[itex]\sum_{i=1}^{n}[i/2^i][/itex] Have looked and looked and cannot find it anywhere. EDITED: To correct mistake.
there is a PF discussion of this series that may help: https://www.physicsforums.com/showthread.php?t=220640
That one's easy. There's nothing inside the sum that depends on i, so your sum is the same as ##\frac n {2^n}\sum_{i=1}^n 1##. Do you mean ##\sum_{i=1}^n \frac i {2^i}## ? Have you tried Wolfram alpha, www.wolframalpha.com ?
Wow. Didn't know Wolfram could do that. Thanks. Here's what it gave me: [itex]\sum_{i=0}^{n} i/2^{-i} = 2^{-n}(-n+2^{n+1} -2)[/itex]
You can solve this by hand by using a neat trick. Form the auxiliary function (*): [tex]F(x)=\sum_{i=1}^{i=n}(\frac{x}{2})^{i}[/tex], that is, F(x) is readily seen to be related to a geometric sum, with alternate expression (**): [tex]F(x)=\frac{1-(\frac{x}{2})^{n+1}}{1-\frac{x}{2}}-1[/tex] Now, the neat trick consists of differentiating (*), and we get: [tex]F'(x)=\sum_{i=1}^{i=n}i*x^{i-1}2^{-i}[/tex] that is, we have: [tex]F'(1)=\sum_{i=1}^{i=n}i*2^{-i}[/tex] which is your original sum!! Thus, you may calculate that sum by differentiating (**) instead, and evaluate the expression you get at x=1