Interesting Telescoping Series - Calc 2

[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

 

[SammyS]

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 ?

 

[danerape]

Problem attached...


It is a telescoping series from shaums outline

Thanks a lot

 

[SammyS]

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.

 

[SammyS]

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.