$\sum_{n=1}^\infty \frac{(-1)^n \pi^{n+1}}{2^n (2n-1)!}$ with complex analysis? How do I go about evaluating this sum using complex analysis techniques? It is clear that it converges thanks to the ratio test, however I am unsure of how to arrive at the following answer. Thank you.
The answer is $-\frac{\pi^{3/2} \sin\big(\sqrt{\frac{\pi}{2}}\big)}{\sqrt{2}}$.
 A: The Taylor series for $\sin(z)$ centered at $0$ is $\sum_{n=0}^\infty \frac{(-1)^n }{ (2n+1)!}z^{2n+1}$.
So for the given series, we have
\begin{aligned}\sum_{n=1}^\infty \frac{(-1)^n \pi^{n+1}}{2^n (2n-1)!}&=\sum_{n=0}^\infty \frac{(-1)^{n+1} \pi^{n+2}}{2^{n+1} (2(n+1)-1)!}
\\&= -\sqrt{\frac{\pi^3}{2}}\sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)!}\left(\sqrt{\frac{\pi}{2}}\right)^{2n+1}
\\&=-\sqrt{\frac{\pi^3}{2}}\sin\left(\sqrt{\frac{\pi}{2}}\right)\end{aligned}
A: Someone has already posted the same answer, but I spent 10 minutes on this, and I'm not about to just delete it.
First of all we recall that, for all $x$, 
$$\sin x=\sum_{n\geq0}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
Hence we have that
$$
\begin{align}
\frac{-\pi^{3/2}}{2^{1/2}}\sin\sqrt{\frac\pi2}=&-\frac{\pi^{3/2}}{2^{1/2}}\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!}\bigg(\frac{\pi^{1/2}}{2^{1/2}}\bigg)^{2n+1}\\
=&\frac{\pi^{3/2}}{2^{1/2}}\sum_{n\geq0}\frac{(-1)^{n+1}}{(2n+1)!}\frac{\pi^{\frac{2n+1}2}}{2^{\frac{2n+1}2}}\\
=&\sum_{n\geq0}\frac{(-1)^{n+1}}{(2n+1)!}\frac{\pi^{\frac{2n+4}2}}{2^{\frac{2n+2}2}}\\
=&\sum_{n\geq0}\frac{(-1)^{n+1}}{(2n+1)!}\frac{\pi^{n+2}}{2^{n+1}}\\
=&\sum_{m\geq1}\frac{(-1)^m}{(2m-1)!}\frac{\pi^{m+1}}{2^m}
\end{align}
$$
This last step coming from the change of index $n=m-1$
A: $$
\eqalign{
  & \sum\limits_{1\, \le \,n} {{{\left( { - 1} \right)^n \pi ^{\,n + 1} } \over {2^{\,n} \left( {2n - 1} \right)!}}}
  = \sum\limits_{0\, \le \,n} {{{\left( { - 1} \right)^{n + 1} \pi ^{\,n + 2} } \over {2^{\,n + 1} \left( {2n + 1} \right)!}}}  =   \cr 
  &  = \pi \sum\limits_{0\, \le \,n} {{{i^{\,2n + 2} \sqrt {\pi /2} ^{\,2n + 2} } \over {\left( {2n + 1} \right)!}}}
  = i\,\pi \sqrt {\pi /2} \sum\limits_{0\, \le \,n} {{{i^{\,2n + 1} \sqrt {\pi /2} ^{\,2n + 1} } \over {\left( {2n + 1} \right)!}}}  =   \cr 
  &  = i\,\pi \sqrt {\pi /2} \sinh \left( {i\sqrt {\pi /2} } \right) =  - \,\pi \sqrt {\pi /2} \sin \left( {\sqrt {\pi /2} } \right) \cr} 
$$
