# Seeking closed form for infinite sum $\sum \limits_{ n=1 }^{ \infty }{ \frac { { \left(n! \right) }^{ 2 } }{ { n }^{ 3 }(2n)! } }$

$\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { \left(n! \right) }^{ 2 } }{ { n }^{ 3 }(2n)! } }$ is approximately $.5229461921333351$ but I've been assured that there is a closed form for this infinite series that involves the TriGamma and Zeta functions but it is not unique. I found one candidate result which is: $$\frac{ \pi \sqrt{3}}{72} \Big[3 \psi_{1} \left(\frac{1}{3} \right) - \psi_{1} \left(\frac{5}{6} \right) \Big]- \frac{4}{3} \zeta(3)$$

My friend (tormentor) has challenged me to find other such expressions. I know how to transform the look of such expressions if the Gamma function is involved by using functional equations like $\Gamma(x) = \Gamma(1+x)/x$ so I suspect he knows a similar result for the PolyGamma functions. I just can not find the information. By trial and error, I did find that $3 \psi_{1}(1)=\psi_{1}(1/2)$

• Probably not the closed-form you want, but using the series for $\arcsin^2(x)$ we can write the sum as the integral: $\sum_{n=1}^\infty\frac{1}{n^3{2n\choose n}} = \int_0^{\frac{1}{2}}\frac{4\arcsin^2\left(x\right)}{x}{\rm d}x$. Putting this integral through Mathematica gives (after simplifying using these identities) gives one possible answer $$-\frac{2}{3} i \pi \text{Li}_2\left(\frac{1+i\sqrt{3}}{2}\right)-\frac{4 \zeta(3)}{3}+\frac{i \pi ^3}{54}$$ Can probably be simplified further. – Winther Jun 21 '16 at 2:53
• By playing around with it I found an identity that you can try to prove which can be used to come up with a different form for the answer you have found: $$2\psi ^{(1)}\left(\frac{1}{6}\right) - 7\psi^{(1)}\left(\frac{1}{3}\right) - 2\psi ^{(1)}\left(\frac{2}{3}\right) + \psi^{(1)}\left(\frac{5}{6}\right) = 0$$ – Winther Jun 21 '16 at 3:20
• Surprisingly, it's very close to the value of $$\frac{\pi}{e}-\frac{\pi^3}{49}$$ – Sophie Agnesi Jun 22 '16 at 7:04
• Agreement to the 8th decimal place is amazingly close !! How did you discover that @Sophie Agnesi ? – Bob Kadylo Jun 22 '16 at 12:15
• @BobKadylo I build a program for that. I'm able to get it at least to 26th decimal place but it's only a good approximation, not a real closed-form. – Sophie Agnesi Jun 23 '16 at 2:02


