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?.

\begin{eqnarray}
E&=&\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\ldots=\sum_{k=1}^\infty\frac{k}{3^k}=\frac{1}{3}\sum_{k=1}^\infty k\cdot\left(\frac{1}{3}\right)^{k-1}\\
&=&\frac{1}{3}\dfrac{d}{dx}\left(\sum_{k=0}^\infty x^k\right)\Big|_{x=\frac{1}{3}}=\frac{1}{3}\frac{d}{dx}\left(\frac{1}{1-x}\right)\Big|_{x=\frac{1}{3}}=\frac{1}{3}\dfrac{1}{(1-x)^2}\Big|_{x=\frac{1}{3}}\\
&=&\frac{1}{3}\cdot\dfrac{1}{\left(1-\frac{1}{3}\right)^2}=\frac{1}{3}\cdot\dfrac{1}{\frac{4}{9}}=\frac{1}{3}\cdot\frac{9}{4}=\frac{3}{4}
\end{eqnarray}
