometric series) Let $r$ be a real number such that $\vert r\vert <1$. Then

$$\{ \sum_{i=1}^n r^{i-1}\} \to {1 \over 1-r}$$ (6.74)

Equation (6.74) is often written in the form

$$\sum_{i=1}^\infty r^{i-1} = {1 \over 1-r} \mbox{ or } \sum_{i=0}^\infty r^i = {1 \over 1-r}$$

Proof:    Let $r$ be a real number such that $\vert r\vert <1$, and for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ let

$$f(n) = \sum_{i=1}^n r^{i-1}.$$

Then by theorem 2.22 we have

$$f(n) = \frac{1-r^n}{1-r},$$

and hence

$$\begin{eqnarray*} \lim\{f(n)\} &=& \lim \left\{ \frac{1-r^n}{1-r} \right\} \\ ... ...\ &=& {1\over 1-r} \left(1 - \lim \left\{ r^n \right\} \right) \end{eqnarray*}$$

Hence by the $n$th power theorem

$$\lim \{ f(n) \} = {1\over 1-r }(1-0) = {1\over 1-r}.\mbox{ $\diamondsuit$}$$

6.75   Exercise. Find the error in the following argument. Let $R$ be a real number with $R\neq 1$, and for $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$, let

$$a_n=1+R+R^2+\cdots +R^{n-1}.$$

Let $L=\lim\{a_n\}$. Then, by the translation rule

$$\begin{eqnarray*} L=\lim\{a_{n+1}\} &=& \lim\{1+R+\cdots +R^n\} \\ &=& \lim\{1+R(1+\cdots +R^{n-1})\}=\lim\{1+Ra_n\}. \end{eqnarray*}$$

Thus by the sum rule and product rule,

$$\begin{eqnarray*} L&=&\lim\{1\}+\lim\{Ra_n\} \\ &=&1 + R\lim\{a_n\} =1+RL. \end{eqnarray*}$$

Now

$$L=1+RL\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}L(1-R)=... ...ox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}L= {1\over {1-R}}.$$

Hence we have shown that

$$\lim\{1+R+R^2+\cdots +R^{n-1}\}={1\over {1-R}}$$

for all $R\in\mbox{{\bf R}}\setminus \{1\}$. (This sort of argument, and the conclusion were regarded as correct in the eighteenth century. At that time the argument perhaps was correct, because the definitions in use were not the same as ours.)

``The clearest early account of the summation of geometric series''[6, page 136] was given by Grégoire de Saint-Vincent in 1647. Grégoire's argument is roughly as follows:

On the line $AZ$ mark off points $B'$, $C'$, $D'$ etc. such that

$$AB' = 1,\;\; B'C' = r,\;\; C'D' = r^2,\;\; D'E' = r^3\cdots$$

On a different line through $A$ mark off points $O$, $B$, $C$, $D$ etc. such that

$$OA = 1,\;\; OB = r,\;\; OC = r^2,\;\;OD = r^3 \cdots$$

Then

$$\begin{eqnarray*} {AB' \over AB} &=& {1 \over 1-r}.\\ {B'C'\over BC} &=& {r \ov... ... \over CD} &=& {r^2 \over r^2 - r^3} = {1\over 1-r}.\\ & etc. & \end{eqnarray*}$$

Now I use the fact that

$${a \over b} = {c \over d} \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}{a+c \over b+d} = {a \over b},$$ (6.76)

(see exercise 6.78), to say that

$$\begin{eqnarray*} {AC'\over AC} &=& {AB'+B'C' \over AB + BC} = {AB' \over AB} = ... ...'+D'E' \over AD + DE} = {AD' \over AD} = {1\over 1-r}\\ & etc.& \end{eqnarray*}$$

It follows that the triangles $BAB'$, $CAC'$, $DAD'$, etc. are all mutually similar, so the lines $BB'$, $CC'$, $DD'$ etc. are all parallel. Draw a line through $O$ parallel to $BB'$ and intersecting $AZ$ at $X$. I claim that

$$AB' + B'C' + C'D' + D'E' + etc. = AX.$$ (6.77)

It is clear that any finite sum is smaller that $AX$, and by taking enough terms in the sequence $A,B,\cdots N$ we can make $ON$ arbitrarily small. Then $XN'$ is arbitrarily small, i.e. the finite sums $AN'$ can be made as close to $AX$ as we please. By similar triangles,

$${1\over 1-r} = {AB' \over AB} = { AX \over AO} = {AX \over 1}$$

so, equation (6.77) says

$$1 + r + r^2 + r^3 + \cdots = {1\over 1-r}.$$

6.78   Exercise. Prove the assertion (6.76).

6.79   Exercise.

a)

Find $${\lim \left\{ 1 + \Big({9\over 10}\Big) + \Big({9\over 10}\Big)^2 + \Big({9\over 10}\Big)^3+ \cdots + \Big({9\over 10}\Big)^{n-1} \right\} }$$.

b)

Find $${\lim \left\{ 1 - \Big({9\over 10}\Big) + \Big({9\over 10}\Big)^2 - \Big({9\over 10}\Big)^3+ \cdots + \Big(-{9\over 10}\Big)^{n-1} \right\} }$$.

c)

For each $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ let

$$a_n = \sum_{j=0}^{\infty} \Big(-{n\over n+1}\Big)^j,$$

(in part (b) you calculated $a_9$). Find a formula for $a_n$, and then find $\lim \{ a_n\}$.

d)

Show that

$$\lim \left\{ \sum_{j=0}^\infty \Big( -{n\over n+1} \Big)^j \... ...{j=0}^\infty \lim \left\{ \Big( -{n\over n+1}\Big)^j \right\}$$ (6.80)

(Thus it is not necessarily true that the limit of an infinite sum is the infinite sum of the limits. The left side of (6.80) was calculated in part c. The right side is $\sum_{j=0}^\infty b_j$, where $b_j = \lim\left\{\Big(-{n\over n+1}\Big)^j\right\}$ depends on $j$, but not on $n$.)