Is $\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}$ convergent? If yes, evaluate it. Is $\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}$ convergent? If yes, evaluate it.
Using the ratio test, I could show that the series is convergent. What is an easy way to evaluate it? I was thinking about estimating it from below and above by series with equal value but coulnd't find any familar series that I could use for this. 

Working it out
in as elementary a way
as possible.
$\begin{array}\\
\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}
&=\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}\\
&=\sum_{n=1}^\infty \frac{n^2}{n!2^n} +\sum_{n=1}^\infty \frac{1}{n!2^n}\\
&=\sum_{n=1}^\infty \frac{n}{(n-1)!2^n} +\sum_{n=0}^\infty \frac{(1/2)^n}{n!}-1\\
&=\sum_{n=0}^\infty \frac{n+1}{n!2^{n+1}} +e^{1/2}-1\\
&=\sum_{n=0}^\infty \frac{n}{n!2^{n+1}}+\sum_{n=0}^\infty \frac{1}{n!2^{n+1}} +e^{1/2}-1\\
&=\sum_{n=1}^\infty \frac{n}{n!2^{n+1}}+\frac12\sum_{n=0}^\infty \frac{1}{n!2^{n}} +e^{1/2}-1\\
&=\sum_{n=1}^\infty \frac{1}{(n-1)!2^{n+1}}+\frac12e^{1/2} +e^{1/2}-1\\
&=\sum_{n=0}^\infty \frac{1}{n!2^{n+2}}+\frac32e^{1/2} -1\\
&=\frac14\sum_{n=0}^\infty \frac{1}{n!2^{n}}+\frac32e^{1/2} -1\\
&=\frac14 e^{1/2}+\frac32e^{1/2} -1\\
&=\frac74 e^{1/2} -1\\
\end{array}
$
