12

How to evaluate

$$S=\sum_{n=1}^\infty\frac{(-4)^nH_{n-1}^3}{{2n\choose n}n^2}\ ?$$ where $H_n$ is the harmonic number.

This problem was posted on a Facebook group with no answer so I am not sure if there is a closed form but seems an interesting problem to try.

I know that from here we have

$$\small{\sum_{n=1}^\infty H_{n-1}^3x^{n-1}= \frac{\operatorname{Li}_3(x)+3\operatorname{Li}_3(1-x)+\frac32\ln x\ln^2(1-x)-3\zeta(2)\ln(1-x)-\ln^3(1-x)-3\zeta(3)}{1-x}}$$

and from here we have

$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2{2n\choose n}}$$

My question is can we exploit the two identities above to get $S$ or is there a different way to get $S$?

Thank you,

I am tagging "integration" as most series can be converted to integrals.

Ali Shadhar
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  • A possible way is to make use of something like $$\sum_{n=1}^\infty f(n)g(n)=\frac{1}{2\pi}\int_0^{2\pi}\sum_{m=1}^\infty e^{imt}f(m)\sum_{n=1}^\infty e^{-int}g(n),dt$$ but the integral seems too intimidating to look at –  Dec 27 '19 at 23:01
  • Thanks but can you find the integral representation of the sum? – Ali Shadhar Dec 28 '19 at 06:54
  • Oh I think David H's treatment is better –  Dec 28 '19 at 08:53

2 Answers2

5

Note: This answer is incomplete. An integral representation for the series is derived below, but it remains to be seen if that integral can be evaluated in turn.

Define the function $f:\mathbb{R}\rightarrow\mathbb{R}$ via the definite integral

$$\begin{align} \mathcal{S} &:=\sum_{n=1}^{\infty}\frac{H_{n-1}^{3}\left(-4\right)^{n}}{n^{2}\binom{2n}{n}}.\tag{1}\\ \end{align}$$

The $n$-th central binomial coefficient is defined through the factorial function for all nonnegative integers by

$$\binom{2n}{n}:=\frac{\left(2n\right)!}{\left(n!\right)^{2}};~~~\small{n\in\mathbb{Z}_{\ge0}}.$$

The reciprocals of the central binomial coefficients can be written in terms of the beta function. For each positive integer $n\in\mathbb{N}$, we have

$$\begin{align} \frac{1}{\binom{2n}{n}} &=\frac{\left(n!\right)^{2}}{\left(2n\right)!}\\ &=\frac{\left[\Gamma{\left(n+1\right)}\right]^{2}}{\Gamma{\left(2n+1\right)}}\\ &=\frac{\left[n\,\Gamma{\left(n\right)}\right]^{2}}{2n\,\Gamma{\left(2n\right)}}\\ &=\frac{n}{2}\cdot\frac{\left[\Gamma{\left(n\right)}\right]^{2}}{\Gamma{\left(2n\right)}}\\ &=\frac{n}{2}\mathrm{B}{\left(n,n\right)}.\\ \end{align}$$

Recall the definition of the beta function via the integral representation

$$\mathrm{B}{\left(a,b\right)}:=\int_{0}^{1}\mathrm{d}t\,t^{a-1}\left(1-t\right)^{b-1};~~~\small{a>0\land b>0}.$$

Using the technique of summing under the integral we can convert the infinite series $\mathcal{S}$ into a definite integral. We obtain

