Geometric Series

[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.

[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?
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 - thats just A times the derivative of a series you know how to sum.