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Is there a simple expression for the power series $$\sum_{n=0}^{\infty} \left[ (4x)^n \frac{(n!)^2}{(2n+1)!} \right]^2?$$ This question came up in a quantum mechanics problem. Mathematica only returns $$ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x^2\right).$$ We know that $$\sum_{n=0}^{\infty} (4x)^n \frac{(n!)^2}{(2n+1)!} = \frac{\arcsin \sqrt x}{\sqrt{x(1-x)}},$$ but it is not clear how to proceed for the other sum.

I will award a 50 point bounty to a correct closed-form expression if desired.

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  • $\begingroup$ The last time I said "I will award a $\ldots$ reps bounty to..." in my OP, I got tremendous downvotes ( つ︣﹏╰) $\endgroup$ – Anastasiya-Romanova 秀 Oct 28 '14 at 13:08
  • $\begingroup$ @Anastasiya-Romanova Really? You didn't understand why you got the downvotes and how this is different? $\endgroup$ – UserX Oct 28 '14 at 13:24
  • $\begingroup$ it gives a hypergeometric series $\endgroup$ – Dr. Sonnhard Graubner Oct 28 '14 at 14:34
  • $\begingroup$ $${_3F_2}\left(\begin{array}c 1,1,1 \\\tfrac32,\tfrac32\end{array}\middle|\,x^2\right) = \frac 1x \int_0^1 \frac{\arcsin(tx)}{\sqrt{(t^2-1)(t^2x^2-1)}} dt,$$ for $x>0$. By the way bounty is not for this. If you use it that way you can avoid wasted reputation. I've "lost" lots of reputation because noone could answer the question after the offer. Maybe this is why someone downvote if you're doing this. $\endgroup$ – user153012 Oct 31 '14 at 23:26
  • $\begingroup$ Furthermore $$\sum_{n=0}^{\infty} \left[ (4x)^n \frac{(n!)^2}{(2n+1)!} \right]^k = {_{k+1}F_k}\left(\begin{array}c 1,1,\dots,1 \\\tfrac32,\tfrac32\dots\tfrac32\end{array}\middle|\,x^k\right),$$ for $k>0$. This is why there is a simple form for $k=1$. $\endgroup$ – user153012 Oct 31 '14 at 23:49
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This isn't quite an answer, but I think it can lead to one: with \begin{align}z:&=_2\!\!\mathrm{F}_1(\tfrac12,\tfrac12;1;x)=\frac2\pi K(\sqrt{x})\\ y:&=\pi\frac{_2\mathrm{F}_1(\tfrac12,\tfrac12;1;1-x)}{_2\mathrm{F}_1(\tfrac12,\tfrac12;1;x)}=\pi\frac{z(1-x)}{z(x)}=\pi\frac{K'(\sqrt{x})}{K(\sqrt{x})},\tag1\\ \sum_{n=0}^{\infty} \left[ (4x)^n \frac{(n!)^2}{(2n+1)!} \right]^2&={_3\mathrm{F}_2}\left(1,1,1;\frac{3}{2},\frac{3}{2};x^2\right)=\frac{4K(x)}{\pi x}\sum_{k\ge1}\frac1{(2k-1)^2 \cosh(k-1/2)y(x^2)}\tag2 \end{align}(from Ramanujan's Notebooks Part III; $(1)$, $(2)$, pgs. 101, 153 resp. [entry 6 in both]) and \begin{align}\int_0^1 u\cos(2k-1)\pi u\,du&=\left[\frac{u\sin(2k-1)\pi u}{(2k-1)\pi}+\frac{\cos(2k-1)\pi u}{(2k-1)^2\pi^2}\right]_0^1\\ &=\frac{-1-1}{(2k-1)^2\pi^2}=\frac{-2}{(2k-1)^2\pi^2}\\ \end{align} $$\frac{Kk}\pi\operatorname{cn}\left(\frac{2Ku}\pi,k\right)=\sum_{n\ge1}\frac{2\cos(2n-1)u}{e^{(n-1/2)y(k^2)}+e^{-(n-1/2)y(k^2)}}=\sum_{n\ge1}\frac{\cos(2n-1)u}{\cosh\pi(n-1/2)K'/K}$$(the last formula is from DLMF 22.11) thus\begin{align}\int_0^1\frac{Kku}\pi\operatorname{cn}\left(2uK,k\right)\, du&=\int_0^1\sum_{n\ge1}\frac{u\cos(2n-1)\pi u}{\cosh\left(\frac{\pi K'}{K}(n-\tfrac12)\right)}\,du\tag3\\ &=-\frac2{\pi^2}\sum_{k\ge1}\frac1{(2n-1)^2\cosh\pi(n-\tfrac12)\frac{K'}K} \end{align} I thought to add a new variable $v$ to the first integral in $(3)$ to make it$$\frac{Kk}\pi\int_0^1u\operatorname{cn}\left(2K(u+v),k\right)\,du\tag4$$I figured since $\operatorname{cn}(2Ku,k)$ has periods $2$ and $1+iK'/K$, so too will the integral (for $v$)...but I'm not completely sure because the simpler $\int_0^1\operatorname{cn}(u+v,k)\,du$ yields a multi-valued $k^{-1}[\sin^{-1}(k\operatorname{sn}(1+v,k))-\sin^{-1}(k\operatorname{sn}(v,k))]$. But if I'm right, then $(4)$ can be studied as an elliptic function in $v$ and so $(2)$ evaluates to a combination of theta and elliptic functions.

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  • $\begingroup$ I came across the OP's series a few years ago considering the odd analog of the classical series $$\sum_{n\ge1}\frac{(2n-1)!!^2}{(2n)!!^2}x^{2n}=\sum_{n\ge1}\frac{(2n-1)!^2}{2^{4n}n!^4}x^{2n}=\frac2\pi K(x)$$ $\endgroup$ – Antonio DJC Jan 15 '17 at 3:55

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