# An infinite series involving harmonic numbers

I am looking for a proof of the following claim:

Let $$H_n$$ be the nth harmonic number. Then, $$\frac{\pi^2}{12}=\ln^22+\displaystyle\sum_{n=1}^{\infty}\frac{H_n}{n(n+1) \cdot 2^n}$$

The SageMath cell that demonstrates this claim can be found here.

• See value of $\text{Li}_2(1/2)$ here: en.wikipedia.org/wiki/Polylogarithm#Particular_values together with the fact that computing $\displaystyle\sum_{n=1}^{\infty}\frac{H_n x^n}{n(n+1)}$ is easily reduced to computing the integral of $\log(1-x)^2/x^2$. Commented Aug 24, 2021 at 6:03

Denoting $$H_0=0$$, we have $$\sum_{n=1}^\infty \frac{H_n}{n(n+1)2^n}=\sum_{n=1}^\infty \left(\frac1n-\frac1{n+1}\right)\frac{H_n}{2^n}=\sum_{n=1}^\infty \frac{1}n\left(\frac{H_n}{2^n}-\frac{H_{n-1}}{2^{n-1}}\right)\\=\sum_{n=1}^\infty\frac1{n^22^n}-\sum_{n=1}^\infty\frac{H_n}{(n+1)2^{n+1}}.$$ It is well known that $$\sum_{n=1}^\infty\frac1{n^22^n}=\frac{\pi^2}{12}-\frac{\log^2 2}2$$ (see the value of $${\rm Li}_2(1/2)$$ here). Thus, it remains to show that $$\sum_{n=1}^\infty\frac{H_n}{(n+1)2^n}=\log^2 2.$$ For this, we take the square of the series $$\log 2=-\log\left(1-\frac12\right)=\frac12+\frac1{2\cdot 2^2}+\frac1{3\cdot 2^3}+\ldots$$ to get $$\log^2 2=\sum_{a,b=1}^\infty \frac1{ab2^{a+b}}=\sum _{a,b=1}^\infty \frac1{(a+b)2^{a+b}}\left(\frac1a+\frac1b\right)=\sum_{n=2}^\infty\frac1{n2^n}2H_{n-1}=\sum_{n=1}^\infty\frac{H_n}{(n+1)2^{n}}.$$