Sum of weird infinite series help

[frankietucci]

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!

 

[Gib Z]

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:

\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{ n 2^n }.

Now consider the following function:

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

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}

So S(x) = - \log (1-x) + C. 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.

 

[frankietucci]

Thanks so much, this makes perfect sense! I sincerely appreciate it!