In this solution we showed that

$$\sum _{n=1}^{\infty } \frac{4^n H_n}{n^2 {2n\choose n}}=6\ln(2)\zeta(2)+\frac72\zeta(3)\tag1$$

using the identity

$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n\choose n}}$$

My question here is can we prove $(1)$ in a different way using elementary methods? Still, don't let my question restrictions limit your approaches, all approaches are appreciated.

The point of this post (challenge) is to learn different techniques if possible and to make this site more entertaining.

Thank you.

  • $\begingroup$ Introducing this particular structure (namely the central binomial coefficient) by other means than the $\arcsin$ series seems to be pretty hard up to nearly impossible. (+1) nevertheless as I'd like to be proven wrong! $\endgroup$ – mrtaurho Aug 20 at 1:06
  • $\begingroup$ @mrtaurho I agree with you and that what makes it a nice challenge . I managed to finish it using beta function but I am giving the readers sometime to approach it their way. $\endgroup$ – Ali Shadhar Aug 20 at 1:08
  • $\begingroup$ Looking forward to your solution! I tried using the Beta Function aswell and some related generating functions but ran into convergence issues which I'm not sure how to overcome. $\endgroup$ – mrtaurho Aug 20 at 1:43
  • $\begingroup$ @mrtaurho There is a nice uncommon integral representation of beta function. I will post solution soon. $\endgroup$ – Ali Shadhar Aug 20 at 2:09

A sketch (for now). Using the identities $$\int_0^1\frac{x^{n-1}}{\sqrt{1-x}}\,{\rm d}x=\frac{4^n}{n\binom{2n}n}\,\,\,\text{and}\,\,\,\sum_{n\ge1}\frac{H_n}nx^n=\operatorname{Li}_2(x)+\frac12\log^2(1-x)$$ gives $$\sum_{n\ge1}\frac{4^nH_n}{n^2\binom{2n}n}=\sum_{n\ge1}\frac{H_n}n\int_0^1\frac{x^{n-1}}{\sqrt{1-x}}\,{\rm d}x=\frac12\int_0^1\frac{2\operatorname{Li}_2(x)+\log^2(1-x)}{x\sqrt{1-x}}\,{\rm d}x$$ Reflecting and enforcing $\sqrt x\mapsto x$ afterwards yields \begin{align*} \frac12\int_0^1\frac{2\operatorname{Li}_2(x)+\log^2(1-x)}{x\sqrt{1-x}}\,{\rm d}x&=\frac12\int_0^1\frac{2\operatorname{Li}_2(1-x)+\log^2(x)}{(1-x)\sqrt{x}}\,{\rm d}x\\ &=\int_0^1\frac{2\operatorname{Li}_2(1-x^2)+4\log^2(x)}{1-x^2}\,{\rm d}x \end{align*} The latter integral evaluates as $7\zeta(3)$ using the geometric series. For the first integral apply IBP two times to obtain \begin{align*} \int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\,{\rm d}x&=-\left[\frac12\operatorname{Li}_2(1-x^2)\log\left(\frac{1-x}{1+x}\right)\right]_0^1+2\int_0^1\frac{x\log x\log\left(\frac{1-x}{1+x}\right)}{1-x^2}\,{\rm d}x\\ &=-\left[\frac12 x\log x\log^2\left(\frac{1-x}{1+x}\right)\right]_0^1+\int_0^1(1+\log x)\log^2\left(\frac{1-x}{1+x}\right)\,{\rm d}x\\ &=\frac{\pi^2}6+\int_0^1\log x\log^2\left(\frac{1-x}{1+x}\right)\,{\rm d}x \end{align*} I am currently not sure how to approach the remaining integral in an elegant way.

Sidenote: Using the usual series expansion of the logarithm and the integral representation of the harmonic numbers leads to an evaluation of the remaining integral. However, this method is rather unelegant and I will see if I can find something more satisfying.

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  • $\begingroup$ our approaches are some what similar. $\endgroup$ – Ali Shadhar Aug 20 at 2:43
  • $\begingroup$ @AliShather I've had a hard time figuring out the first identity (which is nothing else than the one you cited) :D $\endgroup$ – mrtaurho Aug 20 at 2:45
  • $\begingroup$ @AliShather It seems like continuing with the given integral and trying to get rid of the roots leads to a bunch of standard integrals. $\endgroup$ – mrtaurho Aug 20 at 2:59
  • $\begingroup$ just set $x=\frac{1-y}{1+y}$ in $\int_0^1 \frac{x\ln x\ln\left(\frac{1-x}{1+x}\right)}{1-x^2}dx$ and you will come across integrals similar to mine. $\endgroup$ – Ali Shadhar Aug 20 at 5:00

We use the powerful form of the Beta function presented in the book, (Almost) Impossible Integrals, Sums, and Series, $\displaystyle \int_0^1 \frac{x^{a-1}+x^{b-1}}{(1+x)^{a+b}} dx = \operatorname{B}(a,b)$, (see pages $72$-$73$).

