# An Euler type sum: $\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n\cdot 4^n}{2n \choose n}$, where $H_n^{(2)}=\sum\limits_{k=1}^{n}\frac{1}{k^2}$

I've been trying to compute the following series for quite a while :

$$\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n\cdot 4^n}{2n \choose n}$$ where $H_n^{(2)}=\sum\limits_{k=1}^{n}\frac{1}{k^2}$ are the generalized harmonic numbers of order 2.

I've already successfully computed the value of a lot of similar-looking series involving harmonic numbers and central binomial coefficients (by using Abel's transformation, finding the generating function...), and they all had closed forms in terms of the Riemann Zeta function, so I'm confident this one too has a nice closed form ; but it's a tough guy that resists my previous methods.

By numerical test, I suspect that the value is $\frac{3}{2}\zeta(3)$.

Any idea for derivation ?

• What a fitting name! Aug 26, 2018 at 23:10

Since

$$\zeta(2)-H_n^{(2)} = \int_{0}^{1} x^n \frac{-\log x}{1-x}\,dx$$ $$\sum_{n\geq 1}\frac{x^n}{n4^n}\binom{2n}{n} = 2\log(2)-2\log(1+\sqrt{1-x})$$ the computation of your series boils down to the computation of the integral

$$\int_{0}^{1}\frac{\log(x)\log(1+\sqrt{1-x})}{1-x}\,dx=\int_{0}^{1}\frac{\log(1-x)\log(1+\sqrt{x})}{x}\,dx$$ or the computation of the integral $$\int_{0}^{1}\frac{\left[\log(1-x)+\log(1+x)\right]\log(1+x)}{x}\,dx$$ where $$\int_{0}^{1}\frac{\log^2(1+x)}{x}=\frac{\zeta(3)}{4},\qquad \int_{0}^{1}\frac{\log(1-x)\log(1+x)}{x}\,dx =-\frac{5\,\zeta(3)}{8}$$ are simple to prove by Maclaurin series and Euler sums. Your conjecture is correct.

Using the well-known identity

$$\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1$$

Divide both sides by $$x$$ then integrate , we get

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

Then

$$\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=-2\ln(1+\sqrt{1-x})+2\ln2\tag1$$

Multiply both sides of (1) by $$-\frac{\ln(1-x)}{x}$$ then integrate from $$x=0$$ to $$1$$ and use the fact that $$-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$$ we get

\begin{align} \sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}&=2\underbrace{\int_0^1\frac{\ln(1+\sqrt{1-x})\ln(1-x)}{x}dx}_{\sqrt{1-x}=y}-2\ln2\underbrace{\int_0^1\frac{\ln(1-x)}{x}dx}_{-\zeta(2)}\\ &=8\int_0^1\frac{y\ln(1+y)\ln y}{1-y^2}dy+2\ln2\zeta(2)\\ &=4\int_0^1\frac{\ln(1+y)\ln y}{1-y}-4\int_0^1\frac{\ln(1+y)\ln y}{1+y}+2\ln2\zeta(2) \end{align}

where the first integral is

$$\int_0^1\frac{\ln y\ln(1+y)}{1-y}\ dy=\zeta(3)-\frac32\ln2\zeta(2)$$

and the second integral is

$$\int_0^1\frac{\ln y\ln(1+y)}{1+y}\ dy=-\frac12\int_0^1\frac{\ln^2(1+y)}{y}dy=-\frac18\zeta(3)$$

Combine the results of the two integrals we get

$$\boxed{\sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}=\frac92\zeta(3)-4\ln2\zeta(2)}$$

If we differentiate both sides of $$\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$$ we get

$$\int_0^1x^{n-1}\ln x\ln(1-x)dx=\frac{H_n}{n^2}+\frac{H_n^{(2)}-\zeta(2)}{n}\tag2$$

Now multiply both sides of $$(2)$$ by $$\frac{1}{4^n}{2n\choose n}$$ the sum up from $$n=1$$ to $$\infty$$ we get

$$\sum_{n=1}^\infty \frac{H_n}{n^24^n}{2n\choose n}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n4^n}{2n\choose n}-\zeta(2)\sum_{n=1}^\infty \frac{1}{n4^n}{2n\choose n}\\=\int_0^1\frac{\ln x\ln(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n\ dx=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\frac{1}{\sqrt{1-x}}-1\right)\ dx\\=\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{x\sqrt{1-x}}dx}_{\text{Beta function:}7\zeta(3)-6\ln2\zeta(2)}-\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{x}dx}_{\zeta(3)}$$

Substitute $$\sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}=\frac92\zeta(3)-4\ln2\zeta(2)$$ and $$\sum_{n=1}^\infty\frac{1}{n4^n}{2n\choose n}=2\ln2$$ we get

$$\boxed{\sum_{n=1}^\infty\frac{H_n^{(2)}}{n4^n}{2n\choose n}=\frac32\zeta(3)}$$