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?
 A: Start with the binomial theorem:
$$\sum_{k=0}^n \binom{n}{k}x^n=(x+1)^n$$
Substitute $x=y^2$:
$$\sum_{k=0}^n \binom{n}{k}y^{2k}=(y^2+1)^n$$
Integrate both sides:
$$\sum_{k=0}^n \binom{n}{k}\frac{y^{2k+1}}{2k+1}=\int_0^y(t^2+1)^ndt$$
Divide across by $n!$:
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{y^{2k+1}}{2k+1}=\frac{1}{n!}\int_0^y(t^2+1)^ndt$$
Let $y=i$:
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{i(-1)^k}{2k+1}=\frac{1}{n!}\int_0^i(t^2+1)^ndt$$
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{i(-1)^k}{2k+1}=\frac{1}{n!}\int_0^1 i(1-t^2)^ndt$$
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{1}{2 n!}\int_0^1 t^{-1/2}(1-t)^ndt$$
Use Euler's Beta Function:
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{1}{2 n!}\frac{\Gamma(n+1)\Gamma(1/2)}{\Gamma(n+3/2)}$$
$$\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{1}{2 n!}\frac{n!2^{n+1}}{(2n+1)!!}$$
$$\color{green}{\sum_{k=0}^n \frac{1}{k!(n-k)!}\frac{(-1)^k}{2k+1}=\frac{2^{n}}{(2n+1)!!}}$$
A: 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*}
A: You can consider the generating function $\displaystyle f(x)=e^{x^2}\int^x_0e^{-t^2}\,dt$ (that was your start, wasn't it?) and derive the differential equation
$$f'(x)=2\,x\,f(x)+1.$$ The ansatz $$f(x)=\sum^\infty_{n=0}a_n\,x^n$$ gives $a_0=f(0)=0$, $a_1=f'(0)=1$ and $$n\,a_n=2\,a_{n-2}$$ for $n\ge2$. This means $a_{2k}=0$ and $\displaystyle a_{2k+1}=\frac{2^k}{(2k+1)!!}$ for $k\ge0$.
