Sum of $\sum_{n=1}^{\infty}\frac{n}{2^n}$.

Multiply with $(2-1)$: $$ (2-1)\sum_{n=1}^\infty n2^{-n}=\sum_{n=1}^\infty(n2^{1-n}-n2^{-n})=\sum_{n=0}^\infty (n+1)2^{-n}-\sum_{n=1}^\infty n2^{-n}=\sum_{n=0}^\infty2 ^{-n}$$