# Interesting Telescoping Series - Calc 2

1. Sep 16, 2011

### danerape

Interesting Telescoping Series -- Calc 2

My problem with this series arose when I attempted a partial fractions decompisition of the following, (k-1)/(2^(2k+1). I attempted to factor the denominator with 2(2^k) which is right, but where do I go from there? It does not help much to multiply A*(2^k).

My decomposition was {a*2^k+2b}/2(2^k), where did I go wrong?

Thanks,

Dane

2. Sep 16, 2011

### SammyS

Staff Emeritus
Re: Interesting Telescoping Series -- Calc 2

Why try partial fractions? I doubt that it is even applicable for this.

Does your sum start at zero or does it start at 1 ?

3. Sep 16, 2011

### danerape

Re: Interesting Telescoping Series -- Calc 2

Problem attached............

It is a telescoping series from shaums outline

Thanks alot

#### Attached Files:

• ###### Untitled (1).pdf
File size:
58.9 KB
Views:
92
Last edited: Sep 16, 2011
4. Sep 16, 2011

### SammyS

Staff Emeritus
Re: Interesting Telescoping Series -- Calc 2

So, the problem is to evaluate the series:
$\displaystyle \sum_{k=1}^{\infty}\frac{k-1}{2^{k+1}}$​
(There's a typo in your Original Post or in your attached file.)

Note that $\displaystyle \sum_{k=1}^{\infty}\frac{k-1}{2^{k+1}}\ =\ \sum_{k=1}^{\infty}\frac{k}{2^{k+1}}\ -\ \sum_{k=1}^{\infty}\frac{1}{2^{k+1}}$ Added in Edit: I doubt that this helps.

Last edited: Sep 16, 2011
5. Sep 16, 2011

### SammyS

Staff Emeritus
Re: Interesting Telescoping Series -- Calc 2

The k = 1 term is zero.

$\displaystyle \sum_{k=1}^{\infty}\frac{k-1}{2^{k+1}}=0+\sum_{k=2}^{\infty}\frac{k-1}{2^{k+1}}$

Let k = j+1. Rewrite the sum on the right in terms of j.

$\displaystyle \sum_{k=2}^{\infty}\frac{k-1}{2^{k+1}}=\sum_{j=1}^{\infty}\frac{j}{2^{j+2}}$
$\displaystyle =\frac{1}{2}\sum_{j=1}^{\infty}\frac{j-1+1}{2^{j+1}}$

$\displaystyle =\frac{1}{2}\left(\sum_{j=1}^{\infty}\frac{j-1}{2^{j+1}}+\sum_{j=1}^{\infty}\frac{1}{2^{j+1}} \right)$
The first sum in the last line is the same as the beginning series. -- It's just written with j rather than k.

Do some algebra to find that $\displaystyle \sum_{k=1}^{\infty}\frac{k-1}{2^{k+1}}=0+\sum_{j=1}^{\infty}\frac{1}{2^{j+1}}$

The sum with j is a geometric series. It converges to a well known result, but can also be evaluated as a telescoping series in a manner similar to what was done above.