How to Evaluate the Sum $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\left(H_{n}\right)^2}{2n+1}$

Question

How can I evaluate the Sum :

$$S=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\left(H_{n}\right)^2}{2n+1}$$

where $\left(H_{n}\right)^2$ denotes a Harmonic Number Squared.

The sum converges and is approximated by

$0.121017...$

Similar Sum i have encountered :

$$\sum_{n=1}^{\infty} \frac{(-1)^{n}\left(H_{n-1}\right)^2}{n}= -\frac{\zeta(3)}{4}-\frac{\ln(2)^{3}}{3}+\frac{\pi^{2}\ln(2)}{12} $$

I have tried :

Letting :

$$ H_{n} = \int_0^1\frac{1-x^{n}}{1-x}\, dx $$

and :

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

As well as trying to manipulate the sum into an equivalent definition that is more known to me, but still no progress.

Thank you kindly for your help and time.

Answer 1

From this post in Eq $(3)$, we have

$$\sum_{n=1}^\infty H_n^2 x^n=\frac{\ln^2(1-x)}{1-x}+\frac{\text{Li}_2(x)}{1-x}$$

replace $x$ by $-x^2$ then $\int_0^1$

$$\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{2n+1}=\int_0^1\frac{\ln^2(1+x^2)}{1+x^2}dx+\int_0^1\frac{\text{Li}_2(-x^2)}{1+x^2}dx=I_1+I_2$$

where

$$I_1=4\int_0^{\pi/4}\ln^2(\cos u)du=\frac7{48}\pi^3+\frac5{4}\pi\ln^22-2\ln2G-4\text{Im}\operatorname{Li_3}(1+i)$$

is calculated here.

and

$$I_2=\frac{5\pi^3}{48}+\frac{\pi}{4}\ln^22+2\ln2\ G-4\text{Im}\operatorname{Li}_3(1+i)$$

is calculated here.

$$\Longrightarrow \sum_{n=1}^\infty\frac{(-1)^nH_n^2}{2n+1}=\frac{\pi^3}{4}+\frac{3\pi}{2}\ln^22-8\text{Im}\operatorname{Li}_3(1+i)$$