How do I rewrite 2/(n²-1) as a geometric series?

[Rossinole]

Homework Statement

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

Homework Equations

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.

 

[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 $N \rightarrow \infty$

 

[Rossinole]

I see it now. Thank you.

 

[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.

please reply

 

[rock.freak667]

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

please reply

What have you tried so far on it?

 

[Defennder]

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

please reply

Is this related to the original post or something separate?

 

[statdad]

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.

 

[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 - that's just A times the derivative of a series you know how to sum.