What is the sum of $E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$ What is the right way to assess this problem?
$$E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$$
To find the value of the sum I tried to use the fact that it could be something convergent like a geometric series. However it does not seem to be the case as there is no common ratio between the terms. Since the numerator is increasing from $1,2,3,...$ makes it impossible to find a ratio. What should I do?.

$$\sum _{k=1}^{\infty } \frac{k}{3^k}=\frac34$$
To prove this I'll use generating functions
Consider $$f(x)=\sum _{k=1}^{\infty } \frac{ x^k}{3^k}=\sum _{k=1}^{\infty } \left(\frac{ x}{3}\right)^k=\frac{1}{1-x/3}=\frac{3}{3-x}\quad(*)$$
$$f'(x)=\frac{1}{3}\sum _{k=1}^{\infty } k\left(\frac{ x}{3}\right)^{k-1}$$
To make the index back to $k$ multiply both sides by $x$ so that
$$xf'(x)=\frac{x}{3}\sum _{k=1}^{\infty } k\left(\frac{ x}{3}\right)^{k-1}=\sum _{k=1}^{\infty } k\left(\frac{ x}{3}\right)^{k}$$
The sum is then $xf'(x)$ for $x=1$
$$\sum _{k=1}^{\infty } k\,\frac{1}{3^k}$$
On the other side $(*)$ we have $f'(x)=\dfrac{3}{(3-x)^2}$ 
that is $xf'(x)=\dfrac{3 x}{(3-x)^2}$ 
The sum $\sum _{k=1}^{\infty } \frac{k}{3^k}$ is $xf'(x)$ for $x=1$, which gives $\dfrac{3}{4}$
