$\sum\limits_{n=1}^\infty n(\frac{1}{2})^{n}$ I am trying to find the expected value of the number of even numbers rolled before the first odd number when rolling a fair die until an odd number comes up.  I arrived at $\sum\limits_{n=1}^\infty n(\frac{1}{2})^{n+1}$, then I factored out a $\frac{1}{2}$ to give me $\sum\limits_{n=1}^\infty n(\frac{1}{2})^{n}$. What I now lack is knowledge of infinite series.  I know it converges to 2; what I want to know is why?  

Let $S=\sum\limits_{n\geqslant 1}n2^{-n}$. Then $$2S=\sum_{n\geqslant 1}n2^{-(n-1)}=\sum_{n\geqslant 1}(n-1+1)2^{-(n-1)}=\sum_{k\geqslant 0}(k+1)2^{-k}=S+\sum_{k\geqslant 0}2^{-k}=S+2$$
