Sum of reciprocals of squared and factoriels I want to find out the answer to sum of reciprocals of multiplication of $n!$ and $2^n$. 

$$\sum_{n=1}^{\infty} \frac{1}{n!(2^n)}.$$
It is, hopefully, a well known result that
$$ \mathrm{e}^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$$
Your series can be rewritten as
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n! 2^n}
&amp;= \sum_{n=1}^{\infty} \frac{ \left(\frac{1}{2} \right)^n }{n!} \\
&amp;= \left[ 
\frac{ \left(\frac{1}{2}\right)^0 }{0!} \color{red}{- \frac{ \left(\frac{1}{2}\right)^0 }{0!}}\right] + \sum_{n=1}^{\infty} \frac{ \left(\frac{1}{2} \right)^n }{n!} &amp;&amp; (\text{add zero}) \\
&amp;= \color{red}{-1} + \sum_{n=0}^{\infty} \frac{ \left(\frac{1}{2} \right)^n }{n!} \\
&amp;= -1 + \mathrm{e}^{\frac{1}{2}} &amp;&amp; (\text{use the well known result}) \\
&amp;= \sqrt{\mathrm{e}} -1.
\end{align}
Maclauren exspansion of sqrt e is sum(1/(n! * 2n) , 0 &lt; n &lt; infinity. This is an identity. You specified starting at one, the sum of terms lack one.
hence, sqrt e -1 is sum(1/(n! * 2n) , 1 &lt; n &lt; infinity.
ref. and google  https://oeis.org/wiki/Sqrt(e) and Maclauren exspansion.