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:
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 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 2<e,~\bigg|-\frac{2}{e}\bigg|<1.} So,
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=0}^{\infty} (-2)^ne^{-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 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 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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