Oldest problem on Brilliant

Another solution, though it is not very different from the ones provided in the fantastic discussion in the note:

$A=\sum_{n\ge 0}\frac{(-1)^n}{2n+1}\sum_{(2d+1)|2n+1}1\\=\sum_{d\ge 0}\sum_{n:2d+1|2n+1}\frac{(-1)^n}{2n+1}\\=\sum_{d\ge 0}\sum_{l\ge 0}\frac{(-1)^{((2d+1)(2l+1)-1)/2}}{(2d+1)(2l+1)}\\=\sum_{d\ge 0}\sum_{l\ge 0}\frac{(-1)^{d+l}}{(2d+1)(2l+1)}\\=\left(\sum_{d\ge 0}\frac{(-1)^d}{2d+1}\right)^2=\left(\arctan(1)\right)^2=\left(\frac{\pi}{4}\right)^2$ which gives the answer as $\boxed{\frac{\pi^2}{16}}$ .

See https://brilliant.org/discussions/thread/diverges-or-converges-r/ .