Sum of weird infinite series help!

  • #1
Hi all,


SUM of Series from n=2 to infinity of:

1
------------
(2^n) (n-1)

This question is driving me bananas... my tools are Telescoping or Geometric series, but neither seem to work:

I've tried everything to get this into a geometric series form and then using the a/1-r formula, but can't. I've also tried to make partial fractions, but the 2^n term doesn't seem to work there.

Help!
 
Last edited:

Answers and Replies

  • #2
Gib Z
Homework Helper
3,346
5
You'll have to expand your toolbox to some calculus. Change the index of your sum to start at n=1 and you'll have this:

[tex]\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{ n 2^n } [/tex].

Now consider the following function:

[tex] S:(-1,1) \to \mathbb{R}, S(x) = \sum_{n=1}^{\infty} \frac{x^n}{n} [/tex]

[tex] S'(x) = \sum_{n=1}^{\infty} x^{n-1} = \frac{1}{x} \sum_{n=1}^{\infty} x^n= \frac{1}{x} \frac{x}{1-x} = \frac{1}{1-x} [/tex]

So [tex] S(x) = - \log (1-x) + C [/tex]. Let x=0 on both sides, we have C=0. So now let x=1/2, S(x) = ln 2, and your sum is half of that.
 
  • #3
Thanks so much, this makes perfect sense! I sincerely appreciate it!
 

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