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.
