A calculus problem by Priyanshu Mishra

By integrating the geometric series formula, we can attain a closed form (as an integral) for the required summation. Let $\text{S}$ be the sum.

$\displaystyle\sum_{n=1}^\infty x^n=\frac1{1-x}-1\quad\text{Divide by x and integrate:}$

$\displaystyle\int\sum_{n=1}^\infty x^{n-1}dx=\int\frac1{(1-x)(x)}-\frac1{x}dx=\int\frac1{1-x}dx$

$\displaystyle\sum_{n=1}^\infty\frac{x^n}{n}=-\ln{(1-x)}\quad\text{Now repeat, but integrate with limits 0 to .5}$

$\displaystyle\sum_{n=1}^\infty\frac1{2^nn^2}=\int_0^\frac1{2}\sum_{n=1}^\infty\frac{x^{n-1}}{n}dx=-\int_0^\frac1{2}\frac{\ln{(1-x)}}{x}dx$

The above steps could be skipped by seeing directly the power series for $\frac{-\ln(1-x)}{x}$ , but it is a nice process.

$\displaystyle\int_0^\frac1{2}\frac{-\ln{(1-x)}}{x}dx=-\ln{x}\ln{(1-x)}\Big|^\frac1{2}_0+\int_0^\frac1{2}\frac{\ln(x)}{x-1}dx$

$\displaystyle\int_0^\frac1{2}\frac{\ln(x)}{x-1}dx=\int_0^\frac1{2}\sum_{n=0}^\infty\frac{(-1)^n(x-1)^n}{n+1}dx=x-\frac{(x-1)^2}{4}+\frac{(x-1)^3}{9}-...\Big|^\frac1{2}_0=$

$\displaystyle\frac1{2}-\frac1{2^2\times 4}-\frac1{2^3\times 9}-...+\frac1{4}+\frac1{9}+...=-S+\frac{\pi^2}{6}$

Bringing all together: $\displaystyle\text{S}=-\ln^2{\frac1{2}}-\text{S}+\frac{\pi^2}{6}\quad\text{S}=-\ln^2{2}+\frac{\pi^2}{6}$