# Telescoping Series involving 3 parts.

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1. Mar 24, 2013

### B18

1. The problem statement, all variables and given/known data
Determine whether the series is convergent or divergent by expressing s sub n as a telescoping series. If convergent find the sum.
$$\sum_{n=2}^\infty \frac{1}{n^3-1}$$

3. 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.

Last edited: Mar 24, 2013
2. Mar 24, 2013

### 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.

3. Mar 24, 2013

### 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...

4. Mar 24, 2013

### Dick

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?

5. Mar 24, 2013

### B18

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

6. Mar 24, 2013

### Dick

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.

7. Mar 24, 2013

### B18

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