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.
did you try using the ratio test?
wait sorry, haha wrong thing! i'm just learning this section too so i'll try and work it out
i actually think its geometric
Really? It's in a section that says it's telescoping.
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."
hmm well if you write out the first few terms do they cancell out?
No. They just keep getting smaller and besides, they're all positive.
\[\sum_{n=2}^{\infty}=2/(n^2-1)\]
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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