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

Question

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

Christopher Allers · Accepted Answer

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

William Abdalla · Answer

undefined To determine if the series $ \sum_{n=2}^{\infty} \frac{1}{n^2 + 1} $ converges or diverges, we can use the integral test.

1. **Function Selection**: Choose a function $ f(x) $ that is positive, continuous, and decreasing for $ x \geq 2 $, and such that $ f(n) = \frac{1}{n^2 + 1} $ for $ n \geq 2 $.

2. **Integral Setup**: Set up the corresponding integral $ \int_{2}^{\infty} f(x) \, dx $.

3. **Evaluate Integral**: Compute the integral $ \int_{2}^{\infty} \frac{1}{x^2 + 1} \, dx $.

4. **Conclude**: If the integral converges, then the series converges. If the integral diverges, then the series diverges.

In this case, the integral $ \int_{2}^{\infty} \frac{1}{x^2 + 1} \, dx $ can be evaluated using standard techniques. It converges because $ \frac{1}{x^2 + 1} $ approaches $ 0 $ as $ x $ approaches $ \infty $.

Therefore, by the integral test, the series $ \sum_{n=2}^{\infty} \frac{1}{n^2 + 1} $ converges.