Summation of an infinite series $\frac{x}{4^x}$ I've found by the ratio test that the following series is convergent, however I am struggling to find the value of the sum. Could anyone suggest where to go from here to find the sum?
$\sum_{i=0}^\infty \frac{i}{4^i} = $
Thanks in advance.
PS - This is the answer via Wolfram Alpha
 A: METHODOLOGY $1$:  With Calculus
Let $f(x)=\sum_{n=0}^\infty x^n$.  For $|x|<1$, the geometric series converges to $f(x)=\frac{1}{1-x}$.
Note that the series $\sum_{n=0}^\infty nx^{n-1}$, formed by differentiating term-by-term, converges uniformly for all $|x|\le r<1$.  
Since the series representation for $f(x)$ converges, then we have $f'(x)=\frac{1}{(1-x)^2}=\sum_{n=0}^\infty nx^{n-1}$ for all $|x|<1$.   Hence, we have $\sum_{n=0}^\infty nx^n=\frac{x}{(1-x)^2}$
Now, letting $x=\frac14$, we find
$$\sum_{n=0}^\infty \frac{n}{4^n}=\frac{1/4}{(1-(1/4))^2}=\frac{4}{9}$$

METHODOLOGY $2$:  Without Calculus
Note that we can write
$$\begin{align}
\sum_{n=0}^\infty nx^n&=\sum_{n=1}^\infty x^n\sum_{m=1}^{n}(1)\\\\
&=\sum_{m=1}^\infty \sum_{n=m}^{\infty}x^n\\\\
&=\sum_{m=1}^\infty \frac{x^m}{1-x} \\\\
&=\frac{x}{(1-x)^2}
\end{align}$$
A: Hint:

Let's integrate 
$$\int \frac{x}{4^x}dx=\frac{x}{4^x\log 4}-\frac{1}{(\log4)^24^x}$$ $$
\implies\frac{d}{dx}\left(\frac{x}{4^x\log 4}-\frac{1}{(\log 4)^24^x}\right)=\frac{x}{4^x}$$
if $f(x)=\frac{x}{4^x} $ and $g(x)=\sum_{x=0}^\infty \frac{x}{4^x}$ then:
$$g(x)=\frac{g'(x)}{\log 4}-\frac{1}{(\log4)^2}\sum_{x=0}^\infty\frac{1}{4^x}$$
Now it is a differential equation. Solve it.
