<s>[INST] Evaluate $\sum_{n=1}^{\infty}\frac{n-1}{2}\frac{e^{-10}10^n}{n!}$ I need to evaluate $\sum_{n=1}^{\infty}\frac{n-1}{2}\frac{e^{-10}10^n}{n!}$ to solve a statistic problem. Mathematica gives an answer of $\frac{9}{2}+\frac{1}{2e^{10}}$. How can I evaluate explicitly without aid of computer?
 [/INST] Observe
\begin{align}
\sum_{n=1}^{\infty}\frac{n-1}{2}\frac{e^{-10}10^n}{n!}=&\ \frac{e^{-10}}{2}\left(10\sum^\infty_{n=1} n\frac{10^{n-1}}{n!}-\sum^\infty_{n=1} \frac{10^n}{n!} \right)\\
=& \frac{e^{-10}}{2}\left(10\sum^\infty_{n=1} \frac{10^{n-1}}{(n-1)!}-\sum^\infty_{n=1} \frac{10^n}{n!} \right)\\
=& \frac{e^{-10}}{2}\left(10\sum^\infty_{n=0} \frac{10^n}{n!}-\sum^\infty_{n=0} \frac{10^n}{n!}+1 \right)\\
=&\ \frac{e^{-10}}{2}\left(10e^{10}-e^{10}+1 \right).\\
\end{align}
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