1. $\ds{\large\,\mathcal{I}_{1} =\ ?}$. \begin{align} \,\mathcal{I}_{1} & = \begin{array}{|c|}\hline\mbox{}\\ \ds{\quad% \int_{0}^{1}{\ln\pars{x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x\quad} \\ \mbox{}\\ \hline \end{array} = \int_{0}^{1}\ln\pars{x}\ln\pars{1 + x^{3} \over 1 + x}\,{\dd x \over x} \\[4mm] & = \int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x^{3}} \over x}\,\dd x - \int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over x}\,\dd x \\ & = {1 \over 9}\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over x}\,\dd x - \int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over x}\,\dd x = -\,{8 \over 9} \int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over x}\,\dd x \\[4mm] & = {8 \over 9} \int_{0}^{-1}\bracks{-\,{\ln\pars{1 - x} \over x}}\,\ln\pars{-x}\,\dd x = {8 \over 9} \int_{0}^{-1}\,\mathrm{Li}_{2}'\pars{x}\ln\pars{-x}\,\dd x = -\,{8 \over 9} \int_{0}^{-1}{\,\mathrm{Li}_{2}\pars{x} \over x}\,\dd x \\[4mm] & = -\,{8 \over 9}\int_{0}^{-1}\,\mathrm{Li}_{3}'\pars{x}\,\dd x = -\,{8 \over 9}\,\mathrm{Li}_{3}\pars{-1} = -\,{8 \over 9}\pars{\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{3}} - \sum_{n = 1}^{\infty}{1 \over n^{3}} + \sum_{n = 1}^{\infty}{1 \over \pars{2n}^{3}}} \\[4mm] & = {2 \over 3}\sum_{n = 1}^{\infty}{1 \over n^{3}} \end{align} $$\,\mathcal{I}_{1} = \begin{array}{|c|}\hline\mbox{}\\ \ds{\quad% \int_{0}^{1}{\ln\pars{x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x\quad} \\ \mbox{}\\ \hline \end{array} = \begin{array}{|c|}\hline\mbox{}\\ \ds{\quad{2 \over 3}\,\zeta\pars{3}\quad} \\ \mbox{}\\ \hline \end{array}\tag{2}$$
2. $\ds{\large\,\mathcal{I}_{2} =\ ?}$. \begin{align} \,\mathcal{I}_{2} & = \begin{array}{|c|}\hline\mbox{}\\ \ds{\quad% \int_{0}^{1}{\ln\pars{1 - x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x\quad} \\ \mbox{}\\ \hline \end{array} = -\int_{0}^{1}\,\mathrm{Li}_{2}'\pars{x}\ln\pars{x^{2} - x + 1}\,\dd x \\[4mm] & = \int_{0}^{1}\,\mathrm{Li}_{2}\pars{x}\,{2x - 1 \over x^{2} - x + 1}\,\dd x \end{align} The roots of $\ds{x^{2} - x + 1}$ are given by $\ds{r = {1 + \root{3}\ic \over 2} = \expo{\pi\ic/3}}$ and $\ds{\ol{r}}$ such that \begin{align} {2x - 1 \over x^{2} - x + 1} & = 2\pars{{x - 1/2 \over x - r} - {x - 1/2 \over x - \ol{r}}}\,{1 \over r - \ol{r}} = 2\,{1 \over 2\ic\,\Im\pars{r}}\,2\ic\,\Im\pars{x - 1/2 \over x - r} \\[5mm] & = {2 \over \root{3}}\,\Im\pars{2r - 1 \over x - r} = 2\,\Re\pars{1 \over x - r} \end{align}
