Help find closed form for:$\sum_{n=0}^{\infty}{(-1)^n\left({{\pi\over 2}}\right)^{2n}\over (2n+k)!}=F(k)$ What is the closed form for
$$\sum_{n=0}^{\infty}{(-1)^n\left({{\pi\over 2}}\right)^{2n}\over (2n+k)!}=F(k)?$$
My try:
I have found a few values of $F(k)$, but was unable to find a closed form for it.
$F(0)=0$
$F(1)={2\over \pi}$
$F(2)=\left({2\over \pi}\right)^2$
$F(3)=\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^3$
$F(4)={1\over 2}\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^4$
$F(5)={1\over 6}\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^4+\left({2\over \pi}\right)^5$
 A: Note that
$$\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)! } = \cos(x)$$
You can reach the result by integrating $k$-times. 
$$\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+k)! } = \frac{1}{x^{k}}\int^{x}_0 \mathrm{d}t_{k-1}\int^{t_{k-1}}_0 \mathrm{d}t_{k-2} \cdots\int^{t_1}_0\mathrm{d}t_0\cos(t_0) $$
For example when $k=1$
$$\sum_{n=0}^\infty \frac{(-1)^n(\pi/2)^{2n}}{(2n+1)! } = \frac{2}{\pi}\int^{\pi/2}_0 \mathrm{d}t_0 \cos(t_0)\,= \frac{2}{\pi}$$
For $k=2$
\begin{align}
\sum_{n=0}^\infty \frac{(-1)^n(\pi/2)^{2n}}{(2n+2)! } &= \left( \frac{2}{\pi}\right)^2\int^{\pi/2}_0 \mathrm{d}t_{1}\int^{t_1}_0\mathrm{d}t_0\cos(t_0) \\  &=  \left( \frac{2}{\pi}\right)^2\int^{\pi/2}_0  \mathrm{d}t_{1}\sin(t_1)\\
& = \left( \frac{2}{\pi}\right)^2
\end{align}
For 
$$f_k(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+k}}{(2n+k)!}$$
We can define recursively 
$$f_0(x) = \cos(x)$$
$$f_k(x) =\int^x_0 f_{k-1}(t)\,dt$$
By solving the recursive formula 
$$\sum_{n=0}^\infty \frac{(-1)^n x^{2n+k}}{(2n+k)!} = \begin{cases} \sum_{n=0}^{\lceil k/2 \rceil-2}\frac{x^{2n+1}(-1)^{\lceil k/2 \rceil+n}}{(2n+1)!}-(-1)^{\lceil k/2 \rceil}\sin(x) & \text{If $k$ is odd} \\ \sum_{n=0}^{k/2-1}\frac{x^{2n}(-1)^{n+k/2}}{(2n)!}-(-1)^{k/2}\cos(x) & \text{If $k$ is even}\end{cases}$$
Finally we have the closed form 

$$F(k) = \begin{cases}\left( \frac{2}{\pi}\right)^k \sum_{n=0}^{\lceil
 k/2 \rceil-2}\frac{(\pi/2)^{2n+1}(-1)^{\lceil k/2
 \rceil+n}}{(2n+1)!}-\left( \frac{2}{\pi}\right)^k(-1)^{\lceil k/2
 \rceil} & \text{If $k$ is odd} \\ \left(
 \frac{2}{\pi}\right)^k\sum_{n=0}^{k/2-1}\frac{(\pi/2)^{2n}(-1)^{n+k/2}}{(2n)!}&
 \text{If $k$ is even}\end{cases}$$

To check the correctness of the formula 
For $k=5$ we have 
$$F(5)=\left(\frac{2}{\pi}\right)^5 \sum_{n=0}^{1}\frac{(\pi/2)^{2n+1}(-1)^{3+n}}{(2n+1)!}-\left( \frac{2}{\pi}\right)^5(-1)^{3} = {1\over 6}\left({2\over \pi}\right)^2-\left({2\over \pi}\right)^4+\left({2\over \pi}\right)^5$$
A: Your sum is the special value at $x=(-(\pi/2)^2)$ of 
$$\sum_{n=0}^\infty \frac{x^n}{(2n + k)!}=\frac{\, _1F_2\left(1;\frac{k}{2}+\frac{1}{2},\frac{k}{2}+1;\frac{x}{4}\right)}{k!}$$
A: The sum can also be expressed in terms of the Regularized Incomplete Gamma function ($Q(a,z)$).
