Evaluating $\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}$

First, we prove a lemma on the integral representation of $(H_n)^2$. $$I_n=\int_0^1\left(nx^{n-1}\ln^2(1-x)-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)=(H_n)^2$$

Let's prove by induction. $\displaystyle I_0=-\int_0^1\frac{\ln x}{1-x}dx=\zeta(2)=\zeta(2)+(H_0)^2$.\ Assume the equation holds for $n-1$, $$\begin{aligned} I_n&=\int_0^1\left(2(x^n-1)\frac{\ln(1-x)}{1-x}-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)\\ &=I_{n-1}+\int_0^1\left(2(x^n-x^{n-1})\frac{\ln(1-x)}{1-x}-\frac{(x^n-x^{n-1})\ln x}{1-x}\right)d x\\ &=(H_{n-1})^2+\int_0^1\left(-2x^{n-1}\ln(1-x)+x^{n-1}\ln x\right)d x\\ &=\left(H_n-\frac1n\right)^2-\frac1{n^2}+2\cdot\frac{H_n}n=(H_n)^2 \end{aligned}$$ Result Therefore, and by integrating $\displaystyle\sum_{n=1}^\infty\frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1$ from $0$ with respect to $x$, we have $$\begin{aligned} S&=\sum_{n=1}^\infty\frac1n\frac{\binom{2n}n}{4^n}\left(\int_0^1\left(nx^{n-1}\ln^2(1-x)-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)\right)\\ &=\int_0^1\left(\frac{1}{x\sqrt{1-x}}-\frac1x\right)\ln^2(1-x)d x-\int_0^12\ln\frac{2}{1+\sqrt{1-x}}\frac{\ln x}{1-x}d x-2\ln2\zeta(2)\\ &=I_1-I_2-2\ln2\zeta(2) \end{aligned}$$ $I_1=12\zeta(3)$ can be easily deduced by substitution $x\mapsto 1-x^2$. $-2\ln2\zeta(2)+\frac32\zeta(3)$, the value of $I_2$, can also be deduced by the same substitution. By combining these results, $S=\frac{21}2\zeta(3)$.

we have $\quad\displaystyle\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1 \quad$ divide both sides by $x$ then integrate , we get

$$\quad\displaystyle\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=-2 \tanh^{-1}{\sqrt{1-x}}-\ln x+c $$
let $x=0,\ $ we get $C=2\ln2$

then $\quad\displaystyle\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=\color{orange}{-2 \tanh^{-1}{\sqrt{1-x}}-\ln x+2\ln2}$

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also we have $\displaystyle -\int_0^1x^{n-1}\ln(1-x)\ dx=\frac{H_n}{n} \tag{1}$ multiply both sides by $\displaystyle \frac{\binom{2n}n}{n4^n}\quad$ then take the sum, \begin{align} \sum_{n=1}^\infty \frac{H_n}{n^2}\frac{\binom{2n}n}{4^n}&=-\int_0^1\frac{\ln(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^ndx\\ &=-\int_0^1\frac{\ln(1-x)}{x}\left(\color{orange}{-2\tanh^{-1}{\sqrt{1-x}}-\ln x+2\ln2}\right)dx\\ &=\small{2\int_0^1\frac{\ln(1-x)\tanh^{-1}{\sqrt{1-x}}}{x}dx+\int_0^1\frac{\ln x\ln(1-x)}{x}dx-2\ln2\int_0^1\frac{\ln(1-x)}{x}dx}\\ &=2\int_0^1\frac{\ln x\tanh^{-1}{\sqrt{x}}}{1-x}dx+\int_0^1\frac{\ln x\ln(1-x)}{x}dx-2\ln2\int_0^1\frac{\ln x}{1-x}dx\\ &=8\int_0^1 \frac{x\ln x\tanh^{-1}x}{1-x^2}\ dx+\zeta(3)+2\ln2\zeta(2)\\ &=\color{blue}{8I+\zeta(3)+2\ln2\zeta(2)} \end{align}

