# 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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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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