This may be an impossible problem. But I imagine it's worth asking still. What is the closed form of the sum:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+1)!!}$$
Perhaps there isn't a closed form. Double factorials are WAY out of my comfort zone so any tips would be appreciated. Thanks
On
Too long for a comment:
This series converges by the Ratio Test.
I am not going to pretend to know much about the "generalized hypergeometric function". But if you just need the answer and not the derivation, then Wolfram Alpha produces this output as well as using the Fresnel C and S integrals:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+1)!!} = \ _1F_2 \left( 1; \frac{3}{4}; \frac{5}{4}; \frac{-1}{16} \right) \\ = \sqrt{\pi} \ \ C \left( \frac{1}{\sqrt{\pi}} \right) \cos \left(\frac{1}{2} \right) + \sqrt{\pi} \ \ S \left( \frac{1}{\sqrt{\pi}} \right) \sin \left(\frac{1}{2} \right)$$
On
Too long for a comment.
Using the same approach as in @Xoque55's answer, we could go one step further and consider $$f(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+1)!!}x^{4n}=\, _1F_2\left(1;\frac{3}{4},\frac{5}{4};-\frac{x^4}{16}\right)$$ which write $$f(x)=\frac{\sqrt \pi}x \left(C\left(\frac{x}{\sqrt{\pi }}\right) \cos \left(\frac{x^2}{2}\right)+S\left(\frac{x}{\sqrt{\pi }}\right) \sin \left(\frac{x^2}{2}\right) \right)$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{\pars{-1}^n \over \pars{4n + 1}!!}} = \sum_{n = 0}^{\infty}{\pars{-1}^n \over \prod_{k = 0}^{2n}\pars{2k + 1}} = \sum_{n = 0}^{\infty}{\pars{-1}^n \over 2^{2n + 1}\prod_{k = 0}^{2n}\pars{k + 1/2}} \\[5mm] = &\ \sum_{n = 0}^{\infty}{\pars{-1}^n \over 2^{2n + 1}\pars{1/2}^{\overline{2n + 1}}} = \sum_{n = 0}^{\infty}{\pars{-1}^n \over 2^{2n + 1}\bracks{\Gamma\pars{2n + 3/2}/\Gamma\pars{1/2}}} \\[5mm] = &\ \sum_{n = 0}^{\infty}{\pars{-1}^n \over 2^{2n + 1}}\,{1 \over \pars{2n}!}\, {\Gamma\pars{2n + 1}\Gamma\pars{1/2} \over \Gamma\pars{2n + 3/2}} \\[5mm] = &\ {1 \over 2}\sum_{n = 0}^{\infty}{\pars{-1/4}^n \over \pars{2n}!}\, \int_{0}^{1}t^{2n}\pars{1 - t}^{-1/2}\,\dd t \\[5mm] & = {1 \over 2}\int_{0}^{1}\bracks{\sum_{n = 0}^{\infty}{\pars{-t^{2}/4}^n \over \pars{2n}!}}\pars{1 - t}^{-1/2}\,\dd t \\[5mm] = &\ {1 \over 2}\int_{0}^{1} \bracks{\sum_{n = 0}^{\infty}{\pars{\ic t/2}^{2n} \over \pars{2n}!}} \pars{1 - t}^{-1/2}\,\dd t \\[5mm] & = {1 \over 2}\int_{0}^{1} \bracks{\sum_{n = 0}^{\infty}{\pars{\ic t/2}^{n} \over n!}\,{1 + \pars{-1}^{n} \over 2}} \pars{1 - t}^{-1/2}\,\dd t \\[5mm] & = {1 \over 2}\int_{0}^{1} \bracks{\Re\sum_{n = 0}^{\infty}{\pars{\ic t/2}^{n} \over n!}} \pars{1 - t}^{-1/2}\,\dd t = {1 \over 2}\int_{0}^{1}{\cos\pars{t/2} \over \root{1 - t}}\,\dd t \\[5mm] & = \bbx{\root{\pi}\bracks{\mrm{C}\pars{1 \over \root{\pi}}\cos\pars{1 \over 2} + \mrm{S}\pars{1 \over \root{\pi}}\sin\pars{1 \over 2}}} \\[5mm] & \approx 0.9344 \end{align}
$\ds{\mrm{C}\ \mbox{and}\ \mrm{S}}$ are FresnelC Function and FresnelS Function, respectively.
In the last integral, the change $\ds{\pars{\root{1 - t} = x \implies t = 1 - x^{2}}}$ yields inmediately the final solution.