Using the integral test, how do you show whether #sum 1 / (n^2 + 1)# diverges or converges from n=2 to infinity?

1 Answer
Jun 27, 2015

It converges. The improper integral to relate it to is #\int_{2}^{\infty}1/(x^2+1)\ dx#

Explanation:

Let #f(x)=1/(x^2+1)#. The series whose convergence is in question is #\sum_{n=2}^{\infty}f(n)#.

As a function of the continuous variable #x#, #f(x)# is positive and decreasing. Moreover, #\int_{2}^{\infty}f(x)\ dx# converges because it equals #\lim_{b->\infty}(arctan(b)-arctan(2))#. Since #lim_{b->\infty}arctan(b)=pi/2#, it follows that #\int_{2}^{\infty}f(x)\ dx# converges to #\pi/2-arctan(2)#.

The integral test now implies that the series #\sum_{n=2}^{\infty}f(n)=\sum_{n=2}^{\infty}1/(n^2+1)# converges.