Show that $\sum_{n = 1}^\infty\frac{2^{n-1}-1}{2^{n-1}n^2} = (\ln 2)^2$.

Question

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?

Split the sum, the first term is $ \sum_n \frac{1}{n^2}$ and for the other term try to write it in terms of a power series — clark, Jan 08 '18 at 00:55
It is perhaps better to write $\ln (2)^2$ as either $(\ln 2)^2$ or $\ln^2 (2)$. — omegadot, Jan 08 '18 at 05:18
Related: https://math.stackexchange.com/questions/1056052/proof-of-dilogarithm-reflection-formula-zeta2-logx-log1-x-operatorname/1056111#1056111 — Jack D'Aurizio, Jan 08 '18 at 09:49

score 1 · Accepted Answer · answered Jan 08 '18 at 05:17

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 $\ln^2 (2)$.

Show that $\sum_{n = 1}^\infty\frac{2^{n-1}-1}{2^{n-1}n^2} = (\ln 2)^2$.

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