\begin{align} \,\mathcal{I}_{2} & = 2\,\Re\bracks{\int_{0}^{1}{\,\mathrm{Li}_{2}\pars{x} \over x - r}\,\dd x} = -2\,\Re\bracks{\int_{0}^{1} {\,\mathrm{Li}_{2}\pars{x} \over 1 - x/r}\,{\dd x \over r}} = -2\,\Re\bracks{\int_{0}^{\ol{r}} {\,\mathrm{Li}_{2}\pars{rx} \over 1 - x}\,\dd x} \\[5mm] & = 2\,\Re\bracks{\left.\vphantom{\LARGE A}\ln\pars{1 - x}\,\mathrm{Li}_{2}\pars{rx}\, \right\vert_{\ 0}^{\ \ol{r}} - \int_{0}^{\ol{r}}\ln\pars{1 - x}\,\mathrm{Li}_{2}'\pars{rx}r\,\dd x} \\[5mm] & = 2\,\Re\int_{0}^{\ol{r}}\ln\pars{1 - x}\,{\ln\pars{1 - rx} \over x}\,\dd x \\[5mm] & = \Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - x} \over x}\,\dd x + \Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - rx} \over x}\,\dd x - \Re\int_{0}^{\ol{r}}\ln^{2}\pars{1 - x \over 1 - rx}\,{\dd x \over x} \\[5mm] & = \Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - x} \over x}\,\dd x + \int_{0}^{1}{\ln^{2}\pars{1 - x} \over x}\,\dd x - \Re\int_{0}^{\ol{r}}\ln^{2}\pars{1 - x \over 1 - rx}\,{\dd x \over x} \end{align} In the last integral, we'll make the sub$\ds{\ldots {1 - rx \over 1 - x} \mapsto x}$ such that the whole integration is reduced to: \begin{align} \,\mathcal{I}_{2} & = \Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - x} \over x}\,\dd x + \int_{0}^{1}{\ln^{2}\pars{1 - x} \over x}\,\dd x - \bracks{% \Re\int_{1}^{0}{\ln^{2}\pars{x} \over -1 + x}\,\dd x - \Re\int_{1}^{0}{\ln^{2}\pars{x} \over -r + x}\,\dd x} \\[5mm] & = \Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - x} \over x}\,\dd x + \Re\int_{0}^{\ol{r}}{\ln^{2}\pars{rx} \over 1 - x}\,\dd x \end{align} The remaining integrals can be evaluated by 'standard means'. For instance $$\left\lbrace\begin{array}{rcl} \ds{\int{\ln^{2}\pars{1 - x} \over x}\,\dd x} & \ds{=} & \ds{\ln^{2}\pars{1 - x}\ln\pars{x} + 2\ln\pars{1 - x} \,\mathrm{Li}_{2}\pars{1 - x} - 2\,\mathrm{Li}_{3}\pars{1 - x}} \\[3mm] \ds{\int{\ln^{2}\pars{rx} \over 1 - x}\,\dd x} & \ds{=} & \ds{-\ln\pars{1 - x}\ln^{2}\pars{rx} - 2\ln\pars{rx} \,\mathrm{Li}_{2}\pars{x} + 2\,\mathrm{Li}_{3}\pars{x}} \end{array}\right.$$ $$\left\lbrace\begin{array}{rcl} \ds{\Re\int_{0}^{\ol{r}}{\ln^{2}\pars{1 - x} \over x}\,\dd x} & \ds{=} & \ds{-\,{2\pi \over 3}\,\Im\,\mathrm{Li}_{2}\pars{r} - 2\Re\,\mathrm{Li}_{3}\pars{r} + 2\ \overbrace{\mathrm{Li}_{3}\pars{1}}^{\ds{\zeta\pars{3}}}} \\[3mm] \ds{\Re\int_{0}^{\ol{r}}{\ln^{2}\pars{rx} \over 1 - x}\,\dd x} & \ds{=} & \ds{2\,\Re\,\mathrm{Li}_{3}\pars{\ol{r}}} \end{array}\right.$$
With these results: $$\begin{array}{|c|}\hline\mbox{}\\ \ds{\quad\,\mathcal{I}_{2} = 2\zeta\pars{3} - {2 \pi \over 3}\,\Im\,\mathrm{Li}_{2}\pars{r}\,,\qquad r = \half + {\root{3} \over 2}\,\ic = \expo{\pi\ic/3}\quad} \\ \mbox{}\\ \hline \end{array}\tag{3}$$
With $\ds{\pars{1}, \pars{2}\ \mbox{and}\ \pars{3}}$, we'll arrive to the $\ds{\underline{final\ result}}$: $$\begin{array}{|rcl|}\hline\mbox{}\\ \ds{\quad\color{#f00}{\sum _{n = 1}^{\infty}{\pars{n!}^{2} \over n^{3}\pars{2n}!}}} & \ds{=} & \ds{\color{#f00}{{2 \pi \over 3}\,\Im\,\mathrm{Li}_{2}\pars{r} - {4 \over 3 }\,\zeta\pars{3}}}\ds{\,\, \approx\,\, 0.5229\quad} \\[3mm] \ds{r} & \ds{\equiv} & \ds{\expo{\pi\ic/3} = \half + {\root{3} \over 2}\,\ic} \\ \mbox{}\\ \hline \end{array}$$

