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?
 A: Given random variable $X\sim \Pi_{10}$ (Poisson Distribution) with $\mathbb E[X]=10$
one has
$$
\sum_{n=1}^{\infty}\frac{n-1}{2}\frac{e^{-10}10^n}{n!}=\sum_{n=0}^{\infty}\frac{n-1}{2}\mathbb P(X=n) + \frac12 \frac{e^{-10}10^0}{0!}=\mathbb E\left[\frac{X-1}{2}\right]+\frac{1}{2e^{10}}$$ 
$$=
\frac{\mathbb E[X]-1}{2}+\frac{1}{2e^{10}} = \frac{10-1}{2}+\frac{1}{2e^{10}}. 
$$
A: 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}
A: \begin{align*}
\sum\limits_{n=1}^{\infty}\dfrac{n-1}{2}\dfrac{e^{-10}10^n}{n!}&=\dfrac{1}{2}\sum_{n=1}^{\infty}n\dfrac{e^{-10}10^n}{n!}-\dfrac{1}{2}\sum_{n=1}^{\infty}\dfrac{e^{-10}10^n}{n!}\\
&=\dfrac{10e^{-10}}{2}\sum_{n=1}^{\infty}\dfrac{10^{n-1}}{(n-1)!}-\dfrac{e^{-10}}{2}\sum_{n=1}^{\infty}\dfrac{10^{n}}{n!}\\
&=\dfrac{10e^{-10}}{2}\left(1+\dfrac{10}{1!}+\dfrac{10^2}{2!}+\cdots\infty\right)-\dfrac{e^{-10}}{2}\left(\dfrac{10}{1!}+\dfrac{10^2}{2!}+\dfrac{10^3}{3!}+\cdots\infty\right)\\
&=\dfrac{10e^{-10}}{2}\cdot e^{10}-\dfrac{e^{-10}}{2}\left[\left(1+\dfrac{10}{1!}+\dfrac{10^2}{2!}+\dfrac{10^3}{3!}+\cdots\infty\right)-1\right]\\
&=\dfrac{10e^{-10}}{2}\cdot e^{10}-\dfrac{e^{-10}}{2}\cdot\left(e^{10}-1\right)\\
&=\dfrac{10}{2}-\dfrac{1}{2}+\dfrac{e^{-10}}{2}\\
&=\dfrac{9}{2}+\dfrac{1}{2e^{10}}.
\end{align*}
A: \begin{eqnarray}
\sum_{n=1}^\infty \dfrac{n-1}{2}\dfrac{e^{-10}10^n}{n!}&=&
\dfrac{e^{-10}}{2}\sum_{n=1}^\infty \dfrac{(n-1)10^n}{n!}\\
&=&\dfrac{e^{-10}}{2}\sum_{n=1}^\infty\left(\dfrac{n}{n!}-\dfrac{1}{n!}\right)10^n\\
&=&\dfrac{e^{-10}}{2}\sum_{n=1}^\infty\left[\dfrac{1}{(n-1)!}-\dfrac{1}{n!}\right]10^n\\
&=&\dfrac{e^{-10}}{2}\left[\sum_{n=1}^\infty\dfrac{10^n}{(n-1)!}-\sum_{n=1}^\infty\dfrac{10^n}{n!}\right]\\
&=&\dfrac{e^{-10}}{2}\left[1+\sum_{n=0}^\infty\dfrac{10^{n+1}}{n!}-\sum_{n=0}^\infty\dfrac{10^n}{n!}\right]\\
&=&\dfrac{e^{-10}}{2}\left[1+10\sum_{n=0}^\infty\dfrac{10^n}{n!}-\sum_{n=0}^\infty\dfrac{10^n}{n!}\right]\\
&=&\dfrac{e^{-10}}{2}\left[1+9\sum_{n=0}^\infty\dfrac{10^n}{n!}\right]\\
&=&\dfrac{e^{-10}}{2}(1+9e^{10})\\
&=&\dfrac92+\dfrac{e^{-10}}{2}
\end{eqnarray}
