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

Solution 1:

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, $S=\frac{21}2\zeta(3)$.

Solution 2:

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

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}

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 $\quad\displaystyle\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}\frac{\binom{2n}n}{4^n}=\color{red}{4K-8I-2\zeta(3)}$

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}