12
$\begingroup$

$\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { { \left(n! \right) }^{ 2 } }{ { n }^{ 3 }(2n)! } }$ 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 $3 \psi_{1}(1)=\psi_{1}(1/2)$

$\endgroup$
  • 4
    $\begingroup$ 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. $\endgroup$ – Winther Jun 21 '16 at 2:53
  • 1
    $\begingroup$ 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: $$2\psi ^{(1)}\left(\frac{1}{6}\right) - 7\psi^{(1)}\left(\frac{1}{3}\right) - 2\psi ^{(1)}\left(\frac{2}{3}\right) + \psi^{(1)}\left(\frac{5}{6}\right) = 0$$ $\endgroup$ – Winther Jun 21 '16 at 3:20
  • 1
    $\begingroup$ Surprisingly, it's very close to the value of $$\frac{\pi}{e}-\frac{\pi^3}{49}$$ $\endgroup$ – Sophie Agnesi Jun 22 '16 at 7:04
  • $\begingroup$ Agreement to the 8th decimal place is amazingly close !! How did you discover that @Sophie Agnesi ? $\endgroup$ – Bob Kadylo Jun 22 '16 at 12:15
  • $\begingroup$ @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. $\endgroup$ – Sophie Agnesi Jun 23 '16 at 2:02
10
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\sum _{n = 1}^{\infty}{\pars{n!}^{2} \over n^{3}\pars{2n}!}} & = \sum _{n = 1}^{\infty}{1 \over n^{3}}\, {\Gamma\pars{n + 1}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}} = \sum _{n = 1}^{\infty}{1 \over n^{2}}\,\ \overbrace{% {\Gamma\pars{n + 1}\Gamma\pars{n} \over \Gamma\pars{2n + 1}}} ^{\ds{\mathrm{B}\pars{n + 1,n}}} \\[4mm] & = \sum _{n = 1}^{\infty}\ {1 \over n^{2}}\ \overbrace{\int_{0}^{1}x^{n}\pars{1 - x}^{n - 1}\,\dd x} ^{\ds{\mathrm{B}\pars{n + 1,n}}}\ =\ \int_{0}^{1}{1 \over x}\sum_{n = 1}^{\infty} {\bracks{x\pars{1 - x}}^{\, n} \over n^{2}}\,\dd x \\[4mm] & = \int_{0}^{1}{\,\mathrm{Li}_{2}\pars{x\bracks{1 - x}} \over x}\,\dd x = -\int_{0}^{1}\ln\pars{x}\,\mathrm{Li}_{2}'\pars{x\bracks{1 - x}}\pars{1 - 2x} \,\dd x \\[4mm] & = -\int_{0}^{1}\ln\pars{x}\, \braces{-\,{\ln\pars{1 - x\bracks{1 - x}} \over x\pars{1 - x}}} \pars{1 - 2x}\,\dd x \\[4mm] & = \int_{0}^{1}{\ln\pars{x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x - \int_{0}^{1}{\ln\pars{x}\ln\pars{x^{2} - x + 1} \over 1 - x}\,\dd x \\[4mm] & = \underbrace{\int_{0}^{1}{\ln\pars{x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x} _{\ds{\,\mathcal{I}_{1}}}\ -\ \underbrace{\int_{0}^{1}{\ln\pars{1 - x}\ln\pars{x^{2} - x + 1} \over x}\,\dd x} _{\ds{\,\mathcal{I}_{2}}} \\[4mm] & = \,\mathcal{I}_{1} - \,\mathcal{I}_{2}\tag{1} \end{align}


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} \begin{equation} \,\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} \end{equation}
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{equation} \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} \end{equation}
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}

$\endgroup$
  • $\begingroup$ Simply incredible. $\endgroup$ – nbubis Jul 26 '16 at 4:39
  • $\begingroup$ @Felix Marin Direct and clever! (+1) $\endgroup$ – Olivier Oloa Jul 26 '16 at 5:34
  • $\begingroup$ @tired Thanks for holy and for s... $\endgroup$ – Felix Marin Aug 19 '16 at 19:07
  • $\begingroup$ @OlivierOloa Thanks a lot. It was a nightmare. $\endgroup$ – Felix Marin Aug 19 '16 at 19:07
  • $\begingroup$ @nbubis Thanks. A lot of work. $\endgroup$ – Felix Marin Aug 19 '16 at 19:08
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ This is equivalent to the result I had found. Thanks for the confirmation ! $\endgroup$ – Bob Kadylo Jun 22 '16 at 3:55
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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