# Homework Help: Telescoping Series

1. Sep 30, 2006

I want to determine whether $$\sum^{\infty}_{n=2} \frac{2}{n^{2}-1}$$ is convergent or divergent. I did the following:

$$\sum^{\infty}_{n=2} \frac{2}{n^{2}-1} = \sum^{\infty}_{n=2} (\frac{1}{j-1} + \frac{1}{j+1})$$. Writing down some terms, I got:

$$(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5}) + (\frac{1}{4}-\frac{1}{6})...$$. It seemed like all the terms were canceling except $$(1+\frac{1}{2})$$.

But when I looked at the answer, it said the sum was $$s_{n}= 1+\frac{1}{2}-\frac{1}{n-1}-\frac{1}{n}$$.

Can anybody tell me how they got this?

Thanks

2. Sep 30, 2006

### quasar987

S_n is the partial sum, not the whole sum. to get the whole sum, take the limit of S_n as n-->oo and you do get 1+1/2.

3. Sep 30, 2006

### quasar987

btw, it should be

$$\sum^{\infty}_{n=2} \frac{2}{n^{2}-1} = \sum^{\infty}_{n=2} (\frac{1}{j-1} - \frac{1}{j+1})$$

Oh, but you developped the sum correctly, so never mind.

4. Sep 30, 2006

but how did they get the partial sum?

5. Sep 30, 2006

### quasar987

but hum... I do calculate something different for S_n... I get

$$S_n = 1+\frac{1}{2}-\frac{1}{n}+\frac{1}{n-1}-\frac{1}{n+1}$$

6. Sep 30, 2006

### shmoe

It looks like they want:

$$s_n=\sum_{j=2}^{n-1}\frac{1}{j-1}-\frac{1}{j+1}$$

i.e. s_4 is really the first two terms of your sum, that is s_4=(1-1/3)+(1/2-1/4). Then:

$$s_n=\sum_{j=2}^{n-1}\frac{1}{j-1}-\sum_{j=2}^{n-1}\frac{1}{j+1}=\sum_{j=1}^{n-2}\frac{1}{j}-\sum_{j=3}^{n}\frac{1}{j}$$

7. Oct 1, 2006

So lets say I have $$\sum_{n=1}^{\infty} (\frac{3}{n(n+3)}) = (\frac{1}{n}-\frac{1}{n+3})$$. I said that this equals:

$$(1-\frac{1}{4})+(\frac{1}{2}-\frac{1}{5})+(\frac{1}{3}-\frac{1}{6})+(\frac{1}{4}-\frac{1}{7})+...+(\frac{1}{n-1}-\frac{1}{n+2})+(\frac{1}{n}-\frac{1}{n+3})$$. So how do I simplify this sequence of partial sums so that I can take the limit to get the value of the sum

Also if the nth partial sum of a series $$\sum_{n=1}^{\infty} a_{n}$$ is $$s_{n} = 3-\frac{n}{2^{n}}$$ find $$a_{n}$$ and $$\sum_{n=1}^{\infty} a_{n}$$. So would $$a_{n} = s_{n}-s_{n-1} = (\frac{-n}{2^{n}}+\frac{n-1}{2^{n-1}})$$? The sum would be $$\lim_{n\rightarrow \infty} s_{n} = 3$$?

Thanks

Last edited: Oct 1, 2006
8. Oct 1, 2006

### shmoe

The same way I showed you above, break the partial sum into two sums and adjust their variables of summation so the things inside the sums match and you can subtract them (except a few terms on the ends).

Looks fine.