How do we get a closed form for
$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}$$
Answer
Here's another solution. I'll denote various versions of the sum
$$
\sum_{k=1}^\infty\sum_{j=1}^k\frac1j\frac1{k^2}
$$
by an $S$ with two subscripts indicating which parities are included, the first subscript referring to the parity of $j$ and the second to the parity of $k$, with '$\mathrm e$' denoting only the even terms, '$\mathrm o$' denoting only the odd terms, '$+$' denoting the sum of the even and odd terms, i.e. the regular sum, and '$-$' denoting the difference between the even and the odd terms, i.e. the alternating sum. Then
$$
\begin{align}
\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}
&=
2\sum_{n=1}^\infty\sum_{i=1}^n\frac1{2i}\frac1{(2n+1)^2}
\\
&=
2S_{\mathrm{eo}}
\\
&=
2(S_{++}-S_{\mathrm o+}-S_{\mathrm{ee}})
\\
&=
2\left(S_{++}-S_{\mathrm o+}-\frac18S_{++}\right)
\\
&=
2\left(\frac38S_{++}+\left(\frac12S_{++}-S_{\mathrm o+}\right)\right)
\\
&=
\frac34S_{++}+S_{-+}
\\
&=
\frac32\zeta(3)+\sum_{k=1}^\infty\sum_{j=1}^k\frac{(-1)^j}j\frac1{k^2}\;,
\end{align}
$$
where I used the result $\sum_nH_n/n^2=2\zeta(3)$ from the blog post Aeolian linked to and reduced the present problem to finding the analogue of that result with the sign alternating with $j$, which we can rewrite as
$$
\begin{align}
\sum_{k=1}^\infty\sum_{j=1}^k\frac{(-1)^j}j\frac1{k^2}
&=
\sum_{k=1}^\infty\sum_{j=1}^\infty\frac{(-1)^j}j\frac1{k^2}-\sum_{k=1}^\infty\sum_{j=k+1}^\infty\frac{(-1)^j}j\frac1{k^2}
\\
&=
-\zeta(2)\log2+\sum_{j=1}^\infty\frac{(-1)^j}{j+1}\sum_{k=1}^j\frac1{k^2}\;.
\end{align}
$$
This last double sum can be evaluated by the method applied in the blog post, making use of the fact that summing the coefficients of a power series in $x$ corresponds to dividing it by $1-x$:
$$
\begin{align}
\sum_{j=1}^\infty x^j\sum_{k=1}^j\frac1{k^2}=\def\Li{\operatorname{Li}}\frac{\Li_2(x)}{1-x}\;,
\end{align}
$$
where $\Li_2$ is the dilogarithm. Thus
$$
\begin{align}
\sum_{j=1}^\infty\frac{(-1)^j}{j+1}\sum_{k=1}^j\frac1{k^2}
&=
\int_0^1\sum_{j=1}^\infty (-x)^j\sum_{k=1}^j\frac1{k^2}\mathrm dx
\\
&=
\int_0^1\frac{\Li_2(-x)}{1+x}\mathrm dx
\\
&=
\left[\Li_2(-x)\log(1+x)\right]_0^1+\int_0^1\frac{\log^2(1+x)}x\mathrm dx
\\
&=-\frac{\zeta(2)}2\log2+\frac{\zeta(3)}4\;,
\end{align}
$$
where the boundary term is evaluated using $\Li_2(-1)=-\eta(2)=-\zeta(2)+2\zeta(2)/4=-\zeta(2)/2$ and the integral in the second term is evaluated in this separate question. Putting it all together, we have
$$
\begin{align}
\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}
&=
\frac74\zeta(3)-\frac32\zeta(2)\log2
\\
&=
\frac74\zeta(3)-\frac{\pi^2}4\log2\;.
\end{align}
$$
I believe all the rearrangements can be justified, despite the series being only conditionally convergent in $j$, by considering the partial sums with $j$ and $k$ both going up to $M$; then all the rearrangements can be carried out within that finite square of the grid, and the sums of the remaining terms go to zero with $M\to\infty$.
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