Evaluate the infinite sum ∑n=1∞Hn2n\sum_{n=1}^{\infty} \frac{H_n}{2^n}∑n=1∞2nHn, where Hn=∑k=1n1kH_n = \sum_{k=1}^n \frac{1}{k}Hn=∑k=1nk1 is the nnn-th harmonic number.
2ln22 \ln 22ln2
ln2\ln 2ln2
π212\frac{\pi^2}{12}12π2
12ln22\frac{1}{2} \ln^2 221ln22