I want to find a closed form of the following infinite series $$\sum\limits_{n=1}^\infty \frac{H_{n-1}\binom{2n}{n}}{4^n n^2}=?$$ It can be expressed in terms of $\gamma$ and $\pi$? Here $H_n=\sum\limits_{k=1}^n \frac{1}{k}$ is harmonic number.
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Did you prove the sum exists first? Not that I have any idea about a closed form though... – Bill O'Haran Jun 03 '19 at 07:52
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1The sum is convergent by comparison because $\sum_{n=1}^{\infty} \binom{2n}{n}\frac{1}{4^{n} n }= \log(4)$ and $H_{n-1}/n \lt 1$ for $n\ge 1$ – Dr. Wolfgang Hintze Jun 05 '19 at 08:15
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The integral seems to be, $$I = -\frac43\ln^3 2-2\zeta(2)\ln 2+\frac52\zeta(3)$$ – Tito Piezas III Jun 10 '19 at 17:17
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$$S=\sum\limits_{n=1}^\infty \frac{H_{n-1}\binom{2n}{n}}{4^n n^2}=\sum\limits_{n=1}^\infty \frac{H_{n}\binom{2n}{n}}{4^n n^2}-\sum\limits_{n=1}^\infty \frac{\binom{2n}{n}}{4^n n^3}$$
Evaluation of the first sum: Following my solution here, we reached, \begin{align} S_1=\sum\limits_{n=1}^\infty \frac{H_{n}\binom{2n}{n}}{4^n n^2}&=8\int_0^1 \frac{x\ln x\tanh^{-1}x}{1-x^2}\ dx+\zeta(3)+2\ln2\zeta(2) \end{align} Lets calculate the integral: \begin{align} I=\int_0^1 \frac{x\ln x\tanh^{-1}x}{1-x^2}\ dx=-\frac12\int_0^1 \frac{x\ln x}{1-x^2}\ln\left(\frac{1-x}{1+x}\right)\ dx \end{align} Expoilting the identity $\ \displaystyle\frac{\ln\left(\frac{1-x}{1+x}\right)}{1-x^2}=\sum_{n=1}^\infty\left(H_n-2H_{2n}\right)x^{2n-1}$ ( proved here), we get \begin{align} I&=-\frac12\sum_{n=1}^\infty\left(H_n-2H_{2n}\right)\int_0^1x^{2n-1}\ln x\ dx\\ &=\frac12\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{(2n+1)^2}\\ &=\frac12\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}-\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^2}\\ \end{align} For the first sum, Random Variable proved here the following identity :$$\sum_{n=1}^{\infty} \frac{H_{n}}{ (n+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$and by setting $a=1/2$, we obtain $\boxed{\displaystyle\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}=\frac74\zeta(3)-\frac{\pi^2}{4}\ln2}$
As for the second sum: \begin{align} \sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^2}&=\frac12\sum_{n=1}^\infty\frac{H_{n}}{(n+1)^2}(1+(-1)^n)=\frac12\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}(1-(-1)^n)\\ &=\frac12\left(\sum_{n=1}^\infty\frac{H_n}{n^2}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}-\zeta(3)+\operatorname{Li}_3(-1)\right)\\ &=\frac12\left(2\zeta(3)-\left(-\frac58\zeta(3)\right)-\zeta(3)+\left(-\frac34\zeta(4)\right)\right)\\ &\boxed{\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^2}=\frac7{16}\zeta(3)} \end{align} and by combining the boxed results, we get $\ \displaystyle I=\frac7{16}\zeta(3)-\frac{\pi^2}{8}\ln2$ and plugging this result gives: $$\color{blue}{S_1=\frac92\zeta(3)-\frac{2\pi^2}{3}\ln2}$$
Evaluation of the second sum: Using the well known identity $$\quad\displaystyle\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1 \quad$$ multiply both sides by $\ \displaystyle\frac{\ln^2x}{2x}\ $ then integrate from $x=0$ to $x=1,\ $we get \begin{align} S_2&=\sum\limits_{n=1}^\infty \frac{\binom{2n}{n}}{4^n n^3}=\frac12\int_0^1\frac{\ln^2x}{x}\left(\frac{1}{\sqrt{1-x}}-1\right)\ dx, \quad \text{apply IBP}\\ &=-\frac1{12}\int_0^1\ln^3x(1-x)^{-3/2}\ dx\\ &=\frac1{12}\frac{\partial^3}{\partial\alpha^3}\text{B}\left(\alpha,-\frac12\right)_{\large\alpha\ \to\ 1}\\ &\color{blue}{S_2=2\zeta(3)-\frac{\pi^2}{3}\ln2+\frac43\ln^32} \end{align} Finally
$$S=S_1-S_2=\frac52\zeta(2)-\frac{\pi^2}{3}\ln2-\frac43\ln^32$$
Ali Shadhar
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@Shather: The definition of Beta function $B(x,y)$ is defined by $$B(x,y):=\int\limits_{0}^1 t^{x-1}(1-t)^{y-1}dt,\quad {\rm Re}(x),{\rm Re}(y)>0,$$ but you used the $y=-1/2$, why ? – xuce1234 Jul 09 '19 at 01:14
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I am very sorry, I did not understand what you mean. According to the definition of the Beta function, the exponent of $(1-x)$ should be greater than $-1$. – xuce1234 Jul 09 '19 at 02:48
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@xuce1234 $x$ and $y$ are real more than zero when you dont have derivative but once you do, the case will be different. – Ali Shadhar Jul 09 '19 at 16:46