# Geometric Series

1. Jul 16, 2008

### Rossinole

1. The problem statement, all variables and given/known data

Does the series from n=1 to infinity of (2)/(n^2-1) converge or diverge? If it converges, find the sum.

2. Relevant equations

3. The attempt at a solution

I can see right away that the series converges by a limit comparison test by looking at the series. However, to find the sum I have re-write that as a geometric series. There is nothing, at least to me, that gives away how to re-write that as a geometric series. That's where I'm stuck.

Thanks for any help.

2. Jul 16, 2008

### rock.freak667

Why don'y you write $\frac{2}{n^2-1}$ in partial fractions and see if it is a telescoping series?

Find

$$\sum_{n=1} ^{N} \frac{2}{n^2-1}$$

and then check what happens as $N \rightarrow \infty$

3. Jul 16, 2008

### Rossinole

I see it now. Thank you.

4. Aug 25, 2008

### jainal36

what's the summation of the series?

what's the summation of the series?
1.A^1 + 2.A^2 + 3.A^3 + .............+n.A^n.

5. Aug 25, 2008

### rock.freak667

Re: what's the summation of the series?

What have you tried so far on it?

6. Aug 25, 2008

### Defennder

Re: what's the summation of the series?

Is this related to the original post or something separate?

7. Aug 26, 2008

Are you sure you don't mean

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

i.e.- the sum starting at $$n = 2$$? If you try to start at $$n = 1$$ the very first term is undefined (can't divide by zero) so the series would not converge. As I mentioned - this makes the difference between the series converging and not converging, and will influence your value for the sum.

8. Aug 26, 2008

### Gib Z

Yes, he probably meant the sum to start at n=2.

Jainal36- you should really start new threads rather than hijacking others, but ill help anyway - thats just A times the derivative of a series you know how to sum.