Sum of infinite geometric series How do I evaluate this (find the sum)? It's been a while since I did this kind of calculus.
$$\sum_{i=0}^\infty \frac{i}{4^i}$$
 A: For $-1<x<1$, the series $\sum_{i=0}^{\infty}x^i$ converges absolutely to $\frac{1}{1-x}$
$$\sum_{i=0}^{\infty}x^i=\frac{1}{1-x}$$
Then
\begin{align*}
\sum_{i=0}^{\infty}ix^{i-1} &= \frac{d}{dx}\left(\frac{1}{1-x}\right)\\
&=\frac{1}{(1-x)^2}\\
\sum_{i=0}^{\infty}ix^i&=\frac{x}{(1-x)^2}
\end{align*}
Now, by plugging $x=1/4$ into the last equation, we have
$$\sum_{i=0}^{\infty}\frac{i}{4^i}=\frac{1/4}{(1-1/4)^2}=\frac{4}{9}$$
A: Another approach is to write
$$\begin{align}
\sum_{i=0}^{\infty}\frac{i}{4^i}&=\sum_{i=1}^{\infty}\frac{1}{4^i}\left(\sum_{j=1}^{i}1\right)\\\\
&=\sum_{j=1}^{\infty}\sum_{i=j}^{\infty}\frac{1}{4^i}\\\\
&=\sum_{j=1}^{\infty}\frac{1}{4^j}\frac{1}{1-\frac14}\\\\
&=\frac{1/4}{(1-\frac14)^2}\\\\
&=\frac49
\end{align}$$
A: Let me try. 
Set $$S = \sum_{i\geq 0}\frac{i}{4^i}.$$
Then we have $$4S = \sum_{i \geq 0} \frac{i+1}{4^i} = \sum_{i\geq 0}\frac{i}{4^i} + \sum_{i\geq 0}\frac{1}{4^i} = S + \frac{1}{1-\frac{1}{4}}$$
So, $3S = \frac{4}{3}$. It implies that $$S = \frac{4}{9}.$$
