The integral test for convergence states that for the series sum_(n=N)^oof(n), if int_N^oof(x)dx converges to any value then the series converges as well. This only applies when f(n) is positive and increasing for n>=N, which holds for arctann//(n^2+1).
So, let's test the integral for sum_(n=1)^ooarctann/(n^2+1):
int_1^ooarctanx/(x^2+1)dx=lim_(brarroo)int_1^b arctanx/(x^2+1)dx
Let u=arctanx, which implies that du=1/(x^2+1)dx. Don't forget to change the bounds of the integral:
=lim_(brarroo)int_(pi//4)^arctanbucolor(white).du=lim_(brarroo)[1/2u^2]_(pi//4)^arctanb
=lim_(brarroo)1/2arctan^2b-1/2(pi/4)^2
Note that lim_(brarroo)arctanb=pi//2:
=1/2((pi/2)^2-(pi/4)^2)=(3pi^2)/32
Since this integral converges (the value 3pi^2//32 is irrelevant), the series sum_(n=1)^ooarctann/(n^2+1) converges as well.