Set $a=b=n$ we have

$$\int_0^1\frac{2x^{n-1}}{(1+x)^{2n}}dx=\frac{\Gamma^2(n)}{\Gamma(2n)}=\frac{2}{n{2n\choose n}}$$

So $$\frac{1}{n{2n\choose n}}=\int_0^1\frac{x^{n-1}}{(1+x)^{2n}}dx=\int_0^1\frac1x\left(\frac{x}{(1+x)^2}\right)^ndx$$

$$\Longrightarrow \sum_{n=1}^\infty\frac{4^nH_n}{n^2{2n\choose n}}=\int_0^1\frac1x\left(\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{4x}{(1+x)^2}\right)^n\right)dx$$


$$\overset{IBP}{=}\int_0^1\frac{2+2x}{x(1-x)}\ln x\ln\left(\frac{1-x}{1+x}\right)dx$$

$$=\int_0^1\left(\frac2x+\frac{4}{1-x}\right)\ln x\ln\left(\frac{1-x}{1+x}\right)dx$$

$$\small{=2\int_0^1\frac{\ln x\ln(1-x)}{x}dx+4\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{1-x}dx}_{1-x\to x}-2\int_0^1\frac{\ln x\ln(1+x)}{x}dx-4\int_0^1\frac{\ln x\ln(1+x)}{1-x}dx}$$

$$=6\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{x}dx}_{\zeta(3)}-2\underbrace{\int_0^1\frac{\ln x\ln(1+x)}{x}dx}_{-\frac34\zeta(3)}-4\underbrace{\int_0^1\frac{\ln x\ln(1+x)}{1-x}dx}_{\zeta(3)-\frac32\ln(2)\zeta(2)}$$


The latter integral is calculated here.

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  • $\begingroup$ (+1) the given identity dodges dealing with nasty integral quite nicely. $\endgroup$ – mrtaurho Aug 20 at 2:54
  • $\begingroup$ @mrtaurho thank you. But unfortunately this method does not work well for series with higher powers of $n$ and the arcsin series is a good fit. $\endgroup$ – Ali Shadhar Aug 20 at 3:21
  • $\begingroup$ What happened to the second term at the IBP part? $\endgroup$ – Diger Aug 20 at 5:44
  • $\begingroup$ @ Diger they simplify into one integral. $\endgroup$ – Ali Shadhar Aug 20 at 9:45

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{4^{n}H_{n} \over n^{2}{2n \choose n}}} = \int_{0}^{4}\sum_{n = 1}^{\infty}{H_{n} \over n{2n \choose n}} \,x^{n - 1}\,\dd x \\[5mm] = &\ \int_{0}^{4}\sum_{n = 1}^{\infty}H_{n}\, x^{n - 1}\,{\Gamma\pars{n} \Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}}\,\dd x \\[5mm] = &\ \int_{0}^{4}\sum_{n = 1}^{\infty}H_{n}\, x^{n - 1}\int_{0}^{1}t^{n - 1} \pars{1 - t}^{n}\,\dd t\,\dd x \\[5mm] = &\ \int_{0}^{4}\int_{0}^{1}\sum_{n = 1}^{\infty}H_{n} \bracks{xt\pars{1 - t}}^{\, n}\,{\dd t\,\dd x \over tx} \\[5mm] = &\ \int_{0}^{4}\int_{0}^{1}\braces{% -\,{\ln\pars{1 - xt\bracks{1-t}} \over 1 - xt\pars{1-t}}} {\dd t\,\dd x \over tx} \\[5mm] = &\ \int_{0}^{1}{2\ln^{2}\pars{\verts{1 - 2t}} + \mrm{Li}_{2}\pars{4\bracks{1 - t})\, t}\over t}\,\dd t \\[5mm] = &\ 2\int_{-1/2}^{1/2}{2\ln^{2}\pars{\verts{2t}} + \mrm{Li}_{2}\pars{1 - 4t^{2}}\over 1 + 2t}\,\dd t \\[5mm] = &\ 4\int_{0}^{1/2}{2\ln^{2}\pars{2t} + \mrm{Li}_{2}\pars{1 - 4t^{2}} \over 1 - 4t^{2}}\,\dd t \\[5mm] = &\ 2\int_{0}^{1}{2\ln^{2}\pars{t} + \mrm{Li}_{2}\pars{1 - t^{2}} \over 1 - t^{2}}\,\dd t \\[5mm] = &\ 4\ \underbrace{\int_{0}^{1}{\ln^{2}\pars{t} \over 1 - t^{2}}\,\dd t} _{\ds{\color{red}{\LARGE\S}:\ {7 \over 4}\,\zeta\pars{3}}}\ +\ 2\, \underbrace{\int_{0}^{1}{\mrm{Li}_{2}\pars{1 - t^{2}} \over 1 - t^{2}}\,\dd t}_{\ds{\color{red}{\LARGE *}:\ {1 \over 2}\,\pi^{2}\ln\pars{2} - {7 \over 4}\,\zeta\pars{3}}} \\[5mm] = &\ \bbx{6\ln\pars{2}\,\zeta\pars{2} + {7 \over 2}\,\zeta\pars{3}} \\ & \end{align}

$\left\{\begin{array}{lcl} \ds{\color{red}{\LARGE\S}} & \ds{:} & \mbox{First} "Partial\ Fraction\ Split.\ \mbox{Next, integrate}\ twice\ \mbox{by parts.} \\[2mm] \ds{\color{red}{\LARGE *}} & \ds{:} & \mbox{After integration by parts, the final expression seems to be a doable and} \\ && \mbox{known integral.} \end{array}\right.$
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