Does formula exist for this sum?

[db453r]

$\sum_{i=1}^{n}[i/2^i]$

Have looked and looked and cannot find it anywhere.

EDITED: To correct mistake.

 

[jedishrfu]

there is a PF discussion of this series that may help:

https://www.physicsforums.com/showthread.php?t=220640

 

[D H]

$\sum_{i=1}^{n}[n/2^n]$

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 looked and looked and cannot find it anywhere.

Have you tried Wolfram alpha, www.wolframalpha.com ?

 

[db453r]

Oops. Yeah, that's what I meant.

 

[db453r]

Wow. Didn't know Wolfram could do that. Thanks.

Here's what it gave me:

$\sum_{i=0}^{n} i/2^{-i} = 2^{-n}(-n+2^{n+1} -2)$

 

[arildno]

You can solve this by hand by using a neat trick.
Form the auxiliary function (*):
$$F(x)=\sum_{i=1}^{i=n}(\frac{x}{2})^{i}$$, that is, F(x) is readily seen to be related to a geometric sum, with alternate expression (**):
$$F(x)=\frac{1-(\frac{x}{2})^{n+1}}{1-\frac{x}{2}}-1$$
Now, the neat trick consists of differentiating (*), and we get:
$$F'(x)=\sum_{i=1}^{i=n}i*x^{i-1}2^{-i}$$
that is, we have:
$$F'(1)=\sum_{i=1}^{i=n}i*2^{-i}$$
which is your original sum!

Thus, you may calculate that sum by differentiating (**) instead, and evaluate the expression you get at x=1