A series involves harmonic number

Question

How do we get a closed form for $$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}$$

Answer 1

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 $M\to\infty$.

Answer 2

I gave an integral representation for a more general form. Here is an integral representation for your sum

$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}= \frac{1}{4}\,\int_{0}^{1}\!{\frac {\ln \left( 1-z \right) \ln \left( z\right) }{z\sqrt {1-z}}}{dz}= \frac{1}{4}(7\,\zeta \left( 3 \right) -{\pi }^{2}\ln \left( 2 \right))\sim 0.393327464. $$

The above integral can be evaluated through beta function. Here is the technique from previous problems. Basically, you need to consider the integral

$$ \int_{0}^{1} z^s (1-z)^{w-1/2} dz. $$