# Thread: [SOLVED] Partial fractions for series?

1. ## [SOLVED] Partial fractions for series?

Yep, it's me again....

$\displaystyle \sum^{\infty}_{n=2} \frac{2}{n^2 - 1}$

Should I use partial fractions on this one, or is there a better way to find the sum?

2. Yep. It's a telescoping series.

3. Originally Posted by mollymcf2009
Yep, it's me again....

$\displaystyle \sum^{\infty}_{n=2} \frac{2}{n^2 - 1}$

Should I use partial fractions on this one, or is there a better way to find the sum?
to find the sum, partial fractions is a good way to go (think "telescoping series"). or use algebra:

$\displaystyle \frac 2{(n + 1)(n - 1)} = \frac {n + 1 - n + 1}{(n + 1)(n - 1)} = \frac 1{n - 1} - \frac 1{n + 1}$

4. Originally Posted by Jhevon
to find the sum, partial fractions is a good way to go (think "telescoping series"). or use algebra:

$\displaystyle \frac 2{(n + 1)(n - 1)} = \frac {n + 1 - n + 1}{(n + 1)(n - 1)} = \frac 1{n - 1} - \frac 1{n + 1}$

Ok, so when I work this out, I am left with:

$\displaystyle \lim_{n \rightarrow \infty} 1 - \frac{1}{n+1} = 1$

This does not seem right to me.

How high do i need to evaluate this in terms of n? I mean in terms of cancelling out terms?

5. Originally Posted by mollymcf2009
Ok, so when I work this out, I am left with:

$\displaystyle \lim_{n \rightarrow \infty} 1 - \frac{1}{n+1} = 1$

This does not seem right to me.

How high do i need to evaluate this in terms of n? I mean in terms of cancelling out terms?
nope, that's not right. you are missing some guys. for instance, 1/2 doesn't cancel. check again

you need to evaluate the first n terms

try the following just to see a pattern:

1st term + 2nd term + 3rd term + 4th term + 5th term + ...... + (n - 3)th term + (n - 2)th term + (n - 1)th term + nth term

where you see the dots, you skip those, just jump from the 5th term to the (n - 3)th term. our goal is to see the pattern of what cancels and what stays. if you are keen on picking up the pattern early, you might not have to test that many

6. Originally Posted by Jhevon
nope, that's not right. you are missing some guys. for instance, 1/2 doesn't cancel. check again

you need to evaluate the first n terms

Ok, got it! 3/2 Thanks so much for all of your help y'all! I REALLY appreciate it!

7. $\displaystyle \frac{1}{n-1}-\frac{1}{n+1}=\left( \frac{1}{n-1}-\frac{1}{n} \right)+\left( \frac{1}{n}-\frac{1}{n+1} \right),$ and those are telescoping series.