Summation of $n/3^n$ $$\sum_{n=1}^\infty \frac{n}{3^n}$$
How do you find the sum?
I don't know how to start this problem and no other website I found talks about a problem like this.

Approach 1:
$$\sum_{n=1}^\infty \frac{n}{3^n} = \frac{1}{3}\left.\sum_{n=1}^\infty \frac{d}{dx}x^{n}\right|_{x=1/3} = \frac{1}{3} \left.\frac{d}{dx} \frac{x}{1-x} \right|_{x=1/3} = \frac{1}{3} \cdot \frac{1}{(1-1/3)^2}= \frac{3}{4}.$$
Approach 2 (no derivatives):
\begin{align}
S &:= \sum_{n=1}^\infty \frac{n}{3^n}\\
S/3 &= \sum_{n=1}^\infty \frac{n}{3^{n+1}} = \sum_{n=2}^\infty \frac{n-1}{3^n}
\end{align}
Subtract the above two equations.
\begin{align}
2S/3 = S - S/3 &= \frac{1}{3} + \sum_{n=2}^\infty \frac{1}{3^n} = \frac{1}{2}
\\
S &= \frac{3}{4}.
\end{align}
