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anonymous

  • 5 years ago

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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  1. anonymous
    • 5 years ago
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    did you try using the ratio test?

  2. anonymous
    • 5 years ago
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    wait sorry, haha wrong thing! i'm just learning this section too so i'll try and work it out

  3. anonymous
    • 5 years ago
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    i actually think its geometric

  4. anonymous
    • 5 years ago
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    Really? It's in a section that says it's telescoping.

  5. anonymous
    • 5 years ago
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    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."

  6. anonymous
    • 5 years ago
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    hmm well if you write out the first few terms do they cancell out?

  7. anonymous
    • 5 years ago
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    No. They just keep getting smaller and besides, they're all positive.

  8. anonymous
    • 5 years ago
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    \[\sum_{n=2}^{\infty}=2/(n^2-1)\]

  9. anonymous
    • 5 years ago
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    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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