What is the sum of the series with general term $\frac{(-1)^{n-1}\pi^{2n-1}}{(2n-1)!}$ Evaluate
$$\frac 1 {2\pi}\left(\frac{\pi^3}{3\times1!}-\frac{\pi^5}{5\times3!}+\frac{\pi^7}{7\times5!}-\cdots+\frac{(-1)^{n-1}\pi^{2n+1}}{(2n+1)\times(2n-1)!}+\cdots\right)$$
I tried it by using the fact that $\dfrac{(-1)^{n-1}\pi^{2n-1}}{(2n-1)!}$ is a general term of the sine series, but I'm  not getting how to introduce the term $(2n+1)$ in the denominator term?
 A: $$
\sin x=\sum_{k=0}^\infty\frac{(-1)^k x^{2k+1}}{(2k+1)!}\Rightarrow\\ 
\int_0^\pi x\sin x\, dx=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\int_0^\pi x^{2k+2}\, dx=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{x^{2k+3}}{2k+3}\Big|_0^\pi=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{\pi^{2k+3}}{2k+3}.
$$
The computation of $\int_0^\pi x\sin x\, dx$ is left as an exercise.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\bbox[#ffd,10px]{\sum_{n = 1}^{\infty}{\pars{-1}^{n - 1}\pi^{2n - 1} \over \pars{2n - 1}!}} =
\sum_{n = 1}^{\infty}{\pars{-1}^{\bracks{\pars{2n - 1} - 1}/2}\,\pi^{2n - 1} 
 \over \pars{2n - 1}!} 
\\[5px] = &\
\sum_{n = 1}^{\infty}{\pars{-1}^{\pars{n - 1}/2}\,\pi^{n} 
 \over n!}\,{1 - \pars{-1}^{n} \over 2}
\\[5mm] & =
-\,{1 \over 2}\,\ic\bracks{%
\sum_{n = 1}^{\infty}{\pars{\pi\ic}^{n} \over n!} -
\sum_{n = 1}^{\infty}{\pars{-\pi\ic}^{n} \over n!}} =
\Im\sum_{n = 1}^{\infty}{\pars{\pi\ic}^{n} \over n!} =
\Im\pars{\expo{\pi\ic} - 1}
\\[5mm] & = \sin\pars{\pi} = \bbx{0}
\end{align}
