# Writing sigma notation $\sum^n_{i=1} \frac {i}{2^i}$ in closed form

What would be a way to find the closed form of $\frac {1}{2} + \frac {2}{4}+\frac {3}{8}+\cdots+\frac {n}{2^n}=\sum^n_{i=1} \frac {i}{2^i}=s$

I've looked at $\frac {s}{2}=\frac {1}{4} + \frac {2}{8}+\frac {3}{16}+\cdots+\frac {n}{2^{n+1}}$

And then $s-\frac {s}{2}=\frac {s}{2}=\frac {1}{2} + \frac {1}{4}+\frac {1}{8}+\cdots+\frac {1}{2^n}-\frac {n}{2^{n+1}}$

Any idea where to go from here? Please, if you are sure of where I'm trying to go with this, just ask.

• I assume you mean $i/2^i$, or the problem is utterly trivial. For one approach, think about taking a derivative of the function $1 / (1 - x)$; this has been done here many times before. – user296602 Jan 14 '16 at 23:53
• @user Yes, I meant that; thanks for noticing. I have been directed by my professor that he specifically wants it to be done in the fashion that I've began – Yeah.. Jan 14 '16 at 23:56
• You're saying $\frac{s}{2} = \frac{-n}{2^{n+1}} + \sum_{i=1}^n 2^{-i}$. But the latter sum has a formula that you have probably already seen. – Ian Jan 14 '16 at 23:58
• You have also made a mistake in computing s/2. There is a 'n' in the numerator which should be '1'. – Shailesh Jan 14 '16 at 23:59
• @Ian I've seen a formula for that yes, awhile ago though. I'll try that now. – Yeah.. Jan 15 '16 at 0:03

As given in the comments, looking at the derivative of $\frac{1}{1-x}$ gives the answer, but you have mentioned you have been instructed to solve it using this method. Notice that you almost reached a solution! You have $$\frac{s}{2} = \frac{1}{2}+\cdots+\frac{1}{2^n} - \frac{n}{2^{n+1}}$$ The first part of the right hand side is a sum of a geometric sequence. Use this!
$$\sum_{i=1}^n i x^i = \sum_{i=1}^n x \frac d {dx} x^i = x \frac d {dx} \sum_{i=1}^n x^i = x \frac d {dx}\ \frac{x - x^{n+1}}{1-x} = \cdots$$ (Now evaluate the derivative and then substitute $\dfrac 1 2$ for $x$.)