$\sum \limits_{n=1}^{\infty}n(\frac{2}{3})^n$ 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?
 A: 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.
A: Almost the same as Mike's answer:
Let $$
\def\ts{\textstyle}
S_n=\ts{2\over3}+2( {2\over 3})^2 +3 ({2\over3})^3+\cdots+n ({2\over3})^n.$$
Then $$\eqalign{ 
\ts{2\over3}S_n
&=\ts \bigl[\,({2\over3})^2+2  ({2\over 3})^3+3 ({2\over3})^4
               +\cdots+(n-1) ({2\over3})^{n }\,\bigr]+n ({2\over3})^{n+1}\cr
&=\ts S_n-
[{2\over3}+ ({2\over3})^2 +({2\over3})^3 +    \cdots +  ({2\over3})^{n }  ] + n({2\over3})^{n +1} \cr
&=\ts S_n-{2/3 -(2/3)^{n+1}\over 1/3}+ n({2\over3})^{n +1}.
}$$
Whence 
$$S_n={ {2/3 -(2/3)^{n+1}\over 1/3}- n({2\over3})^{n +1}\over 1/3}.$$
Taking the limit as $n$ tends to infinity gives
$$
S_n={2-0\over 1/3}=6.
$$
This method was observed (for the general differentiated geometric series $\sum n r^n$) by Roger B. Nelsen, who has another lovely proof  here.
A: I've read about doing it like this:
$y=\sum \limits_{n=1}^\infty n(\frac23)^n$
$\frac23y=\sum \limits_{n=1}^\infty n(\frac23)^{n+1}=\sum \limits_{n=2}^\infty (n-1)(\frac23)^n=\sum \limits_{n=1}^\infty (n-1)(\frac23)^n$, since $n=1$ yields $0$.
$y-\frac23y=\sum \limits_{n=1}^\infty [n-(n-1)](\frac23)^n$
$\frac13y=\sum \limits_{n=1}^\infty (\frac23)^n$
This sum you may be more familiar with.  If not, you can solve it similarly to obtain
$\frac13y=\frac{\frac23}{1-\frac23}=2$
$y=6$
A: $$
\begin{align*}
\sum_{n=1}^\infty n(2/3)^n &=
\sum_{m=1}^\infty \sum_{n=m}^\infty (2/3)^n \\ &=
\sum_{m=1}^\infty \frac{(2/3)^m}{1-2/3} \\ &=
\frac{2/3}{(1-2/3)^2} = 6.
\end{align*}
$$
A: You can do this with power series.  If you let $f(x) := \sum \limits_{n=1}^{\infty} nx^n$ and restrict the domain of $f$ to the interval $|x|<1$ then 
$$\begin{align} f(x) &= x \sum_{n=1}^{\infty} nx^{n-1} \\&= x \sum_{n=1}^{\infty} \frac{d}{dx} x^n \\&= x\frac{d}{dx}(\sum_{n=1}^{\infty} x^n) \\&= x \frac{d}{dx} \bigg( \frac{1}{1-x}\bigg) \\&= \frac{x}{(1-x)^2}
\end{align}
$$
and substituting $x=2/3$ gives $f(x)=6$.
A: Just a curioursity.
Using Fourier series one comes to:
$$\sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n\ \cos nx = 6\ \frac{13 \cos x-12}{(13-12\cos x)^2}$$
hence:
$$\sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n = \sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n\ \cos nx \Bigg|_{x=0} =6\ \frac{13 \cos x-12}{(13-12\cos x)^2}\Bigg|_{x=0} =6\; .$$
