7.6 Geometric Series

7.63   Theorem ( $${ \left\{r^{{1\over n}}\right\}\to 1}$$.)If $r\in\mbox{${\mbox{{\bf R}}}^{+}$}$, then $${\{r^{1\over n}\}\to 1}$$.

Proof: $\;$

Case 1:

[$r \geq 1$]. By the formula for factoring $s^n - a^n$ (3.78), we have for all $n\in\mbox{{\bf Z}}_{\geq 1}$ and all $s \geq 1$

$$(s^n-1)=(s-1)\sum_{j=0}^{n-1}s^j\geq(s-1)\sum_{j=0}^{n-1} 1^j =n(s-1)$$

so

$$(s-1)\leq{1\over n}(s^n-1).$$

If we let $s=r^{{1\over n}}$ in this formula, we get

$$\vert r^{1\over n}-1\vert=r^{1\over n}-1\leq{1\over n}(r-1).$$

Since $${\left\{ {{r-1}\over n}\right\}}$$ is a null sequence, it follows from the comparison theorem for null sequences that $\{r^{1/n}-1\}\to 0$; i.e., $${ \left\{r^{{1\over n}}\right\}\to 1}$$.

Case 2:

[$0<r<1$.] Let $${R={1\over r}}$$. Then $R>1$, so by Case 1, $${\left\{R^{{1\over n}}\right\}\to 1}$$. By the reciprocal theorem $${\left\{ {1\over {R^{1\over n}}}\right\}\to 1}$$; i.e., $${ \left\{r^{{1\over n}}\right\}\to 1}$$.

We have shown that the theorem holds in all cases. $\mid\!\mid\!\mid$

7.64   Theorem (Convergence of geometric sequences.) Let $\alpha\in\mbox{{\bf C}}$. Then

$$\begin{eqnarray*} &\;&\{\alpha^n\}\to 0 \mbox{ if } \vert\alpha\vert<1 \\ &\;&\... ... diverges if } \vert\alpha\vert\geq 1 \mbox{ and } \alpha\neq 1. \end{eqnarray*}$$

Proof: The last assertion was shown in theorem 7.7, and the second statement is clear, and it is also clear that $\{\alpha^n\}\to 0$ if $\alpha=0$.

Suppose that $0<\vert\alpha\vert<1$. I will show that

$$\vert\alpha^k\vert \leq {1\over 2} \mbox{ for some } k \in\mbox{{\bf N}}.$$ (7.65)

It will then follow that

$$\vert\alpha^n\vert^k = (\vert\alpha\vert^k)^n \leq \left({1\o... ...}\right)^n = {1\over 2^n} \mbox{ for all }n \in \mbox{{\bf N}}.$$

Since $\{{1\over 2^n}\}$ is a null sequence, it follows from the comparison theorem for null sequences that $\{ \vert\alpha^n\vert^k\}$ is a null sequence, and then by the root theorem for null sequences (Theorem 7.19), it follows that $\{\alpha^n\}$ is a null sequence.

To prove (7.65), let $N$ be a precision function for $${\left\{\Big({1\over 2}\Big)^{1\over n}-1 \right\}}$$, and let $k = N(1-\vert\alpha\vert)$. Then $${\left\vert\Big( {1\over 2}\Big)^{1\over k} - 1 \right\vert < 1 - \vert\alpha\vert}$$, so $${1 - \Big({1\over 2}\Big) ^{1\over k} < 1 - \vert\alpha\vert}$$, so
$${\vert\alpha\vert < \Big( {1\over 2}\Big)^{1\over k}}$$. and hence $\vert\alpha^k\vert\leq {1\over 2}$, which is what we wanted to show. $\mid\!\mid\!\mid$

7.66   Theorem (Geometric series.) Let $\alpha\in\mbox{{\bf C}}$. If $\vert\alpha\vert<1$, then the geometric series

$$g_\alpha\colon n\mapsto \sum_{j=0}^n\alpha^j$$

converges to $${ {1\over {1-\alpha}}}$$. If $\vert\alpha\vert\geq 1$, then $g_{\alpha}$ diverges.

Proof: We saw in theorem 3.71 that $${ g_{\alpha}(n)=\sum_{j=0}^n\alpha^j={{1-\alpha^{n+1}}\over {1-\alpha}}}$$ for all $\alpha\neq 1$. If $\alpha=1$, $g_\alpha (n)=n+1$. This sequence diverges, since it is not bounded. If $\vert\alpha\vert<1$, then by the previous theorem $\{\alpha^n\}\to 0$, so

$$\{g_{\alpha}(n)\}=\left\{ {1\over {1-\alpha}}-{\alpha\over {1... ...1-\alpha}}-{\alpha\over {1-\alpha}}\cdot 0={1\over {1-\alpha}}.$$

Suppose now $\vert\alpha\vert\geq 1$ and $\alpha\neq 1$. Then for all $n\in\mbox{{\bf N}}$ we have

$$\begin{eqnarray*}\alpha^n &=& {1\over \alpha} \cdot \alpha^{n+1} = {1\over \alph... ... \alpha} \right)\\ &=& {1 - (1-\alpha)g_\alpha(n) \over \alpha}. \end{eqnarray*}$$

