009C Sample Final 1, Problem 2

Find the sum of the following series:

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

b) $\sum _{n=1}^{\infty }{\bigg (}{\frac {1}{2^{n}}}-{\frac {1}{2^{n+1}}}{\bigg )}$

Foundations:
Review geometric series.
Review telescoping series.

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,
$\sum _{n=0}^{\infty }(-2)^{n}e^{-n}={\frac {1}{1+{\frac {2}{e}}}}={\frac {1}{\frac {e+2}{e}}}={\frac {e}{e+2}}$.

(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,
$\sum _{n=1}^{\infty }{\bigg (}{\frac {1}{2^{n}}}-{\frac {1}{2^{n+1}}}{\bigg )}=\lim _{k\rightarrow \infty }s_{k}=\lim _{k\rightarrow \infty }{\frac {1}{2}}-{\frac {1}{2^{k+1}}}={\frac {1}{2}}$

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

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