In order to see the relation with the Trigamma Function $\ds{\Psi\, '}$, we can exploit the 'symmetries' of $\ds{\expo{\pi\ic/3}}$. Namely,

\begin{align} \Im\,\mathrm{Li}_{2}\pars{r} & = \Im\,\mathrm{Li}_{2}\pars{\expo{\pi\ic/3}} = \Im\sum_{n = 1}^{\infty}{\expo{n\pi\ic/3} \over n^{2}} = \Im\sum_{n = 0}^{\infty}\bracks{% {\expo{\pars{3n + 1}\pi\ic/3} \over \pars{3n + 1}^{2}} + {\expo{\pars{3n + 2}\pi\ic/3} \over \pars{3n + 2}^{2}} + {\expo{\pars{3n + 3}\pi\ic/3} \over \pars{3n + 3}^{2}}} \\[5mm] & = {\root{3} \over 2}\bracks{% \sum_{n = 0}{\pars{-1}^{n} \over \pars{3n + 1}^{2}} + {\pars{-1}^{n} \over \pars{3n + 2}^{2}}} \end{align} However, \begin{align} \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{3n + z}^{2}} & = \sum_{n = 0}^{\infty}{1 \over \pars{6n + z}^{2}} - \sum_{n = 0}^{\infty}{1 \over \pars{6n + z + 2}^{2}} \\[5mm] & = {1 \over 36}\sum_{n = 0}^{\infty}{1 \over \pars{n + z/6}^{2}} - {1 \over 36}\sum_{n = 0}^{\infty}{1 \over \pars{n + z/6 + 1/3}^{2}} \\[5mm] & = {1 \over 36}\,\Psi\, '\pars{z \over 6} - {1 \over 36}\,\Psi\, '\pars{{z \over 6} + {1 \over 3}} \end{align} Then, \begin{align} \Im\,\mathrm{Li}_{2}\pars{r} & = {\root{3} \over 72}\bracks{% \Psi\, '\pars{1 \over 6} -\ \underbrace{\Psi\,'\pars{\half}}_{\ds{\pi^{2} \over 2}}\ +\ \Psi\, '\pars{1 \over 3} - \Psi \, '\pars{2 \over 3}} \end{align}

• Simply incredible. – nbubis Jul 26 '16 at 4:39
• @Felix Marin Direct and clever! (+1) – Olivier Oloa Jul 26 '16 at 5:34
• @tired Thanks for holy and for s... – Felix Marin Aug 19 '16 at 19:07
• @OlivierOloa Thanks a lot. It was a nightmare. – Felix Marin Aug 19 '16 at 19:07
• @nbubis Thanks. A lot of work. – Felix Marin Aug 19 '16 at 19:08

There is a few identities you could use $$\psi ^{(1)}\left(z\right)=\zeta \left(2,z\right)$$ $$\psi ^{(1)}(1-z)+\psi ^{(1)}(z)=\pi ^2 \csc ^2(\pi z)$$ $$\psi ^{(1)}(z+1)=\psi ^{(1)}(z)-\frac{1}{z^2}$$See here.

The series can be expressed in the form \begin{align} \sum_{n=1}^{\infty} \frac{(n!)^2}{n^3 \, (2n)!} &= \frac{1}{2} \, {}_{4}F_{3}\left(1,1,1,1; 2,2,\frac{3}{2}; \frac{1}{4} \right) \\ &= \frac{1}{2} \, \left[ - \frac{8}{3} \, \zeta(3) + \frac{\pi}{4 \sqrt{3}} \, \psi^{(1)}\left(\frac{1}{3}\right) - \frac{\pi}{12 \sqrt{3}} \, \psi^{(1)}\left(\frac{5}{6}\right) \right] \end{align}

• This is equivalent to the result I had found. Thanks for the confirmation ! – Bob Kadylo Jun 22 '16 at 3:55

I am French and I research expressions of trigamma zeta 3 Catalan ... it's very difficult but for your example, I propose $\frac{\pi\sqrt{3}\Psi\left(1,\frac{1}{3}\right)}{9}-\frac{2\sqrt{3}\pi^3}{27}-4\frac{\zeta(3)}{3}$. So I confirm the expression with MAPLE.