Find the sum of the following series:
(a) ∑ n = 0 ∞ ( − 2 ) n e − n {\displaystyle \sum _{n=0}^{\infty }(-2)^{n}e^{-n}}
(b) ∑ n = 1 ∞ ( 1 2 n − 1 2 n + 1 ) {\displaystyle \sum _{n=1}^{\infty }{\bigg (}{\frac {1}{2^{n}}}-{\frac {1}{2^{n+1}}}{\bigg )}}
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