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?.
 A: Hint: Try looking at $3E - E$.
A: You can split it up into a sum of geometric progressions:
\begin{align} 3^{-1} + 2 \cdot 3^{-2} + 3 \cdot 3^{-3} + \dotsb = 3^{-1} + 3^{-2} + 3^{-3} + \dotsb \\
+ 3^{-2} + 3^{-3} + \dotsb \\
+ 3^{-3} + \dotsb \\
\ddots \\ 
= 1(3^{-1}+ 3^{-2} + 3^{-3} + \dotsb) \\
+3^{-1}(3^{-1}+ 3^{-2} + \dotsb) \\
+ 3^{-2}(3^{-1} + \dotsb) \\
\ddots \\
= (1+3^{-1}+3^{-2}+\dotsb)(3^{-1}+3^{-2}+3^{-3}+\dotsb)
\end{align}
A: \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}
A: $$\sum _{k=1}^{\infty } \frac{k}{3^k}=\frac34$$
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 $\dfrac{3}{4}$
