$\sum\limits_{n=1}^\infty n(\frac{1}{2})^{n}$ 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?  

Consider the following matrix of infinite dimension:
$$
\left[\begin{matrix}
\frac 12&\frac 1{2^2}&\dots&\frac 1{2^n}&\dots\\
0 &\frac 1{2^2}&\dots&\frac 1{2^n}&\dots\\
0& 0&\ddots&&\dots\\
0&0&0&\frac 1{2^n}&\dots\\
\vdots&\vdots&\vdots&\vdots&\vdots\\
\end{matrix}\right].
$$
Suppose that we are interested in the sum of all elements of the matrix. One way is to sum the columns first and then sum them up which gives us $\sum\limits_{i=1}^\infty n(\frac{1}{2})^{n}$. However another way is to sum elements of each row first and then sum them up which is the following:
$$
\sum\limits_{i=1}^\infty n(\frac{1}{2})^{n}=\\
\begin{matrix}
\frac 12+&\frac 1{2^2}+&\dots+&\frac 1{2^n}+&\dots&\color{red}{=1}\\
+0 +&\frac 1{2^2}+&\dots+&\frac 1{2^n}+&\dots&\color{red}{=\frac 12}\\
+0+& 0+&\ddots+&&\dots&\color{red}{=\vdots}\\
+0+&0+&0+&\frac 1{2^n}+&\dots&\color{red}{=\frac 1{2^{n-1}}}\\
+\vdots+&\vdots+&\vdots+&\vdots+&\vdots&\color{red}{=\vdots}\\
\end{matrix}\\
\color{red}{=1+\frac 12+\dots=2.}
$$
