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?
On
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
$$ \sin x=\sum_{k=0}^\infty\frac{(-1)^k x^{2k+1}}{(2k+1)!}\Rightarrow\\ \int_0^\pi x\sin x\, dx=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\int_0^\pi x^{2k+2}\, dx=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{x^{2k+3}}{2k+3}\Big|_0^\pi=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{\pi^{2k+3}}{2k+3}. $$
The computation of $\int_0^\pi x\sin x\, dx$ is left as an exercise.