In fact, premised that for a general function of a integer $f(k)$ we have
$$
\begin{gathered}
  \frac{{\left( {i^{\,k}  + \left( { - i} \right)^{\,k} } \right)}}
{2}f(k) = \left( {\frac{{e^{\,i\,k\frac{\pi }
{2}}  + e^{\, - \,i\,k\frac{\pi }
{2}} }}
{2}} \right)f(k) = \cos \left( {k\frac{\pi }
{2}} \right)f(k) =  \hfill \\
   = \left[ {k = 2j} \right]\left( { - 1} \right)^{\,j} f(2j)\quad \left| {\;k,j\; \in \;\;\mathbb{Z}\,} \right. \hfill \\ 
\end{gathered} 
$$
and that the Lower Incomplete Gamma function can be expressed as:
$$
\begin{gathered}
  \gamma (s,z) = \int_{\,0\,}^{\,z\,} {t^{\,s - 1} \;e^{\, - \,t} dt}  =  \hfill \\
   = z^{\,s} \;e^{\,\, - z} \;\Gamma (s)\;\sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }}
{{\Gamma (s + k + 1)}}}  = \frac{{z^{\,s} \;e^{\,\, - z} }}
{s}\sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }}
{{\left( {s + 1} \right)^{\,\overline {\,k\,} } \,}}}  =  \hfill \\
   = z^{\,s} \;e^{\,\, - z} \sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }}
{{s^{\,\overline {\,k + 1\,} } \,}}}  = z^{\,s} \sum\limits_{0\, \leqslant \,j} {\frac{{\left( { - 1} \right)^{\,j} }}
{{\,\left( {s + j} \right)}}\frac{{z^{\,j} }}
{{j!}}}  \hfill \\ 
\end{gathered} 
$$
we can then write
$$
\begin{gathered}
  F(x,m) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{x^{\,2k} }}
{{\left( {2k + m} \right)!}}}  =  \hfill \\
   = \frac{1}
{2}\left( {\sum\limits_{0\, \leqslant \,k} {\frac{{\left( {i\,x} \right)^{\,k} }}
{{\left( {k + m} \right)!}}}  + \sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - \,i\,x} \right)^{\,k} }}
{{\left( {k + m} \right)!}}} } \right) =  \hfill \\
   = \frac{1}
{{2\,\Gamma (m)}}\left( {\frac{{e^{\,\,i\,x} \,\gamma (m,i\,x)}}
{{\left( {i\,x} \right)^{\,m} }} + \frac{{e^{\, - \,i\,x} \,\gamma (m, - i\,x)}}
{{\left( { - i\,x} \right)^{\,m} }}} \right) =  \hfill \\
   = \frac{1}
{{\left| x \right|^{\,\,m} }}\,\operatorname{Re} \left( {e^{\,\,i\,x} e^{\,\, - \,i\,m\,\left( {sign(x)\pi /2} \right)} \,\frac{{\gamma (m,i\,x)}}
{{\Gamma (m)}}} \right) =  \hfill \\
   = \frac{1}
{{\left| x \right|^{\,\,m} }}\,\operatorname{Re} \left( {e^{\,\,i\,x} e^{\,\, - \,i\,m\,\left( {sign(x)\pi /2} \right)} \,\left( {1 - Q(m,i\,x)} \right)} \right) \hfill \\ 
\end{gathered} 
$$
So, for $x=\pi /2$ we get
$$
\begin{gathered}
  F(\pi /2,m) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }}
{{\left( {2k + m} \right)!}}}  =  \hfill \\
   = \left( {\frac{2}
{\pi }} \right)^{\,\,m} \,\operatorname{Re} \left( {e^{\,\, - \,i\,\left( {m - 1} \right)\,\left( {\pi /2} \right)} \,\frac{{\gamma (m,i\,\pi /2)}}
{{\Gamma (m)}}} \right) =  \hfill \\
   = \left( {\frac{2}
{\pi }} \right)^{\,\,m} \,\operatorname{Re} \left( {e^{\,\, - \,i\,\left( {m - 1} \right)\,\left( {\pi /2} \right)} \,\left( {1 - Q(m,i\,\pi /2)} \right)} \right) \hfill \\ 
\end{gathered} 
$$
example with $m=5$
$$
\begin{gathered}
  F(\pi /2,5) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }}
{{\left( {2k + 5} \right)!}}}  =  \hfill \\
   = \left( {\frac{2}
{\pi }} \right)^{\,\,5} \,\operatorname{Re} \left( {e^{\,\, - \,i\,4\,\left( {\pi /2} \right)} \,\left( {1 - Q(5,i\,\pi /2)} \right)} \right) =  \hfill \\
   = \left( {\frac{2}
{\pi }} \right)^{\,\,5} \,\operatorname{Re} \,\left( {1 - \frac{1}
{{24}}\left( {12\,\pi  - \frac{1}
{2}\pi ^{\,3}  + i\left( {3\pi ^{\,2}  - \frac{1}
{{16}}\pi ^{\,4}  - 24} \right)} \right)} \right) =  \hfill \\
   = \left( {\frac{2}
{\pi }} \right)^{\,\,5} \left( {1 - \,\frac{\pi }
{2} + \frac{1}
{{48}}\pi ^{\,3} } \right) = \left( {\frac{2}
{\pi }} \right)^{\,\,5}  - \,\left( {\frac{2}
{\pi }} \right)^{\,\,4}  + \frac{1}
{6}\left( {\frac{2}
{\pi }} \right)^{\,2}  \hfill \\ 
\end{gathered} 
$$
which matches with the value given by Zaid Alyafeai,
 as well as with the fomula indicated by Igor Rivin
$$
\begin{gathered}
  F(\pi /2,5) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }}
{{\left( {2k + 5} \right)!}}}  = \left( {\frac{2}
{\pi }} \right)^{\,\,5}  - \,\left( {\frac{2}
{\pi }} \right)^{\,\,4}  + \frac{1}
{6}\left( {\frac{2}
{\pi }} \right)^{\,2}  =  \hfill \\
   = 0.007860176 \cdots  =  \hfill \\
   = \frac{1}
{{5!}}{}_1F_2 \left( {1\;;\;\frac{5}
{2} + \frac{1}
{2},\;\frac{5}
{2} + 1\;;\; - \frac{1}
{4}\left( {\frac{\pi }
{2}} \right)^{\,2} } \right) \hfill \\ 
\end{gathered} 
$$
and concerning the latter, a computer calculation over multiple values of $m$ and $x$ shows a full match.