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differentiate $(1)$ w.r.t $\ n\ $, we get $\quad\displaystyle \int_0^1x^{n-1}\ln x\ln(1-x)dx=\frac{H_n}{n^2}+\frac{H_n^{(2)}-\zeta(2)}{n}$

multiply both sides by $\ \displaystyle \frac{\binom{2n}n}{4^n}$ then take the sum, we get, \begin{align} \sum_{n=1}^\infty \frac{H_n}{n^2}\frac{\binom{2n}n}{4^n}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}\frac{\binom{2n}n}{4^n}-\zeta(2)\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}&=\int_0^1\frac{\ln x\ln(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n\ dx\\ \color{blue}{8I+\zeta(3)+2\ln2\zeta(2)}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}\frac{\binom{2n}n}{4^n}-2\ln2\zeta(2)&=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\frac{1}{\sqrt{1-x}}-1\right)\ dx\\ &=\int_0^1\frac{\ln x\ln(1-x)}{x\sqrt{1-x}}\ dx-\zeta(3)\\ &=\int_0^1\frac{\ln(1-x)\ln x}{(1-x)\sqrt{x}}\ dx-\zeta(3)\\ &=4\int_0^1\frac{\ln(1-x^2)\ln x}{1-x^2}\ dx-\zeta(3)\\ &=4K-\zeta(3) \end{align} rearranging the terms, we have $\quad\displaystyle\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}\frac{\binom{2n}n}{4^n}=\color{red}{4K-8I-2\zeta(3)}$

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using the second derivative of beta function, we have $\quad\displaystyle\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2}{n}+\frac{H_n^{(2)}}{n}$

multiply both sides by $\ \displaystyle \frac{\binom{2n}n}{4^n} $ then take the sum, we get, \begin{align} \sum_{n=1}^\infty \frac{H_n^{2}}{n}\frac{\binom{2n}n}{4^n}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}\frac{\binom{2n}n}{4^n}&=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n\ dx\\ \sum_{n=1}^\infty \frac{H_n^{2}}{n}\frac{\binom{2n}n}{4^n}+\color{red}{4K-8I-2\zeta(3)}&=\int_0^1\frac{\ln^2(1-x)}{x}\left(\frac{1}{\sqrt{1-x}}-1\right)\ dx\\ &=\int_0^1\frac{\ln^2(1-x)}{x\sqrt{1-x}}\ dx-\int_0^1\frac{\ln^2(1-x)}{x}\ dx\\ &=\int_0^1\frac{\ln^2x}{(1-x)\sqrt{x}}\ dx-\int_0^1\frac{\ln^2x}{1-x}\ dx\\ &=8\int_0^1\frac{\ln^2x}{1-x^2}\ dx-2\zeta(3)\\ &=8\left(\frac74\zeta(3)\right)-2\zeta(3)\\ &=12\zeta(3) \end{align} rearranging the terms, we have $\quad\displaystyle\sum_{n=1}^\infty \frac{H_n^2}{n}\frac{\binom{2n}n}{4^n}=14\zeta(3)+8I-4K$

by applying IBP for$\ I$, we get $\quad\displaystyle8I=4K+4\int_0^1\frac{\ln(1-x^2)\tanh^{-1}x}{x}dx$

then \begin{align} \sum_{n=1}^\infty\frac{H_n^2}{n}\frac{\binom{2n}n}{4^n}&=14\zeta(3)+4\int_0^1\frac{\ln(1-x^2)\tanh^{-1}x}{x}dx\\ &=14\zeta(3)+2\int_0^1\frac{\left[\ln(1+x)+\ln(1-x)\right]\left[\ln(1+x)-\ln(1-x)\right]}{x}dx\\ &=14\zeta(3)+2\int_0^1\frac{\ln^2(1+x)-\ln^2(1-x)}{x}dx\\ &=14\zeta(3)+2\left(\frac14\zeta(3)-2\zeta(3)\right)\\ &=\frac{21}{2}\zeta(3) \end{align}

This is not a complete solution but some first steps.

