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In continuation of this MO question: The sum of an hydrogen atom related infinite series.

Can the sum $$\sum\limits_{n=1}^{\infty}\frac{n-\frac{1}{2}}{n}\left[\frac{\Gamma(n)}{\Gamma\left(n+\frac{1}{2}\right)}\right]^4 \tag{1}$$ be calculated explicitly?

In the case of previous series, besides the explicit formula for the partial sums given by Robert Israel, the following simple reasoning works. Let $$a_n=\frac{1}{n+\frac{1}{2}}\left[\frac{\Gamma(n)}{\Gamma\left(n+\frac{1}{2}\right)}\right]^2.$$ Then $$\frac{a_n}{a_{n-1}}=\frac{(n-1)^2}{n^2-\frac{1}{4}}=\frac{4(n-1)^2}{4n^2-1}.$$ That is $$4n^2a_n=4(n-1)^2a_{n-1}+a_n.$$ Applying this recurrence relation again and again, we get $$4n^2a_n=4a_1+a_2+a_3+\ldots+a_n.$$ Therefore $$s_n=\sum\limits_{i=1}^na_i=4n^2a_n-3a_1,$$ and because $\lim\limits_{n\to\infty}n^2a_n=1$ and $a_1=\frac{8}{3\pi}$, we get $$s=\lim\limits_{n\to\infty}s_n=4-\frac{8}{\pi}.$$ However, in the case of (1), I was unable to elaborate the similar method.

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  • $\begingroup$ Maple gives an answer using Meijer G-function so I'm not sure it is possible to have a more explicit answer. $\endgroup$ – Héhéhé Apr 6 '17 at 9:58
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    $\begingroup$ Mathematica evaluates this sum for arbitrary integer powers $p$ of the ratio of $\Gamma$ functions, with an answer in terms of hypergeometric functions $_{p+1}F_p$ and $_{p+2}F_{p+1}$, but only for $p=2$ does this reduce to an explicit constant. $\endgroup$ – Carlo Beenakker Apr 6 '17 at 10:19
  • $\begingroup$ Zurab: what makes you choose the new sum, in its displayed format? $\endgroup$ – T. Amdeberhan Apr 7 '17 at 2:06
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    $\begingroup$ $a_n$ are variational energy levels of the hydrogen atom (up to some irrelivent factor) if you use Gaussian trial function $r^l e^{-\alpha r^2}Y_l^m(\theta,\phi)$, as did Friedmann and Hagen in arxiv.org/abs/1510.07813. If you instead use Lorentz trial function $$\frac{r^l}{(a^2+r^2)^{l+1}}Y_l^m(\theta,\phi),$$ you get summands of (1) as the variational energy levels of the hydrogen atom. $\endgroup$ – Zurab Silagadze Apr 7 '17 at 4:13
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    $\begingroup$ Thank you. Your question might be enriched if you put your reply into the question box itself. $\endgroup$ – T. Amdeberhan Apr 7 '17 at 23:36

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