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Consider the following series: $$S=\sum_{n=1}^\infty\frac{(-1)^n\ \Gamma\left(\frac{5}{4}+n\right)}{n^2\ \Gamma(n)}.$$ It can be expressed in terms of a hypergeometric function: $$S=-\frac{5}{16}\Gamma\left(\frac{1}{4}\right)\ { _3F_2}\left(1,1,\frac{9}{4};2,2;-1\right).$$ I tried to find an expression of $S$ using elementary functions and ended up with this conjecture: $$S\stackrel{?}{=}\frac{\Gamma\left(\frac{1}{4}\right)}{16}\Bigg(8\sqrt[4]{8}-16-\ln\frac{24\sqrt{69708+49291\sqrt{2}}+3168\sqrt{2}+4481}{4096}\\\\+\arctan\frac{24\sqrt{49291\sqrt{2}-13260}}{6913}\Bigg).$$My derivation of this formula is quite long and uses a non-rigorous, heuristic approach involving heavy use of techniques like guessing sequence formulas using Mathematica command FindSequenceFunction and OEIS superseeker, and recognizing approximate numeric quantities using RootApproximant, TranscendentalRecognize, Inverse Symbolic Calculator and Mathematica command

WolframAlpha[ToString[value], IncludePods -> "PossibleClosedForm", TimeConstraint -> ∞]

The formula holds with at least $1800$ decimal digits of precision.

Now I am looking for a rigorous way to prove this formula. Can you suggest any ideas how to do that?

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This might be worth a look: – Steve Huntsman May 29 '13 at 20:29
up vote 16 down vote accepted

Use formula 16.5.2 from DLMF: $$_3F_2\left(1,1,\frac94;2,2;-1\right)=\int_0^1{_2F_1}\left(1,\frac94;2;-t\right)dt=\frac45\int_0^1\frac{1-(1+t)^{-5/4}}tdt=\\\\\frac{2}{5}\left(8-4\sqrt[4]{8}+\pi-6\ln2+4 \ln (1+\sqrt[4]{2})-4 \arctan\sqrt[4]{2}+2\ln(1+\sqrt2)\right).$$ Some elementary transformations show that this result is equivalent to your conjecture.

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