How do you find the infinite series of 2/(n^2-1)? Apparently, it's telescoping and the answer is 3/2 but I don't know how to get there.

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- anonymous

- chestercat

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- anonymous

did you try using the ratio test?

- anonymous

wait sorry, haha wrong thing! i'm just learning this section too so i'll try and work it out

- anonymous

i actually think its geometric

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- anonymous

Really? It's in a section that says it's telescoping.

- anonymous

The directions say "Determine whether the series is convergent or divergent by expressing s(subn) as a telescoping sum. If it is convergent, find its sum."

- anonymous

hmm well if you write out the first few terms do they cancell out?

- anonymous

No. They just keep getting smaller and besides, they're all positive.

- anonymous

\[\sum_{n=2}^{\infty}=2/(n^2-1)\]

- anonymous

I can't see any way you could manipulate that to a telescoping series, but it does seem to converege to zero

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