1. ## [SOLVED] Series Question

I'm trying to find the sum of this infinite series (with n=2) of
6/(n^2-1)

for some reason my answer isn't matching up with the answer choices.

What I did:
1/(n+1)-1/(n-1)
Sn=(1/3-1)+(1/4-1/2)+(1/5 - 1/3)+.....+(1/(n-1)-1/(n-3))+(1/n - 1/(n-2))+(1/(n+1)- 1/(n-1))

after canceling things out and taking the limit I ended up with -1-(1/2)=-3/2

Can somebody please tell me what I'm doing wrong. Any help is appreciated, thank you!

2. Originally Posted by yzobel
I'm trying to find the sum of this infinite series (with n=2) of
6/(n^2-1)

for some reason my answer isn't matching up with the answer choices.

What I did:
1/(n+1)-1/(n-1)
Sn=(1/3-1)+(1/4-1/2)+(1/5 - 1/3)+.....+(1/(n-1)-1/(n-3))+(1/n - 1/(n-2))+(1/(n+1)- 1/(n-1))

after canceling things out and taking the limit I ended up with -1-(1/2)=-3/2

Can somebody please tell me what I'm doing wrong. Any help is appreciated, thank you!
The correct expression $3 \left( \frac{1}{n - 1} - \frac{1}{n+1}\right)$. Now, if you do the arithmetic correctly you should get the correct answer.