24
$\begingroup$

I have the following infinite sum that can be expressed in terms of the generalized hypergeometric function: $$\sum_{n=1}^\infty\frac{(2n-1)!!\ (2n+1)!!}{4^n\ (n+2)\ (n+2)!^2}=\frac{31}8-4\times{_4F_3}\left(-\frac12,\frac12,1,1;\ 2,2,2;\ 1\right)\\\ \\\approx0.008749644047541935203478962326551903908774780849356243615274...$$ I wonder if it can be expressed in terms of simpler functions and well-known mathematical constants.

$\endgroup$
3
  • 1
    $\begingroup$ Is there any motivation that led you to think that it has a closed form? $\endgroup$ May 26, 2013 at 20:13
  • 1
    $\begingroup$ @sos440 No particular reason to expect it exists. I just wonder if there is one, because it would be nice. $\endgroup$ May 26, 2013 at 20:38
  • $\begingroup$ My answer to this post instantly solves the problem. $\endgroup$ Sep 21, 2020 at 9:49

3 Answers 3

26
$\begingroup$

Ookay... Consider just the hypergeometric function and its closed form.

Take 16.5.2 from DLMF (with $a_0=1/2$, $b_0=2$) and write (I used Mathematica to substitute the special form for the hypergeometric function in the integrand; I don't really know how to do it by hand): $$ F(-1/2,1/2,1,1;2,2,2;1) = \int_0^1 \frac{8 \sqrt{1-t} \left(4-4 \sqrt{1-t}+t\sqrt{1-t}-\log8+3 \log(1+\sqrt{1-t})\right)}{9 \pi t^{3/2}}\,dt. $$

Mathematica can then do the integral in closed form to give $$\frac{1}{9 \pi }8 \left(\frac{70}{3}+\frac{11 i \pi ^2}{4}+\pi \left(-7+\log 512-6\log\left(1+\frac{1+i}{\sqrt{2}}\right)\right)+24 i \mathrm{Li}_2\left(-\frac{1+i}{\sqrt{2}}\right)-24 i \mathrm{Li}_2\left(1-\frac{1+i}{\sqrt{2}}\right)\right),$$ where $\mathrm{Li}_2$ is the polylogarithm.

Of this we take the real part only and do some more FunctionExpand: $$-\frac{56}{9}+\frac{560}{27 \pi }-\frac{16 C}{3 \pi }+\frac{64 \Im\left(\mathrm{Li}_2(1-(-1)^{1/4})\right)}{3 \pi }+\frac{8 \log512}{9}-\frac{8}{3} \log\left(\frac{1}{2}+\left(1+\frac{1}{\sqrt{2}}\right)^2\right)+\frac{\psi_1\left(\frac{1}{8}\right)}{3 \sqrt{2} \pi }+\frac{\psi_1\left(\frac{3}{8}\right)}{3 \sqrt{2} \pi }-\frac{\psi_1\left(\frac{5}{8}\right)}{3 \sqrt{2} \pi }-\frac{\psi_1\left(\frac{7}{8}\right)}{3 \sqrt{2} \pi }, $$ where $\psi_1$ is the polygamma function and $C$ is the Catalan constant.

Now, the polylogarithm term there can be simplified using the identity (DLMF 25.12.6) $$ \mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\frac{\pi^2}{6}-\log x\log(1-x), $$ because $\mathrm{Li}_2((-1)^{1/4})$ is simpler.

After a further FunctionExpand (which gets rid of the polygamma functions also), ComplexExpand to get the real part, and FullSimplify to simplify the expression, the answer is $$ -\frac{56}{9}+\frac{560}{27 \pi }-\frac{32C}{3 \pi }+\frac{16}{3} \log2$$

$\endgroup$
6
  • 7
    $\begingroup$ It is really amazing! I wonder if we ever see the time when CAS would be able to do such simplifications without human aid... $\endgroup$ May 27, 2013 at 3:16
  • 5
    $\begingroup$ Is there any catalog of special values of hypergeometric function where we could submit such discoveries? $\endgroup$ May 27, 2013 at 3:18
  • 1
    $\begingroup$ I did this by (systematic) trial and error: I looked through mathworld and DLMF to see if I could find any identity that would let me simplify ${}_4F_3$ into a ${}_3F_2$ or ${}_2F_1$, and eventually I found 16.5.2 which happens to do the trick with the right choice of which coefficients $a_0,b_0$ to use (with $a_0=1$, $b_0=2$ you get integrals of elliptic integral functions, which is a dead end). Without a CAS this is basically impossible because one's motivation would run out far earlier than one arrives at an answer. $\endgroup$
    – Kirill
    May 27, 2013 at 3:23
  • $\begingroup$ DLMF 16.4, and also mathworld list some special values of generalized hypergeometric functions, especially with argument 1. It would definitely be useful as a lookup table for a CAS (not for a human, given how rarely any given special value appears). $\endgroup$
    – Kirill
    May 27, 2013 at 3:32
  • 1
    $\begingroup$ The hypergeometric function in this question has a complicated closed form in terms of standard functions and $F(\frac32,\frac32,\frac32;\frac52,\frac52;\frac{1+\sqrt{1-x}}{2})$, $0<x\leq 1$, and that hypergeometric is special because the differences between top and bottom parameters are integers, so it is a (repeated) integral of a much simpler hypergeometric. $\endgroup$
    – Kirill
    May 27, 2013 at 3:40
16
$\begingroup$

I can give a partial answer:

My program based on the TranscendentalRecognize algorithm using a wide set of commonly occuring constants after several hours of work discovered the following inequality:

$$\Bigg|\frac{16\ln2}{3}-\frac{32C}{3\pi}+\frac{560}{27\pi}-\frac{56}{9}-{_4F_3}\left(-\frac{1}{2},\frac{1}{2},1,1;2,2,2;1\right)\Bigg|<10^{-1000},$$ where $C$ is the Catalan constant.

