- #1
courtrigrad
- 1,236
- 2
I want to determine whether [tex] \sum^{\infty}_{n=2} \frac{2}{n^{2}-1} [/tex] is convergent or divergent. I did the following:
[tex] \sum^{\infty}_{n=2} \frac{2}{n^{2}-1} = \sum^{\infty}_{n=2} (\frac{1}{j-1} + \frac{1}{j+1}) [/tex]. Writing down some terms, I got:
[tex] (1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5}) + (\frac{1}{4}-\frac{1}{6})... [/tex]. It seemed like all the terms were canceling except [tex] (1+\frac{1}{2}) [/tex].
But when I looked at the answer, it said the sum was [tex] s_{n}= 1+\frac{1}{2}-\frac{1}{n-1}-\frac{1}{n} [/tex].
Can anybody tell me how they got this?Thanks
[tex] \sum^{\infty}_{n=2} \frac{2}{n^{2}-1} = \sum^{\infty}_{n=2} (\frac{1}{j-1} + \frac{1}{j+1}) [/tex]. Writing down some terms, I got:
[tex] (1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5}) + (\frac{1}{4}-\frac{1}{6})... [/tex]. It seemed like all the terms were canceling except [tex] (1+\frac{1}{2}) [/tex].
But when I looked at the answer, it said the sum was [tex] s_{n}= 1+\frac{1}{2}-\frac{1}{n-1}-\frac{1}{n} [/tex].
Can anybody tell me how they got this?Thanks