009C Sample Final 1, Problem 2

Find the sum of the following series:

(a) $\sum _{n=0}^{\infty }(-2)^{n}e^{-n}$

(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)}

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

Solution:

(a)

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

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

(b)

Step 1:
This is a telescoping series. First, we find the partial sum of this series.
Let $s_{k}=\sum _{n=1}^{k}{\bigg (}{\frac {1}{2^{n}}}-{\frac {1}{2^{n+1}}}{\bigg )}.$
Then,
$s_{k}={\frac {1}{2}}-{\frac {1}{2^{k+1}}}.$

Step 2:
Thus,
${\begin{array}{rcl}\displaystyle {\sum _{n=1}^{\infty }{\bigg (}{\frac {1}{2^{n}}}-{\frac {1}{2^{n+1}}}{\bigg )}}&=&\displaystyle {\lim _{k\rightarrow \infty }s_{k}}\\&&\\&=&\displaystyle {\lim _{k\rightarrow \infty }{\frac {1}{2}}-{\frac {1}{2^{k+1}}}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\\\end{array}}$

Final Answer:
(a) ${\frac {e}{e+2}}$
(b) ${\frac {1}{2}}$

Return to Sample Exam