Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 13

sum_(n=1)^ooarctan(n)/(n^2+1)
Integral test is applicable if f is positive, continuous and decreasing function on infinite interval [k,oo) where k>=1 and a_n=f(x) . Then the series sum_(n=1)^ooa_n converges or diverges if and only if the improper integral int_1^oof(x)dx converges or diverges.
For the given series a_n=arctan(n)/(n^2+1)
Consider f(x)=arctan(x)/(x^2+1)
Refer the attached graph of the function. From the graph, we observe that the function is positive , continuous and decreasing for x>=1
We an apply integral test as the function satisfies all the conditions for the integral test.
Now let's determine whether the corresponding improper integral int_1^ooarctan(x)/(x^2+1)dx converges or diverges.
int_1^ooarctan(x)/(x^2+1)dx=lim_(b->oo)int_1^b arctan(x)/(x^2+1)dx
Let's first evaluate the indefinite integral intarctan(x)/(x^2+1)dx ,
Apply integral substitution:u=arctan(x)
du=1/(x^2+1)dx
=intudu
Apply power rule,
=u^2/2
Substitute back u=arctan(x)
=1/2(arctan(x))^2+C where C is a constant
int_1^ooarctan(x)/(x^2+1)dx=lim_(b->oo)[1/2(arctan(x))^2]_1^b
=lim_(b->oo)1/2[(arctan(b))^2-(arctan(1))^2]
=1/2[(pi/2)^2-(pi/4)^2]
=1/2(pi^2/4-pi^2/16)
=1/2((4pi^2-pi^2)/16)
=3/32pi^2
Since the integral int_1^ooarctan(x)/(x^2+1)dx converges, we conclude from the integral test that the series converges.