<s>[INST] Show that $\sum_{n = 1}^\infty\frac{2^{n-1}-1}{2^{n-1}n^2} = (\ln 2)^2$. Show that $\sum_{n = 1}^\infty\frac{2^{n-1}-1}{2^{n-1}n^2} = (\ln 2)^2$.
I've been trying to find a solution this for a while now, but I can't make any progress. The problem comes from a math competition, so there should be an elementary solution to this. Can someone please help me?
 [/INST] As the series converges absolutely we can write
$$S = \sum_{n = 1}^\infty \left [\frac{1}{n^2} - \frac{2}{2^n n^2} \right ] = \sum_{n = 1}^\infty \frac{1}{n^2} - 2 \sum_{n = 1}^\infty \frac{(1/2)^n}{n^2}.$$
The first sum to the right corresponds to the Riemann zeta function $\zeta (s)$, namely
$$\zeta (s) = \sum_{n = 1}^\infty \frac{1}{n^s},$$
when $s = 2$ while the second corresponds to the dilogarithm function $\text{Li}_2 (x)$, namely
$$\text{Li}_2 (x) = \sum_{n = 1}^\infty \frac{x^n}{n^2},$$ 
when $x = 1/2$. Thus
$$S = \zeta(2) - 2\text{Li}_2 \left (\frac{1}{2} \right ).$$
Now using the (reasonably?) well-known result for the value of dilogarithm function when its argument is equal to one half, namely
$$\text{Li}_2 \left (\frac{1}{2} \right ) = \frac{1}{2} \zeta (2) - \frac{1}{2} \ln^2 (2),$$
we immediately see the series sums to $\ln^2 (2)$.  
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