Telescoping Series involving 3 parts.

[B18]

Homework Statement

Determine whether the series is convergent or divergent by expressing s sub n as a telescoping series. If convergent find the sum.
\sum_{n=2}^\infty \frac{1}{n^3-1}

The Attempt at a Solution

First thing I did was partial fraction decomposition.
Resulting in: Ʃ (n=2 to ∞) (-1/n+(1/2)/(n+1)+(1/2)/(n-1))
What I am confused with is when I try to find the partial sum nothing is canceling out. I tried to making it into this: Ʃ(n=2 to ∞) -1/n+(1/2)*Ʃ(n=2 to ∞) (1/n-1)+(1/n+1) but this didn't click with me either.

 

[SithsNGiggles]

$\displaystyle \sum_{n=2}^\infty \frac{1}{n^3-n}$,

which you found to be equivalent to

$\displaystyle \sum_{n=2}^\infty \left(-\frac{1}{n}+\frac{1}{2}\left(\frac{1}{n+1}+\frac{1}{n-1}\right)\right)$

Or, equivalently,

$\displaystyle \sum_{n=2}^\infty \left(\frac{n}{n^2-1}-\frac{1}{n}\right)$

I'm not sure about the specifics of convergence/divergence for telescoping series, but maybe writing the series in this way will help.

 

[B18]

The part I am currently struggling with is when i find the partial sum and plug in some numbers nothing is canceling out. Still haven't figured this out yet...

 

[Dick]

The part I am currently struggling with is when i find the partial sum and plug in some numbers nothing is canceling out. Still haven't figured this out yet...

n=2 gives -1/2+(1/6+1/2)
n=3 gives -1/3+(1/8+1/4)
n=4 gives -1/4+(1/10+1/6)
n=5 gives -1/5+(1/12+1/8)
...

Now notice that the 1/6 in the first line plus the 1/6 in the third line gives you 1/3 which cancels the -1/3 in the second line. Do you see the cancellation pattern now?

 

[B18]

So would the sum be :-1/2+1/2+1/4=1/4 If I am understanding the cancellation pattern correctly?

 

[Dick]

So would the sum be :-1/2+1/2+1/4=1/4 If I am understanding the cancellation pattern correctly?

Yes, it should. Writing out a lot of terms and staring at it will usually show the pattern, plus the factor of (1/2) on some of them will suggest you need to combine some in pairs.

 

[B18]

Thank you sith and dick for your help I understand the problem now!