EDIT 12.04.19 23:20

Much simpler single integral derived.

Original post

The sum in question is

$$s = \sum_{n=1}^\infty a_n\tag{1}$$

with

$$a_n = \frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}\tag{2}$$

1. Representation as a single integral

1.1

Let us replace just one harmonic number in $a_n$.

Using the definition

$$H_n = \sum _{k=1}^{\infty } \frac{n}{k (k+n)}\tag{3}$$

and writing

$$\frac{1}{n+k}=\int_0^1 x^{n+k-1}\,dx\tag{4}$$

gives for the n-sum

$$\sum_{n=1}^{\infty } \frac{\binom{2 n}{n} H_n x^n}{4^n}=\frac{\partial}{\partial{c}} \left( {_2}F{_1} \left( \frac{1}{2},1,c,x\right)\right)|_{ c \to 1}\tag{5}$$

The remaining k-sum is easily done

$$-\sum _{k=1}^{\infty } \frac{x^{k-1}}{k} =\frac{\log (1-x)}{x} $$

Hence $s$ can be expressed as

$$s_1 = \frac{\partial}{\partial{c}} i(c)|_{ c \to 1} \tag{6a}$$

with

$$i(c) = \int_0^1 \frac{\log (1-x)}{x} {_2}F{_1} \left( \frac{1}{2},1,c,x\right)\,dx\tag{6b}$$

Here ${_2}F{_1}$ is the hypergeometric function.

Numerically, we find in this form

$$s = 12.6216...$$.

1.2 Simpler single integral

The expression derived in the previous paragraph is correct but not very useful because it contains the hypergeometric function. Here we derive the following simpler formula with an elementary integrand.

$$s_2 = \int_0^\infty \frac{v}{\sinh \left(\frac{v}{2}\right)} \left(\frac{v}{\sqrt{2-e^{-v}}}-2 \log \left(\frac{\sqrt{2-e^{-v}}+1}{e^{-\frac{v}{2}}+1}\right)\right)\,dv\tag{7}$$

This is a well converging integral, suited for numerical evaluation. The integrand is depicted here

The derivation starts with replacing both $H_n$ by (3) and (4).

This gives the integral

$$s = \int_0^1 \int_0^1 \frac{\log(1-x) \log(1-y)}{2(1-x y )^{\frac{3}{2}}}\,dx\,dy\tag{8}$$

Transforming $x\to 1-e^{-u}$, $y\to 1-e^{-v}$ leads to

$$s = \int_0^\infty \int_0^v (u v ) \frac{e^{\frac{u+v}{2}}}{(e^u + e^v -1 )^{\frac{3}{2}}}\,du\,dv\tag{8}$$

Here we have made use of the symmetry of the integrand to restrict the integration region to $u\le v$ (and applying a factor 2). Luckily the $u$-integral can be done with the result (7).

2. Sum with asymptotic summands

An attempt to get a feeling for the ingredients of a possible closed form.

The leading asymptotic term of $a_n$ is

$$a_n \simeq b_n = \frac{(\log (n)+\gamma )^2}{\sqrt{\pi } n^{3/2}}\tag{1}$$

The sum of $b_n$ instead of $a_n$ gives

$$s \simeq \sum_{n=1}^\infty b_n = \frac{1}{\sqrt{\pi }}\left(\zeta ''\left(\frac{3}{2}\right)-2 \gamma \zeta '\left(\frac{3}{2}\right)+\gamma ^2 \zeta \left(\frac{3}{2}\right)\right)\simeq 12.0733\tag{2}$$

Here $\zeta(x)$ is the Riemann zeta function (and it derivatives), and $\gamma$ is the Euler-Mascheroni constant.

Notice that the numerical value is close to the one mentioned in the OP. Taking higher terms in the asymptotic expansion of $a_n$ leads to slightly higher numerical values.