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

Solution 1:

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.

Solution 2:

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

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