$$\begin{align} \mathcal{S} &=\sum_{n=1}^{\infty}\frac{H_{n-1}^{3}\left(-4\right)^{n}}{n^{2}\binom{2n}{n}}\\ &=-2\sum_{n=1}^{\infty}\frac{H_{n-1}^{3}\left(-4\right)^{n-1}}{n}\cdot\frac{2}{n\binom{2n}{n}}\\ &=-2\sum_{n=1}^{\infty}\frac{H_{n-1}^{3}\left(-4\right)^{n-1}}{n}\mathrm{B}{\left(n,n\right)}\\ &=-2\sum_{n=1}^{\infty}\frac{H_{n-1}^{3}\left(-4\right)^{n-1}}{n}\int_{0}^{1}\mathrm{d}t\,t^{n-1}\left(1-t\right)^{n-1}\\ &=-2\sum_{n=0}^{\infty}\frac{H_{n}^{3}\left(-4\right)^{n}}{n+1}\int_{0}^{1}\mathrm{d}t\,t^{n}\left(1-t\right)^{n}\\ &=-2\sum_{n=0}^{\infty}\frac{H_{n}^{3}\left(-1\right)^{n}}{n+1}\int_{0}^{1}\mathrm{d}t\,\left[4t\left(1-t\right)\right]^{n}\\ &=-4\sum_{n=0}^{\infty}\frac{H_{n}^{3}\left(-1\right)^{n}}{n+1}\int_{0}^{\frac12}\mathrm{d}t\,\left[4t\left(1-t\right)\right]^{n};~~~\small{symmetry}\\ &=-4\sum_{n=0}^{\infty}\frac{H_{n}^{3}\left(-1\right)^{n}}{n+1}\int_{0}^{1}\mathrm{d}x\,\frac{x^{n}}{4\sqrt{1-x}};~~~\small{\left[t=\frac{1-\sqrt{1-x}}{2}\right]}\\ &=-\sum_{n=0}^{\infty}\int_{0}^{1}\mathrm{d}x\,\frac{H_{n}^{3}\left(-x\right)^{n}}{\sqrt{1-x}}\cdot\frac{1}{n+1}\\ &=-\sum_{n=0}^{\infty}\int_{0}^{1}\mathrm{d}x\,\frac{H_{n}^{3}\left(-x\right)^{n}}{\sqrt{1-x}}\int_{0}^{1}\mathrm{d}y\,y^{n}\\ &=-\sum_{n=0}^{\infty}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{H_{n}^{3}\left(-xy\right)^{n}}{\sqrt{1-x}}\\ &=-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\sum_{n=0}^{\infty}\frac{H_{n}^{3}\left(-xy\right)^{n}}{\sqrt{1-x}}\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{1}{\sqrt{1-x}}\int_{0}^{1}\mathrm{d}y\,\sum_{n=1}^{\infty}H_{n}^{3}\left(-xy\right)^{n}\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{1}{x\sqrt{1-x}}\int_{-x}^{0}\mathrm{d}t\,\sum_{n=1}^{\infty}H_{n}^{3}t^{n};~~~\small{\left[y=-x^{-1}t\right]}.\tag{2}\\ \end{align}$$

It can be demonstrated that

$$\begin{align} \sum_{n=1}^{\infty}H_{n}^{3}z^{n} &=-\frac{3S_{1,2}{\left(z\right)}-\operatorname{Li}_{3}{\left(z\right)}+3\ln{\left(1-z\right)}\operatorname{Li}_{2}{\left(z\right)}+\ln^{3}{\left(1-z\right)}}{\left(1-z\right)};~~~\small{z\in\left(-1,1\right)}.\tag{3}\\ \end{align}$$

Substituting this expression for the infinite series in the last line of $(3)$ above, and then using integration by parts to integrate the various terms of the inner-integral, we arrive at a single-variable definite integral representation for $\mathcal{S}$:

$$\begin{align} \mathcal{S} &=-\int_{0}^{1}\mathrm{d}x\,\frac{1}{x\sqrt{1-x}}\int_{-x}^{0}\mathrm{d}t\,\sum_{n=1}^{\infty}H_{n}^{3}t^{n}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{x\sqrt{1-x}}\int_{-x}^{0}\mathrm{d}t\,\frac{3S_{1,2}{\left(t\right)}-\operatorname{Li}_{3}{\left(t\right)}+3\ln{\left(1-t\right)}\operatorname{Li}_{2}{\left(t\right)}+\ln^{3}{\left(1-t\right)}}{\left(1-t\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{x\sqrt{1-x}}\int_{-x}^{0}\mathrm{d}t\,\bigg{[}\frac{3S_{1,2}{\left(t\right)}}{1-t}-\frac{\operatorname{Li}_{3}{\left(t\right)}}{1-t}+\frac{3\ln{\left(1-t\right)}\operatorname{Li}_{2}{\left(t\right)}}{1-t}+\frac{\ln^{3}{\left(1-t\right)}}{1-t}\bigg{]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{x\sqrt{1-x}}\bigg{[}\int_{-x}^{0}\mathrm{d}t\,\frac{3S_{1,2}{\left(t\right)}}{1-t}-\int_{-x}^{0}\mathrm{d}t\,\frac{\operatorname{Li}_{3}{\left(t\right)}}{1-t}\\ &~~~~~+\int_{-x}^{0}\mathrm{d}t\,\frac{3\ln{\left(1-t\right)}\operatorname{Li}_{2}{\left(t\right)}}{1-t}+\int_{-x}^{0}\mathrm{d}t\,\frac{\ln^{3}{\left(1-t\right)}}{1-t}\bigg{]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{x\sqrt{1-x}}\bigg{[}-3\operatorname{Li}_{1}{\left(-x\right)}\,S_{1,2}{\left(-x\right)}+3\int_{-x}^{0}\mathrm{d}t\,\frac{\ln^{3}{\left(1-t\right)}}{2t}\\ &~~~~~+\operatorname{Li}_{1}{\left(-x\right)}\operatorname{Li}_{3}{\left(-x\right)}+\int_{-x}^{0}\mathrm{d}t\,\frac{\operatorname{Li}_{1}{\left(t\right)}\operatorname{Li}_{2}{\left(t\right)}}{t}\\ &~~~~~+\frac32\operatorname{Li}_{1}{\left(-x\right)}^{2}\operatorname{Li}_{2}{\left(-x\right)}-\frac32\int_{-x}^{0}\mathrm{d}t\,\frac{\ln^{3}{\left(1-t\right)}}{t}\\ &~~~~~+\frac14\ln^{4}{\left(1+x\right)}\bigg{]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{x\sqrt{1-x}}\bigg{[}3\ln{\left(1+x\right)}\,S_{1,2}{\left(-x\right)}-\ln{\left(1+x\right)}\operatorname{Li}_{3}{\left(-x\right)}\\ &~~~~~+\frac32\ln^{2}{\left(1+x\right)}\operatorname{Li}_{2}{\left(-x\right)}+\frac14\ln^{4}{\left(1+x\right)}-\frac12\operatorname{Li}_{2}{\left(-x\right)}^{2}\bigg{]}.\tag{4}\\ \end{align}$$

