Is there any procedure to simplify summation of a series, such as,
$$\begin{align}S &=1-\frac 1 3+\frac 1 {10}-\frac 1 {42}+\frac 1 {210}-\frac 1 {1320}+.....=\sum_{n=0}^\infty\frac{(-1)^n}{n!(2n+1)}\end{align}$$
using fourier series expansion, or any criteria should also be given?
If this type of questions has already been asked, please send the link.
Any help will be appreciated, Thank you.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$
\begin{align} &\bbox[5px,#ffd]{% \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n!\pars{2n + 1}}} = \sum_{n = 0}^{\infty} {\pars{-1}^{n} \over n!}\int_{0}^{1}x^{2n}\,\dd x \\[5mm] = &\ \int_{0}^{1}\sum_{n = 0}^{\infty} {\pars{-x^{2}}^{n} \over n!}\,\dd x = \int_{0}^{1}\expo{-x^{2}}\,\dd x \\[5mm] = &\ \bbx{{\root{\pi} \over 2}\on{erf}\pars{1}} \approx 0.7468 \\ & \end{align}
$\ds{\on{erf}}$ is the Error Function.