find the $\sum\limits_{k=0}^n \frac{(-1)^k}{k! (2k+1)} \frac{1}{(n-k)!}$ I'm stuck on computing the sum of
\begin{align*}
\sum\limits_{k=0}^n \frac{(-1)^k}{k! (2k+1)} \frac{1}{(n-k)!}
\end{align*}
I tried some manipulations which include
\begin{align*}
\frac{1}{n!} \binom{n}{k} = \frac{1}{k! (n-k)!}
\end{align*}
but still that $2k+1$ at the denominator complicates things. By the way, wolframalpha says that
\begin{align*}
\sum\limits_{k=0}^n \frac{(-1)^k}{k! (2k+1)} \frac{1}{(n-k)!} = \frac{\sqrt{\pi}}{2(n+\frac{1}{2})!}
\end{align*}
for $n\geq 1$.
Can anyone help me?

Use
\begin{eqnarray*}
\int_{0}^{1} x^{2k} dx =\frac{1}{2k+1}.
\end{eqnarray*}
interchange the order the integral and plum 
\begin{eqnarray*}
\sum_{k=0}^{n} \frac{(-1)^k}{(2k+1) k! (n-k)!} &=& \frac{1}{n!} \int_{0}^{1} \sum_{k=0}^{n}x^{2k}  \binom{n}{k} \\
&=& \frac{1}{n!} \int_{0}^{1}  (1-x^2)^n dx \\
\end{eqnarray*}
It is well known that 
\begin{eqnarray*}
\int_{0}^{1} (1-x^{2})^n dx =\frac{(2n)!!}{(2n+1)!!}.
\end{eqnarray*}
Substituting this gives
\begin{eqnarray*}
\sum_{k=0}^{n} \frac{(-1)^k}{(2k+1) k! (n-k)!} =  \color{red}{\frac{2^{2n} n! }{(2n+1)!}}.
\end{eqnarray*}
