$\sum \limits_{n=1}^{\infty}n(\frac{2}{3})^n$ Evalute Sum 
Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$ 

How can you compute the limit of 
$\sum \limits_{n=1}^{\infty} n(2/3)^n$
Evidently it is equal to 6 by wolfram alpha but how could you compute such a sum analytically?

Toss a coin that has probability $1/3$ of landing "heads" until we get a head.  Let $X$ be the number of tosses required.  We find the mean of $X$ in two different ways. Let our sum be $S$.
Note that $P(X=1)=1/3$, $P(X=2)=(2/3)(1/3)$, $P(X=3)=(2/3)^2(1/3)$, and so on. It follows that 
$$E(X)=1\cdot\left(\frac{1}{3}\right)+ 2\cdot\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)+ 3\cdot\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)+ 4\cdot\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)+\cdots.$$
Thus 
$$E(X)=\frac{1}{2}\left[1\cdot\left(\frac{2}{3}\right)+ 2\cdot\left(\frac{2}{3}\right)^2+ 3\cdot\left(\frac{2}{3}\right)^3+ 4\cdot\left(\frac{2}{3}\right)^4+\cdots\right]=\frac{S}{2}.$$
If the first toss is a head, then $X=1$. If that the first toss is a tail,  we have used up $1$ toss, and the game begins again. By the Law of Total Expectation,
$$E(X)=1\cdot\frac{1}{3} +(1+E(X))\cdot\frac{2}{3}.$$
Solve for $E(X)$. We get $E(X)=3$, and therefore $S=6$.
Comment: There is a very nice book on bijective arguments called Proofs that Really Count. Maybe one should start collecting Mean Proofs.
