Summation of $n/3^n$

Question

$$\sum_{n=1}^\infty \frac{n}{3^n}$$

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.

Answer 1

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?

Answer 2

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}

Answer 3

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 $f'(1)$.

Answer 4

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 $$\sum_{n=1}^\infty nx^{n-1}=\frac1{(1-x)^2}, \quad\text{whence }\enspace x\sum_{n=1}^\infty nx^{n-1}=\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}.$$