Question on a tricky Arithmo-Geometric Progression:: 
$$\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+\dfrac{5}{64}+\cdots\infty$$

This summation was irritating me from the start,I don't know how to attempt this ,tried unsuccessful attempts though.
 A: The series is 
$$
\frac{1}{4}\sum_{n\ge1}nx^{n-1}
$$
where $x=1/2$. Does this remind you about something?
A: One may start with the standard evaluation,
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ one gets
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}-\frac{(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by multiplying by $x^2$ and by making $n \to +\infty$ in $(2)$, using $|x|<1$, one has

$$
\sum_{n=1}^\infty nx^{n+1}=\frac{x^2}{(1-x)^2}. \tag3
$$ 

Then one may put $x=\dfrac12$ to obtain an answer to the given sum.
A: $$\begin{align}
S&=\qquad \frac 14+\frac 28+\frac 3{16}+\frac 4{32}+\frac 5{64}+\cdots\tag{1}\\
2S&=\frac 12+\frac 24+\frac 38+\frac 4{16}+\frac 5{32}\cdots\tag{2}\\
(2)-(1):\qquad\\
S&=\frac 12+\frac 14+\frac 18+\frac 1{16}+\frac 1{32}\cdots\\
&=\frac {\frac 12}{1-\frac 12}\\
&=\color{red}1
\end{align}$$
A: Apply $(1-x)^{-2}=1+2x+3x^2+4x^3+\cdots\infty$.
Now, 
$\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+\dfrac{5}{64}+\cdots\infty\\
=\dfrac{1}{4}\left(1+\dfrac{2}{2}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+\cdots\infty\right)\\
=\dfrac{1}{4}\left(1-\dfrac{1}{2}\right)^{-2}\hspace{25pt}\text{ here }x=\dfrac{1}{2}.\\
=\dfrac{1}{4}\times4=1.$
