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I am trying to find the closed form for this sum, but facing a problem. I can't complete it completely.

$$\sum_{n=0}^{\infty}\frac{32^{-n}{4n \choose 2n}}{(4n+1)(4n+3)(4n+5)\cdots[4n+(4k-3)]}=2^{2k-1}\sqrt{2}\sin\left(\frac{\pi}{8}\right)F(k)\tag1$$

Where $k\ge1$

$F(k)=1,2,3,4,...$ has values of $1$, $\frac{1}{63}$, $\frac{1}{16065}$, $\frac{1}{9237375}$... respectively.

I can't seem to see a pattern.

Does $F(k)$ have a closed form?

Leonhard Euler
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Endgame
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    What is the origin of the problem? – Anubhab Jan 20 '19 at 09:05
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    I guess what Anubhab and other users who downvoted (and voted to close) would like to see how you obtained the sine on the right hand side (or why that is desirable). I personally feel it would be a shame if this question post gets closed. As a reminder, thinking that "those who cannot see the series pattern in the summand that leads to sine are just not my target reader (so leave my post alone)" would not go over very well on this site. Please click the edit button to add context (and don't just comment). – Lee David Chung Lin Jan 20 '19 at 11:52
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    How have you computed the given values of $F$? – user Jan 20 '19 at 12:10
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    What makes you believe that this series has a closed form? Where does this problem come from? What motivates it? – Xander Henderson Apr 19 '19 at 16:37

2 Answers2

2

For any $x\in[0,1)$ we have $$\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)}x^n=\frac{\arcsin(2\sqrt{x})}{2\sqrt{x}}\tag{1}$$ $$\sum_{n\geq 0}\frac{\binom{4n}{2n}}{(4n+1)}x^{2n}=\frac{\arcsin(2\sqrt{x})}{4\sqrt{x}}+\frac{\text{arcsinh}(2\sqrt{x})}{4\sqrt{x}}\tag{2}$$ hence by evaluating $(2)$ at $x=\frac{1}{4\sqrt{2}}$ we get $$\sum_{n\geq 0}\frac{\binom{4n}{2n}}{32^n(4n+1)}=\frac{\text{arccsc}(2^{1/4})+\text{arccsch}(2^{1/4})}{2^{3/4}}\tag{3} $$ and the other values can be computed in a similar way. For instance, multiplying both sides of $(1)$ by $x$, then replacing $x$ with $z^2$ and integrating, we get $$ \sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)(2n+3)}x^n =\frac{2\sqrt{x(1-4x)}+(8x-1)\arcsin(2\sqrt{x})}{32 x^{3/2}} \tag{4}$$ allowing us to derive the value of the hypergeometric series $$ \sum_{n\geq 0}\frac{\binom{4n}{2n}}{32^n(4n+1)(4n+3)} $$ from a discrete Fourier transform applied to $(4)$.

Jack D'Aurizio
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1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\left.\sum_{n = 0}^{\infty} {32^{-n}{4n \choose 2n} \over \pars{4n + 1}\pars{4n + 3}\ldots\bracks{4n + \pars{4k - 3}}} \,\right\vert_{\ k\ \geq\ 1}} = \sum_{n = 0}^{\infty} {32^{-n}{4n \choose 2n} \over \prod_{\ell = 0}^{2k - 2}\pars{4n + 2\ell + 1}} \\[5mm] = &\ \sum_{n = 0}^{\infty} {32^{-n}{4n \choose 2n} \over 2^{2k - 1}\prod_{\ell = 0}^{2k - 2}\pars{\ell + 2n + 1/2}} = {1 \over 2^{2k - 1}}\sum_{n = 0}^{\infty} {32^{-n}{4n \choose 2n} \over \pars{2n + 1/2}^{\overline{2k - 1}}} \\[5mm] = &\ {1 \over 2^{2k - 1}}\sum_{n = 0}^{\infty} {32^{-n}{4n \choose 2n} \over \Gamma\pars{2n + 2k - 1/2}/\Gamma\pars{2n + 1/2}} \\[5mm] = &\ {1 \over 2^{2k - 1}\pars{2k - 2}!}\sum_{n = 0}^{\infty} {4n \choose 2n}32^{-n}\, {\Gamma\pars{2n + 1/2}\Gamma\pars{2k - 1} \over \Gamma\pars{2n + 2k - 1/2}} \\[5mm] = &\ {1 \over 2^{2k - 1}\pars{2k - 2}!}\sum_{n = 0}^{\infty} {4n \choose 2n}32^{-n}\int_{0}^{1}t^{2n - 1/2}\pars{1 - t}^{2k - 2}\dd t \\[5mm] = &\ {1 \over 2^{2k - 1}\pars{2k - 2}!}\int_{0}^{1}t^{-1/2}\pars{1 - t}^{2k - 2}\sum_{n = 0}^{\infty} {4n \choose 2n}32^{-n}\, t^{2n}\dd t \end{align}

Note that $\ds{{4n \choose 2n} = {-1/2 \choose 2n}16^{n}}$

Then, \begin{align} &\bbox[10px,#ffd]{\left.\sum_{n = 0}^{\infty} {32^{-n}{4n \choose 2n} \over \pars{4n + 1}\pars{4n + 3}\ldots\bracks{4n + \pars{4k - 3}}} \,\right\vert_{\ k\ \geq\ 1}} \\[5mm] = &\ {1 \over 2^{2k - 1}\pars{2k - 2}!}\int_{0}^{1}t^{-1/2}\pars{1 - t}^{2k - 2}\sum_{n = 0}^{\infty} {-1/2 \choose 2n}\,\pars{t \over \root{2}}^{2n}\dd t \\[5mm] = &\ {1 \over 2^{2k - 1}\pars{2k - 2}!}\int_{0}^{1}t^{-1/2} \pars{1 - t}^{2k - 2}\sum_{n = 0}^{\infty} {-1/2 \choose n}\,\pars{t \over \root{2}}^{n} \,{1 + \pars{-1}^{n} \over 2}\dd t \\[5mm] = &\ {1 \over 2^{2k}\pars{2k - 2}!}\left[% \int_{0}^{1}t^{-1/2}\pars{1 - t}^{2k - 2}\bracks{1 -\pars{-2^{-1/2}}t}^{-1/2}\dd t\right. \\[2mm] & \phantom{{1 \over 2^{2k}\pars{2k - 2}!}}+ \left.\int_{0}^{1}t^{-1/2}\pars{1 - t}^{2k - 2} \pars{1 -2^{-1/2}t}^{-1/2}\dd t\right] \end{align}

The remaining integrals are Euler Type $\ds{_{2}\mrm{F}_{1}}$ Hypergeometric Function.

Then, \begin{align} &\bbox[10px,#ffd]{\left.\sum_{n = 0}^{\infty} {32^{-n}{4n \choose 2n} \over \pars{4n + 1}\pars{4n + 3}\ldots\bracks{4n + \pars{4k - 3}}} \,\right\vert_{\ k\ \geq\ 1}} \\[5mm] = &\ \bbx{% \root{\pi}\,{\mbox{}_{2}\mrm{F}_{1}\pars{1/2,1/2;2k - 2,-\root{2}/2} + \mbox{}_{2}\mrm{F}_{1}\pars{1/2,1/2;2k - 2,\root{2}/2} \over 2^{2k}\Gamma\pars{2k - 1/2}}} \end{align}

Felix Marin
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