<s>[INST] Evaluating $\sum_{n=0}^\infty \frac{n}{3^n}$ How can i prove that $ \sum\limits_{n=0}^\infty \frac{n}{3^n}=\frac{3}{4}$?
I tried to find some pattern but didn't succeed 
 [/INST] Another approach:  As the series converges, let L denote its limit.  Now you have $$L=\sum_{n=0}^\infty \frac{n}{3^n}=\sum_{n=1}^\infty  \frac{n}{3^n}= \frac{1}{3}\sum_{n=1}^\infty  \frac{n}{3^{n-1}}=  \frac{1}{3}\sum_{n=0}^\infty  \frac{n+1}{3^{n}}= \frac{1}{3}\left[\sum_{n=0}^\infty  \frac{n}{3^{n}}+\sum_{n=0}^\infty  \frac{1}{3^{n}}\right] =\frac{1}{3}\left[L+ \sum_{n=0}^\infty  \frac{1}{3^{n}}\right].$$
Now just use the usual formula for the limit of a geometric series, and solve algebraically for $L$.  
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