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.
 A: Hint: Consider the power series
$$
f(x)=\sum_{n=1}^{\infty}nx^n,
$$
so that your series if $f(\frac{1}{3})$.
For that power series, if you factor out an $x$ you get
$$
f(x)=x\sum_{n=1}^{\infty}nx^{n-1}.
$$
Does this suggest a relationship to any other power series that you already know?
A: 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}
A: For all $x\in\mathbb{R}$ such that $|x|<3$, let define:
$$f(x):=\sum_{n=0}^{+\infty}\frac{x^n}{3^n}=\frac{1}{1-x/3}.$$
Since $f$ converges uniformly on every compact of $\{x;|x|<3\}$, one has: $$f'(x)=\sum_{n=1}^{+\infty}\frac{n}{3^n}x^{n-1}$$
Therefore, the value you are looking for is $f'(1)$.
A: Start from the geometric series:
$$\sum_{n= 0}^\infty x^n=\frac1{1-x}\quad \text{for all }\; x\;\text{such that }\;\lvert x\rvert <1.$$
Differentiating, you obtain
$$\sum_{n=1}^\infty nx^{n-1}=\frac1{(1-x)^2}, \quad\text{whence }\enspace  x\sum_{n=1}^\infty nx^{n-1}=\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}.$$