Hence for all $L\in\mbox{{\bf C}}$ we have

$$\{g_\alpha(n)\} \to L \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\{\alpha^n\} \to {1-(1-\alpha)L\over \alpha}.$$

By theorem 7.7, if $\vert\alpha\vert\geq 1$ and $\alpha\neq 1$, then $\{\alpha^n\}$ diverges, and hence $\{g_\alpha (n)\}\to L$ is false for all $L\in\mbox{{\bf C}}$; i.e., $g_\alpha$ diverges. $\mid\!\mid\!\mid$

7.67   Notation. If $\{a_j\}_{j\geq 1}$ is a sequence of digits, then we denote $${\sum_{j=1}^n{{a_j}\over {10^j}}}$$ by $.a_1a_2\cdots a_n$. Thus

$$.14159={1\over {10}}+{4\over {10^2}}+{1\over {10^3}}+{5\over {10^4}}+{9\over {10^5}}$$

and

$$\begin{eqnarray*} \lefteqn{.351351351}\\ &=&\left({3\over {10}}+{5\over {100}}+... ...ght] \\ &=& {{351}\over {1000}}\sum_{j=0}^2 {1\over {10^{3j}}}. \end{eqnarray*}$$

7.68   Example. Let $a,b,c$ be digits, and let

$$\{x_n\}=\left\{ {{abc}\over {1000}}\sum_{j=0}^n {1\over {10^{3j}}}\right\}$$

so informally, $${ x_n=\underbrace{.abcabc\cdots abc}_{3(n+1)\mbox{ digits }}}$$. Then $\{x_n\}$ is a convergent sequence, and

$$\{x_n\}\to{{abc}\over {1000}}\cdot {1\over {1-{1\over {1000}}}}={{abc}\over {999}}.$$

As an example, we have

$$\{.351,.351351,.351351351,\cdots\}\to{{351}\over {999}}={{39}\over {111}}={{13}\over {37}}.$$

7.69   Exercise. A Let

$$\{a_n\}=\{.672,.67272,.6727272,.672727272,\cdots\}_{n\geq 1}.$$

Show that $\{a_n\}$ converges to a rational number.

7.70   Exercise. A

a)

Let $${\{a_n\}=\left\{ \sum_{j=0}^n \left(\left({3\over 5}\right)^j+\left({4\over 5}\right)^ji\right)\right\}}$$. Does $\{a_n\}$ converge? If it does, find $\lim\{a_n\}$ in the form $a+bi$ where $a,b\in\mbox{{\bf R}}$.

b)

Let $${\{b_n\}=\left\{\sum_{j=0}^n\left( {{3+4i}\over 5}\right)^j\right\}}$$. Does $\{b_n\}$ converge? If it does, find $\lim\{b_n\}$ in the form $a+bi$ where $a,b\in\mbox{{\bf R}}.$

c)

Let $${\{c_n\}=\left\{\sum_{j=0}^n\left(\left({3\over 5}\right)^j +\left({{4i}\over 5}\right)^j\right)\right\}}$$. Does $\{c_n\}$ converge? If it does, find $\lim\{c_n\}$ in the form $a+bi$ where $a,b\in\mbox{{\bf R}}$.

7.71   Exercise. Show that the sequences $${\left\{\sum_{j=0}^n\left({{1+i}\over 2}\right)^j\right\}}$$ and $${\left\{\sum_{j=0}^n\left({{2+i}\over 3}\right)^j\right\}}$$ (which are drawn above in figure b 7.1) converge, and that the limits appear to be in agreement with Figure b above.

7.72   Entertainment (Snowflakes) Let $E$ be an equilateral triangle with

Snowflakes

area $A$, and side $s$. Note that an equilateral triangle with side $${{s\over 3}}$$ has area $${{A\over 9}}$$. Starting with $E$, we will now construct a sequence $\{S_n\}$ of polygons. $S_n$ will have $4^n\cdot 3$ sides, all having length $${{s\over {3^n}}}$$. We let $S_0=E$ (so $S_0$ has $4^0\cdot 3$ sides of length $${{s\over {3^0}}}$$). To construct $S_{n+1}$ from $S_n$ we attach an equilateral triangle with side of length $${{1\over 3}\cdot\mbox{ side }(S_n)}$$ to the middle third of each side of $S_n$.

A horizontal side of $S_n$ might be replaced by . Each side of $S_n$ is replaced by 4 sides of length $${{1\over 3}\left({s\over {3^n}}\right)}$$, so $S_{n+1}$ will have $4\cdot(4^n3)=4^{n+1}\cdot 3$ sides of length $${ {s\over {3^{n+1}}}}$$. The figure shows some of these polygons. I will call the polygons $S_n$ snowflake polygons. We have $S_n\subset S_{n+1}$ for all $n$. The snowflake $S$ is the union of all of the sets $S_n$; i.e., a point $x$ is in $S$ if and only if it is in $S_n$ for some $n\in\mbox{{\bf N}}$.

Find the area of $S_n$ (in terms of the area $A$ of $E$), for example

$$\mbox{\rm area}{(S_1)=A+3\left({A\over 9}\right)={4\over 3}A}.$$

Then find the area of $S$ in terms of $A$. Make any reasonable assumptions that you need. What is the perimeter of $S$?