David H
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    Thanks David for the efforts I really appreciate it (+1). At least you got the integral representation. I'll try to work on the rest and see what I get. – Ali Shadhar Dec 28 '19 at 08:37
  • What $S_{1,2}(x)$ stands for ? – Ali Shadhar Dec 28 '19 at 14:27
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    @AliShather $S_{1,2}$ is a Nielsen generalized polylogarithm, and is defined by the integral representation seen here. – David H Dec 28 '19 at 14:32
  • I see now, so $$S_{1,2}(x)=\frac12\int_0^1\frac{\ln^2(1-xt)}{t}dt=\frac12\int_0^x\frac{\ln^2(1-y)}{y}dy$$

    $$=\zeta(2)\ln(1-x)-\frac12\ln x\ln^2(1-x)-\ln(1-x)\operatorname{Li}_2(x)-\operatorname{Li}_3(1-x)+\zeta(3)$$

     – Ali Shadhar Dec 28 '19 at 14:53
  • @AliShather Exactly. More generally, for any positive integer $p$ the term $S_{1,p}(z)$ is reducible to a sum of standard polylogarithms. – David H Dec 28 '19 at 15:02
  • yes, also $S_{1,2}(x)$ can be written as $\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}x^n$ as we know that $\frac12\ln^2(1-y)=\sum_{n=1}^\infty \frac{H_{n-1}}{n}y^n$. – Ali Shadhar Dec 28 '19 at 15:09
3

For sake of completing my comment above, I hereby present my unsuccessful attempt:

Use the formula I commented, which is easily provable from the Kronecker delta function $\delta_{mn}=\frac{1}{2\pi}\int_0^{2\pi}e^{\pm i(m-n)t}dt$, and use the two formulae the OP provided, we get: \begin{align*} S&=\frac{1}{2\pi}\int_0^{2\pi}\left(\sum_{m=1}^\infty\frac{\left(4e^{-it}\right)^m}{m^2\binom{2m}{m}}\right)\left(\sum_{n=1}^\infty\left(-e^{it}\right)^nH_{n-1}^3\right)dt\\ &=-\frac{1}{\pi}\int_0^{2\pi}\arcsin^2\left(e^{-\frac{it}{2}}\right)e^{it}\frac{F\left(-e^{it}\right)}{1+e^{it}}dt \end{align*} where $$F(x)=\operatorname{Li}_3(x)+3\operatorname{Li}_3(1-x)+\frac32\ln x\ln^2(1-x)-3\zeta(2)\ln(1-x)-\ln^3(1-x)-3\zeta(3)$$ Then, by a variable change $x=e^{\frac{it}{2}}$ and noting that the integrand is an even function with respect to $x$, the integral reduces to \begin{align*} S&=\frac{4}{\pi}\Im\int_0^1x\arcsin^2\left(\frac{1}{x}\right)\frac{F\left(-x^2\right)}{1+x^2}dx\\ &=-2\int_0^1\log\left(\frac{1+\sqrt{1-x}}{\sqrt{x}}\right)\frac{F\left(-x\right)}{1+x}dx \end{align*} as confirmed by Mathematica numerically.