Geometric Series (College Board AP® Calculus BC): Revision Note

Take note of the lower limit of a summation.

If it starts with 1 rather than 0, then you can rewrite it so that it starts at 0.

For example, $\sum _{n=1}^{\infty }\left(\frac{3}{5^n}\right)=\sum _{n=0}^{\infty }\frac{3}{5}·\left(\frac{1}{5^n}\right)$.

However, you only need to know the first term and the common ratio in order to use the formula. In the above example, $a=\frac{3}{5}$ and $r=\frac{1}{5}$.

Consider the infinite series $\sum _{n=1}^{\infty }\frac{1}{2^n}$. Show that the series converges and determine the value of the sum.

Answer:

Rewrite the series so that it has the standard form of a geometric series

$\begin{array}{rcl}\sum _{n=1}^{\infty }\frac{1}{2^n}& =& \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\\ & =& \frac{1}{2}·{\left(\frac{1}{2}\right)}^0+\frac{1}{2}·{\left(\frac{1}{2}\right)}^1+\frac{1}{2}·{\left(\frac{1}{2}\right)}^2+..\\ & =& \sum _{n=0}^{\infty }\frac{1}{2}·{\left(\frac{1}{2}\right)}^n\end{array}$

Note that we've changed the sum to start with $n=0$

That is a geometric series with $a=\frac{1}{2}$ and $r=\frac{1}{2}$

Show that it satisfies the convergence condition

That is a geometric series with $\left|r\right|=\left|\frac{1}{2}\right|=\frac{1}{2}<1$ so it converges

Use the formula $\sum _{n=0}^{\infty }a{r}^n=\frac{a}{1-r}$

So $\sum _{n=1}^{\infty }\frac{1}{2^n}=\sum _{n=0}^{\infty }\frac{1}{2}·{\left(\frac{1}{2}\right)}^n=\frac{{\displaystyle \frac{1}{2}}}{1-{\displaystyle \frac{1}{2}}}=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}}=1$

$\sum _{n=1}^{\infty }\frac{1}{2^n}=1$

Consider the number $0.\dot{6}2\dot{1}=0.621621621621...$.

(a) Write the number in the form of a geometric series.

(b) Use the result from part (a) to find the value of $0.\dot{6}2\dot{1}$ as a fraction in lowest terms.

Answer:

(a)

Use the fact that $0.\dot{6}2\dot{1}=0.621+0.000621+0.000000621+...$

$\begin{array}{rcl}0.\dot{6}2\dot{1}& =& \frac{621}{1000}+\frac{621}{1000000}+\frac{621}{1000000000}+...\\ & =& \frac{621}{1000}+\frac{621}{{\left(1000\right)}^2}+\frac{621}{{\left(1000\right)}^3}+...\\ & =& \frac{621}{1000}·{\left(\frac{1}{1000}\right)}^0+\frac{621}{1000}·{\left(\frac{1}{1000}\right)}^1+\frac{621}{1000}·{\left(\frac{1}{1000}\right)}^2+...\\ & =& \sum _{n=0}^{\infty }\frac{621}{1000}·{\left(\frac{1}{1000}\right)}^n\end{array}$

$\begin{array}{rcl}0.\dot{6}2\dot{1}& =& \sum _{n=0}^{\infty }\frac{621}{1000}·{\left(\frac{1}{1000}\right)}^n\end{array}$

(b)

$\begin{array}{rcl}& & \sum _{n=0}^{\infty }\frac{621}{1000}·{\left(\frac{1}{1000}\right)}^n\end{array}$ is a geometric series with $a=\frac{621}{1000}$ and $r=\frac{1}{1000}$

First show that the series converges

For that geometric series, $\left|r\right|=\left|\frac{1}{1000}\right|=\frac{1}{1000}<1$ so it converges

Now you can use the formula $\sum _{n=0}^{\infty }a{r}^n=\frac{a}{1-r}$

Therefore

$\begin{array}{rcl}\sum _{n=0}^{\infty }\frac{621}{1000}·{\left(\frac{1}{1000}\right)}^n& =& \frac{{\displaystyle \raisebox{1ex}{$621$}\!\left/ \!\raisebox{-1ex}{$1000$}\right.}}{1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$1000$}\right.}}\\ & =& \frac{{\displaystyle \raisebox{1ex}{$621$}\!\left/ \!\raisebox{-1ex}{$1000$}\right.}}{{\displaystyle \raisebox{1ex}{$999$}\!\left/ \!\raisebox{-1ex}{$1000$}\right.}}\\ & =& \frac{621}{999}\\ & =& \frac{69}{111}\\ & =& \frac{23}{37}\end{array}$

$\begin{array}{rcl}0.\dot{6}2\dot{1}& =& \frac{23}{37}\end{array}$