# 009C Sample Final 1, Problem 2

Find the sum of the following series:

a) ${\displaystyle \sum _{n=0}^{\infty }(-2)^{n}e^{-n}}$
b) ${\displaystyle \sum _{n=1}^{\infty }{\bigg (}{\frac {1}{2^{n}}}-{\frac {1}{2^{n+1}}}{\bigg )}}$
Foundations:
Recall:
1. For a geometric series ${\displaystyle \sum _{n=0}^{\infty }ar^{n}}$ with ${\displaystyle |r|<1,}$
${\displaystyle \sum _{n=0}^{\infty }ar^{n}={\frac {a}{1-r}}.}$
2. For a telescoping series, we find the sum by first looking at the partial sum ${\displaystyle s_{k}}$
and then calculate ${\displaystyle \lim _{k\rightarrow \infty }s_{k}.}$

Solution:

(a)

Step 1:
First, we write
${\displaystyle {\begin{array}{rcl}\displaystyle {\sum _{n=0}^{\infty }(-2)^{n}e^{-n}}&=&\displaystyle {\sum _{n=0}^{\infty }{\frac {(-2)^{n}}{e^{n}}}}\\&&\\&=&\displaystyle {\sum _{n=0}^{\infty }{\bigg (}{\frac {-2}{e}}{\bigg )}^{n}.}\\\end{array}}}$
Step 2:
Since ${\displaystyle 2 So,
${\displaystyle {\begin{array}{rcl}\displaystyle {\sum _{n=0}^{\infty }(-2)^{n}e^{-n}}&=&\displaystyle {\frac {1}{1+{\frac {2}{e}}}}\\&&\\&=&\displaystyle {\frac {1}{\frac {e+2}{e}}}\\&&\\&=&\displaystyle {{\frac {e}{e+2}}.}\\\end{array}}}$

(b)

Step 1:
This is a telescoping series. First, we find the partial sum of this series.
Let ${\displaystyle s_{k}=\sum _{n=1}^{k}{\bigg (}{\frac {1}{2^{n}}}-{\frac {1}{2^{n+1}}}{\bigg )}.}$
Then,
${\displaystyle s_{k}={\frac {1}{2}}-{\frac {1}{2^{k+1}}}.}$
Step 2:
Thus,
${\displaystyle {\begin{array}{rcl}\displaystyle {\sum _{n=1}^{\infty }{\bigg (}{\frac {1}{2^{n}}}-{\frac {1}{2^{n+1}}}{\bigg )}}&=&\displaystyle {\lim _{k\rightarrow \infty }s_{k}}\\&&\\&=&\displaystyle {\lim _{k\rightarrow \infty }{\frac {1}{2}}-{\frac {1}{2^{k+1}}}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\\\end{array}}}$
(a) ${\displaystyle {\frac {e}{e+2}}}$
(b) ${\displaystyle {\frac {1}{2}}}$