is the sum from k=1 to infinity of 1/(k^2+1) convergent?
i said it was because the integral is convergent
2 answers
I agree
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
so the only integral you really have to do is dx/x^2
which is really easy. being -1/x