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