Finding the value of $\sum k/3^k$ Possible Duplicate:
  How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$  




please help me with this 
$$\sum\limits_{k=0}^{\infty}\frac{k}{3^k}$$

I need just a hint, not a full answer. Thanks!
There are several ways to do this.  One involves the fact that
$$
\frac{k}{3^k} = kx^k = x\cdot kx^{k-1} = x\cdot \frac{d}{dx} x^k
$$
where $x=1/3$, and $\displaystyle\sum_{k=0}^\infty x\cdot \frac{d}{dx} x^k$ can be found.

Another looks like this:
$$
\sum_{k=0}^\infty \frac{k}{3^k} = \sum_{k=1}^\infty \frac{k}{3^k}
= \left\{\begin{array}{cccccccccccccccc} 
&amp; &amp; 1/3 \\
&amp; + &amp; 1/9 &amp; + &amp; 1/9 \\
&amp; + &amp; 1/27 &amp; + &amp; 1/27 &amp; + &amp; 1/27 \\
&amp; + &amp; 1/81 &amp; + &amp; 1/81 &amp; + &amp; 1/81 &amp; + &amp; 1/81 \\
&amp; + &amp; \cdots
\end{array}\right.
$$

Then you can sum the first column, then the second column, etc., and finally find the sum of all of the column sums.
$$\sum_{k=0}^{\infty}{kr^k}=\frac{r}{(r-1)^2}$$
In your case $r=\frac{1}{3}$