I am trying to solve an infinite series of the form: $$\sum_{n=1}^{+\infty} 2^{-n} \cdot \frac{1}{n^2}$$
I know about the basel problem and its result: $$\sum_{n=1}^{+\infty} \frac{1}{n^2} = \frac{\pi^2}{6} $$
And wolfram alpha actually tells me that: $$\sum_{n=1}^{+\infty} 2^{-n} \cdot \frac{1}{n^2} = \frac{\pi^2}{12} - \frac{\log^2(2)}{2}$$
But I can't find the steps to this solution, can someone give more details to the solution?
