calculus

is the sum from k=1 to infinity of 1/(k^2+1) convergent?
i said it was because the integral is convergent

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asked by sarah
  1. I agree

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    posted by Damon
  2. by the way, every term is smaller than 1/k^2
    so the only integral you really have to do is dx/x^2
    which is really easy. being -1/x

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    posted by Damon

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