2.4 Finite Geometric Series

For each $n$ in $\mbox{{\bf Z}}^+$ let $B_n$ denote the box

$$B_n=B\left( {1\over {2^{n-1}}}, {2\over {2^{n-1}}}\colon 0, {1\over {2^{n-1}}}\right),$$

and let

$$S_n=B_1\cup B_2 \cup\cdots\cup B_n=\bigcup_{j=1}^n B_j.$$

I want to find the area of $S_n$. I have

$$\mbox{\rm area}(B_n)=\left( {2\over {2^{n-1}}}-{1\over {2^{n-... ...={1\over {2^{n-1}}}\cdot {1\over {2^{n-1}}}={1\over {4^{n-1}}}.$$

Since the boxes $B_i$ intersect only along their boundaries, we have

$$\mbox{\rm area}(S_n) = \mbox{\rm area}(B_1)+\mbox{\rm area}(B_2)+\cdots +\mbox{\rm area}(B_n) \mbox{{}}$$

$$= 1 +{1\over 4}+\cdots +{1\over {4^{n-1}}}.$$ (2.20)

Thus

$$\mbox{\rm area}(S_1) = 1,$$

$$\mbox{\rm area}(S_2) = 1+{1\over 4}={5\over 4},$$

$$\mbox{\rm area}(S_3) = {5\over 4}+{1\over 16}={{20}\over {16}}+{1\over {16}}={{21}\over {16}}={{21}\over {4^2}},$$

$$\mbox{\rm area}(S_4) = {{21}\over {16}}+{1\over {64}}={{84}\over {64}}+{1\over {64}}={{85}\over {64}}={{85}\over {4^3}}.$$ (2.21)

You probably do not see any pattern in the numerators of these fractions, but in fact $\mbox{\rm area}(S_n)$ is given by a simple formula, which we will now derive.

2.22   Theorem (Finite Geometric Series.) Let $r$ be a real number such that $r\neq 1$. Then for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$

$$1+r+r^2+\cdots + r^{n-1}={{1-r^n}\over {1-r}}.$$ (2.23)

Proof: Let

$$S=1+r+r^2+\cdots +r^{n-1}.$$

Then

$$rS=r+r^2+\cdots +r^{n-1}+r^n.$$

Subtract the second equation from the first to get

$$S(1-r)=1-r^n,$$

and thus

$$S={{1-r^n}\over {1-r}}.$$

Remark: Theorem 2.22 is very important, and you should remember it. Some people find it easier to remember the proof than to remember the formula. It would be good to remember both.

If we let ${r={1\over 4}}$ in (2.23), then from equation (2.20) we obtain

$$\mbox{\rm area}(S_n) = 1+{1\over 4}+\cdots +{1\over {4^{n-1}}}$$

$$= {{1-{1\over {4^n}}}\over {1-{1\over 4}}} = {4\over 3}\left(1-{1\over {4^n}}\right)$$ (2.24)

$$= \frac{4^n-1}{3\cdot 4^{n-1}}.$$

As a special case, we have

$$\mbox{\rm area}(S_4)=\frac{4^4-1}{3\cdot 4^3}=\frac{256-1}{3\cdot 4^3} =\frac{255} {3\cdot 4^3}=\frac{85}{4^3}$$

which agrees with equation (2.21).

2.25   Entertainment (Pine Tree Problem.) Let $T$ be the subset of $\mbox{{\bf R}}^2$ sketched below:

Here $P = (0,4)$, $B_0 = (1,0)$, $A_1 = (2,0)$, and $B_1$ is the point where the line $B_0P$ intersects the line $y=1$. All of the points $A_j$ lie on the line $PA_1$, and all of the points $B_j$ lie on the line $PB_0$. All of the segments $A_iB_{i-1}$ are horizontal, and all segments $A_jB_j$ are parallel to $A_1B_1$. Show that the area of $T$ is ${44\over 7}$. You will probably need to use the formula for a geometric series.

2.26   Exercise.     
(a) Find the number

$$1+{1\over 7}+{1\over {7^2}}+{1\over {7^3}}+\cdots +{1\over {7^{100}}}$$

accurate to 8 decimal places.

(b) Find the number

$$1+{1\over 7}+{1\over {7^2}}+{1\over {7^3}}+\cdots +{1\over {7^{1000}}}$$

accurate to 8 decimal places.

(You may use a calculator, but you can probably do this without using a calculator.)

2.27   Exercise. A Let

$$\begin{eqnarray*} a_1 &=& .027\\ a_2 &=& .027027\\ a_3 &=& .027027027\\ & etc. & \end{eqnarray*}$$

Use the formula for a finite geometric series to get a simple formula for $a_n$. What rational number should the infinite decimal $.027027027\cdots$ represent? Note that

$$a_3 = .027(1.001001) = .027(1 + {1 \over 1000} + {1 \over 1000^2}).$$

The Babylonians[45, page 77] knew that

$$1 + 2 + 2^2 + 2^3 + \cdots + 2^n = 2^n + (2^n - 1),$$ (2.28)

i.e. they knew the formula for a finite geometric series when $r=2$.

Euclid knew a version of the formula for a finite geometric series in the case where $r$ is a positive integer.

Archimedes knew the sum of the finite geometric series when $r={1\over 4}$. The idea of Archimedes' proof is illustrated in the figure.

If the large square has side equal to $2$, then

$$\begin{array}{rll} A& & = A \hspace{1em} = 3\\ \frac{1}{4}A... ...[1ex] (\frac{1}{4})^3 A & = \frac{1}{4} C & = D.\\ \end{array}$$

Hence

$$\begin{eqnarray*}{\textstyle (1 + \frac{1}{4} + (\frac{1}{4})^2 + (\frac{1}{4})^... ...({1\over 8})^2 = 4 - (\frac{1}{4})^3 = 4( 1 - ({1\over 4})^4). } \end{eqnarray*}$$

i.e.

$$\textstyle{ (1 + \frac{1}{4} + (\frac{1}{4})^2 +(\frac{1}{4})^3)\cdot 3 = 4(1 - (\frac{1}{4})^4).}$$

For the details of Archimedes' argument see [2, pages 249-250].

2.29   Exercise. Explain why formula (2.28) is a special case of the formula for a finite geometric series.