# Prove that ${\sum\limits_{n=1}^{\infty}}(-1)^{n-1} \frac{H_n}{n} = \frac{\pi^2}{12} - \frac12\ln^2 2$

We know that $H_n = \sum_{j=1}^{n}{1 \over j}$. Article in The Sum of Certain Series Related To Harmonic Numbers of Omran Kolba, we have proof of this identity which involves some advanced concepts.

I tried to turn the sum into a definite integral and could not. I appreciate any help.

$$\sum_{n=1}^{\infty}(-1)^{n-1} \dfrac{H_n}{n} = \sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}\sum_{i=2}^{n+1}\dfrac{1}{i+1} = \sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}\sum_{i=2}^{n+1}\int_{0}^{1}x^{i}dx = ?$$

• interestingly enough, $\sum_{n = 1}^{\infty}\frac{1}{2^nn^2}$ is equal to the same value. I wonder if there's a simple way to prove the two are equal. – recursive recursion Jul 12 '14 at 21:33
• @recursiverecursion That serie is $\displaystyle{\large{\rm Li}_{\,\,2}\left(1 \over 2\right)}$. See my answer below. – Felix Marin Jul 12 '14 at 23:07
• Check this. – Mhenni Benghorbal Jul 13 '14 at 2:14

You may consider the standard identity $$\sum_{n=1}^{\infty}H_n x^{n-1} = -\dfrac{\ln(1-x)}{x(1-x)} \quad -1 < x<1,\,x\neq0.$$ Then integrate from $x=-1$ to $x=0$ to obtain easily$$\sum_{n=1}^{\infty}(-1)^{n-1} \dfrac{H_n}{n} \!= -\! \int_{-1}^{0}\dfrac{\ln(1-x)}{x(1-x)} dx = -\!\int_{-1}^{0}\left(\dfrac{\ln(1-x)}{x}\! + \!\dfrac{\ln(1-x)}{1-x}\right) \! dx=\dfrac{\pi^2}{12} - \dfrac{1}{2}\ln^2 2.$$
Then, \begin{align} &\bbox[10px,#ffd]{\sum_{n = 1}^{\infty} \pars{-1}^{n - 1}\,{H_{n} \over n}} = -\int_{0}^{1}\ln\pars{1 - t} \sum_{n = 1}^{\infty}\pars{-t}^{n - 1}\,\dd t \\[5mm] = &\ -\int_{0}^{1}{\ln\pars{1 - t} \over 1 + t}\,\dd t = -\int_{0}^{1}{\ln\pars{t} \over 2 - t}\,\dd t =-\int_{0}^{1/2}{\ln\pars{2t} \over 1 - t}\,\dd t \\[5mm] = &\ -\int_{0}^{1/2}{\ln\pars{1 - t} \over t}\,\dd t = \int_{0}^{1/2}{{\rm Li}_{1}\pars{t} \over t}\,\dd t =\int_{0}^{1/2}{\rm Li}_{2}'\pars{t}\,\dd t \\[5mm] = &\ {\rm Li}_{2}\pars{\half} = \bbox[10px,border:1px groove navy] {{\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}} \approx 0.5822 \end{align}
• +1. BTW, I think the OP relates to $$\sum_{n=1}^\infty\frac{1}{2^n\ n^2}.$$ – Tunk-Fey Jul 14 '14 at 19:14
• @Tunk-Fey That was the coincidence which $\color{#88f}{\tt\mbox{@recursive recursion}}$ already commented above. Thanks. – Felix Marin Jul 14 '14 at 20:15