Calculate the result of the following sequence. I'm stuck with this sequence.I can't calculate the result.
Any help would be greatly appreciated.
$$A=\frac{1}{2} + \frac{2}{4} + \frac{3}{16} + \frac{4}{32} + ... $$
Please feel free to edit the tags if you wish.Thanks
Please do not edit the question I'm sure I've written it correctly.
EDIT:The nth number of this sequence is evaluated by the formula: 
$\displaystyle\frac{n}{2^{n- 2([\frac{n}{2}])+1}}$ if  i'm not mistaken.Because the denominator is multiplied by 2 at first and then by 4 and then again by 2 and ... goes on like that.
EDIT2:I've made a big mistake!The denominator itself is evaluated by :
$\displaystyle d_n=2^{n- 2([\frac{n}{2}])+1} \times d_{n-1}$ where $d_i$ is the $i$ th denominator.
EDIT3:Can anything be done using $2A-A=A$ ?I think this should get me somewhere but I can't figure it out.
 A: It seem that you could express this series
$$C_n = \frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^4}+\frac{4}{2^5}+\frac{5}{2^7}+\frac{6}{2^8}+\dots$$
as the sum of two other series
$$C_n=A_n+B_n$$
where
$$A_n=\frac{1}{2^1}+\frac{3}{2^4}+\frac{5}{2^7}+\frac{7}{2^{10}}+\dots$$
and
$$B_n=\frac{2}{2^2}+\frac{4}{2^5}+\frac{6}{2^8}+\frac{8}{2^{11}}+\dots$$
Hint:
$8A_n - A_n = $ ?
A: Hint:
$$
\frac{d}{dx}\frac{1}{1-x}=\frac{d}{dx} (1+x+x^2+\dots)
$$
A: Hint
In the same spirit as Ma Ming's answer, consider $$A=\sum_{i=1}^{\infty} i x^i=x\sum_{i=1}^{\infty} i x^{i-1}$$ where $x=\frac 12$. As written, $A$ looks like the derivative of a function which itself is the development of a quite classical expression, isn't it ?
I am sure that you can take from here.
A: I think the result is as follows:
$A=\frac{1}{2}+\frac{2}{4}+\frac{3}{16}+...$
This is equal to:
$\sum_{i=1}^{\infty}\frac{i}{2^i} - \sum_{i=1}^{\infty}\frac{3i}{2^{3i}} - \sum_{i=1}^{\infty}\frac{i}{2^{3i+1}}+\frac{i}{2^{3i+2}}$
And you can probably evaluate this in some way. 
btw, the answer is $\frac{68}{49}$
