Prove or disprove equality of two sums By playing with maxima I found that the sums 
$$\sum_{n=1}^\infty \frac{1}{2^n n^2}$$
and 
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n}$$
are numerically equal, where $H_n=\sum\limits_{k=1}^n\frac1k$. I'm not sure if it's just a coincidence or we can prove the equality. Any clarification is welcome.
 A: The harmonic number can be represented by the integral 
$$H_n=\int_0^1\frac{1-x^n}{1-x}\,dx \tag 1$$
Then, using $(1)$ reveals
$$\begin{align}
\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n}&=\int_0^1 \frac{1}{1-x}\sum_{n=1}^\infty \frac{(-1)^{n+1}(1-x^n)}{n}\,dx \tag 2\\\\
&=\int_0^1 \frac{\log(2)-\log(1+x)}{1-x}\,dx \tag 3\\\\
&=-\int_{0}^{1/2} \frac{\log(1-x)}{x}\,dx \tag 4\\\\
&=\text{Li}_2(1/2)\\\\
&=\sum_{n=1}^\infty \frac{1}{2^n\,n^2}
\end{align}$$
as was to be shown.
In going from $(2)$ to $(3)$, we noted that $\log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}$.
In going from $(3)$ to $(4)$, we enforced the substitution $\frac{1+x}{2}\to 1-x$.
A: Here's a start.
These sums are a
value of the dilogarithm
defined by
$Li_2(z)
=\sum_{k=1}^{\infty} \dfrac{z^k}{k^2}
=-\int_0^z \dfrac{\ln(1-u)du}{u}
$.
See here:
https://en.wikipedia.org/wiki/Spence's_function
Your first sum is
$Li_2(\frac12)
=\dfrac{\pi^2}{12}-\dfrac{\ln^2(2)}{2}
$.
I don't know how to prove
that the second sum
equals the first,
but I am sure someone here does.
