2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user296602 Jan 14 '16 at 23:53
  • $\begingroup$ @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 $\endgroup$ – Yeah.. Jan 14 '16 at 23:56
  • $\begingroup$ 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. $\endgroup$ – Ian Jan 14 '16 at 23:58
  • $\begingroup$ You have also made a mistake in computing s/2. There is a 'n' in the numerator which should be '1'. $\endgroup$ – Shailesh Jan 14 '16 at 23:59
  • $\begingroup$ @Ian I've seen a formula for that yes, awhile ago though. I'll try that now. $\endgroup$ – Yeah.. Jan 15 '16 at 0:03
1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

$$ \sum_{i=1}^n i x^i = \sum_{i=1}^n x \frac d {dx} x^i = x \frac d {dx} \sum_{i=1}^n x^i = x \frac d {dx}\ \frac{x - x^{n+1}}{1-x} = \cdots $$ (Now evaluate the derivative and then substitute $\dfrac 1 2$ for $x$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.