I have no idea if the actual difference is exact $0$, or how to (dis-)prove it.

$\endgroup$
3
  • $\begingroup$ The actual difference is $0$, see the other answer! WOW! $\endgroup$
    – Pedro
    May 27, 2013 at 3:13
  • 3
    $\begingroup$ I think this is the correct closed form. $\endgroup$
    – Kirill
    May 27, 2013 at 3:14
  • $\begingroup$ Can your TranscendentalRecognize algorithm find some values for this expression? math.stackexchange.com/questions/2368960/… $\endgroup$
    – Wolfgang
    Jul 24, 2017 at 9:52
3
$\begingroup$

Those calculations are really addictive.. Now, this sum is doable because the term in the sum can be written as a product of a binomial factor, a beta function and a rational function . Let me show how it goes. Firstly we have: \begin{eqnarray} (2n-1)!!&=& 2^n \frac{(n-\frac{1}{2})!}{(-\frac{1}{2})!}\\ (2n+1)!!&=& 2^n \frac{(n+\frac{1}{2})!}{(+\frac{1}{2})!} \end{eqnarray} Therefore the sum in question reads: \begin{eqnarray} &&\sum\limits_{n=1}^\infty \frac{(2n-1)!! (2n+1)!!}{4^n (n+2)[(n+2)!]^2}=\\ &&\sum\limits_{n=1}^\infty \frac{(n-\frac{1}{2})!}{(-\frac{1}{2})!} \frac{(n+\frac{1}{2})!}{(+\frac{1}{2})!} \cdot \frac{1}{(n+2)[(n+2)!]^2}\\ &&=\sum\limits_{n=1}^\infty \binom{n-\frac{1}{2}}{n} \cdot \frac{(n+\frac{1}{2})! (+\frac{1}{2})!}{[(+\frac{1}{2})!]^2(n+2)!} \cdot \frac{1}{(n+2)^2 (n+1)}\\ &&= \sum\limits_{n=1}^\infty (-1)^n \binom{-\frac{1}{2}}{n} \cdot \frac{1}{[(+\frac{1}{2})!]^2} \cdot \int\limits_0^1 t^{n+\frac{1}{2}} (1-t)^{\frac{1}{2}} dt \cdot \frac{1}{(n+2)^2 (n+1)}\\ &&= \frac{16}{3 \pi} \sum\limits_{n=1}^\infty (-1)^n\binom{\frac{3}{2}}{n+2} \cdot \int\limits_0^1 t^{n+\frac{1}{2}} (1-t)^{\frac{1}{2}} dt \cdot \underbrace{\frac{1}{(n+2)}}_{\int\limits_0^1 \theta^{n+1} d\theta}\\ &&=\frac{16}{3 \pi} \int \limits_0^1 t^{\frac{1}{2}} (1-t)^{\frac{1}{2}} \cdot \int\limits_0^1 \theta\underbrace{\left\{ \sum\limits_{n=1}^\infty \binom{\frac{3}{2}}{n+2} (-\theta t)^n\right\}}_{\frac{-3 t^2 \theta ^2+12 t \theta -8 t \theta \sqrt{1-t \theta }+8 \sqrt{1-t \theta }-8}{8 t^2 \theta ^2}} d\theta dt\\ &&= \int\limits_0^1 (1-t)^{\frac{1}{2}} \cdot \\ &&\left( \frac{\left(-9 t-32 \sqrt{1-t}+72\right) t+32 \left(4 \sqrt{1-t}-4+\log (8)\right)-48 \log (t)-96 \tanh ^{-1}\left(\sqrt{1-t}\right)}{9 \pi t^{3/2}}\right) dt =\\ &&\frac{31}{8}-\frac{128}{27 \pi} + \int\limits_0^1 (1-t)^{\frac{1}{2}}\cdot\left(\frac{32 \left(4 \sqrt{1-t}-4+\log (8)\right)-48 \log (t)-96 \tanh ^{-1}\left(\sqrt{1-t}\right)}{9 \pi t^{3/2}} \right)dt=\\ &&\frac{1303}{72}-\frac{1664}{27 \pi} + \int\limits_0^1 (1-t)^{\frac{1}{2}}\cdot \left( \frac{-48 \log (t)-96 \tanh ^{-1}\left(\sqrt{1-t}\right)+32 \log (8)}{9 \pi t^{3/2}}\right) dt=\\ &&\frac{1303}{72}-\frac{1664}{27 \pi} + \frac{32 (4 C-2+\pi -\pi \log (4))}{3 \pi }=\\ &&\frac{128 C}{3 \pi }+\frac{2071}{72}-\frac{2240}{27 \pi }-\frac{32 \log (4)}{3} \end{eqnarray} which matches the answer given by Kirill.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .