mathimatikoi.org

Let $\mathcal{H}_n$ denote the $n$-th harmonic number. Evaluate the sum:

$$\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)}$$

Recall the generating function for harmonic numbers:

$$\sum_{n=1}^\infty\,H_nx^n = - \frac{1}{1-x}\,\ln(1-x).$$

Integrating this yields

$$\sum_{n=1}^\infty\,\frac{H_n}{n+1}x^{n+1} = \frac{1}{2}\,\ln^2(1-x).$$

This can be rewritten as

$$\sum_{n=1}^\infty\,\frac{H_n}{n}x^{n} = \frac{1}{2}\,\ln^2(1-x) + Li_2(x),\qquad(1)$$

where $Li_2$ is the polylogarithm function. Thus, it follows from (1) that

$$\sum_{n=0}^\infty\,\frac{H_{2n+1}}{2n+1}x^{2n+1} = \frac{1}{4}\,\left(\ln^2(1-x)-\ln^2(1+x)\right) + \frac{1}{2}\,(Li_2(x) - Li_2(-x)).\qquad(2)$$

Setting $x = \sqrt{2}/2$ in (2), after some simplification, yields

$$\sum_{n=0}^\infty\,\frac{H_{2n+1}}{2^n(2n+1)} = -\frac{\sqrt{2}}{4}\ln2\ln(3-2\sqrt{2}) + \sum_{n=0}^\infty\,\frac{1}{2^n(2n+1)^2}.$$

Unfortunately, I can not find the exact value of the last series.

Hello mathofusva. Well I left it the way it was in equation $(2)$ of yours. That is why I said the final expression may contain polylogs.

mathofusva wrote: $$\sum_{n=0}^\infty\,\frac{H_{2n+1}}{2^n(2n+1)} = -\frac{\sqrt{2}}{4}\ln2\ln(3-2\sqrt{2}) + \sum_{n=0}^\infty\,\frac{1}{2^n(2n+1)^2}.$$

Unfortunately, I can not find the exact value of the last series.

That is because Lerch Transcendent does not behave well in this case. See after your simplifications we get that:

$$\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)} = - \frac{\sqrt{2}}{4} \ln 2 \ln \left ( 3-2\sqrt{2} \right ) + \frac{1}{4}\Phi \left ( \frac{1}{2}, 2, \frac{1}{2} \right )$$

I am pleased with either formula. More information on Lerch Transcendent may be found here.

You know where this came from? Well it arose while playing around with this little fellow:

$$\mathcal{J}=\int_0^1 \frac{\ln (1-x)}{2-x^2} \, {\rm d}x$$

Well,

\begin{align*}
\int_{0}^{1}\frac{\ln (1-x)}{2-x^2} \, {\rm d}x &=\frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^n} \int_{0}^{1}x^{2n} \ln (1-x) \, {\rm d}x \\
&\overset{(*)}{=} -\frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n} \frac{\mathcal{H}_{2n+1}}{2n+1}\\
&=-\frac{1}{2}\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)}
\end{align*}

$(*)$ since $\displaystyle \int_{0}^{1}x^n \ln (1-x) \, {\rm d}x = -\frac{\mathcal{H}_{n+1}}{n+1}$.

It seems that there is a closed form for the even index. In fact, we have
$$\sum_{n=1}^\infty\,\frac{H_{2n}}{(2n)\,2^n} = \frac{1}{4}\left(\ln^2\left(\frac{2 -\sqrt{2}}{2}\right)+ \ln^2\left(\frac{2 +\sqrt{2}}{2}\right)\right) + \frac{\pi^2}{48} - \frac{1}{8}\,\ln^22.$$
To this end, as we did above, we have
$$\sum_{n=1}^\infty\,\frac{H_{2n}}{2n}\,x^{2n} = \frac{1}{4}\,(\ln^2(1-x)+ \ln^2(1+x)) + \frac{1}{2}\,(\mbox{Li}_2(x) + \mbox{Li}_2(-x)) \tag{1} $$
In view of the facts that $\mbox{Li}_2(x) + \mbox{Li}_2(-x) = \frac{1}{2}\,\mbox{Li}_2(x^2)$ and
$$\mbox{Li}_2(1/2) = \frac{1}{12}\,\pi^2 - \frac{1}{2}\,\ln^22,$$
Letting $x = 1/\sqrt{2}$ in (1) yields the claimed result.