Question

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

Answer 1

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.