Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
This might be worth a look: math.upenn.edu/~wilf/Downld.html –  Steve Huntsman May 29 '13 at 20:29
add comment

1 Answer

up vote 14 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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.