Using the integral test, how do you show whether #sum1/[(n^2)+4)# diverges or converges from n=1 to infinity?

Question

Christopher Allers · Accepted Answer

Intuitively, #\sum_{n=1}^{\infty}1/(n^2+4)# should converge since it is "like" the #p#-series #\sum_{n=1}^{\infty}1/n^2# which converges since #p=2>1#. In fact, you can even use the comparison test directly with this series to show that #\sum_{n=1}^{\infty}1/(n^2+4)# converges.

But now on to the integral test as requested:

The function #f(x)=1/(x^2+4)# is continuous, positive, and decreasing for all #x\geq 1#. Moreover, #\int_{1}^{\infty}f(x)\ dx=\frac{1}{4}\int_{1}^{\infty}\frac{1}{1+(x/2)^2}#

#=lim_{b->\infty}(\frac{1}{2}arctan(x/2))|_{1}^{b}=\frac{1}{2}\cdot \frac{\pi}{2}-\frac{1}{2}\cdot arctan(1/2)#

In other words, the improper integral converges.

The integral test now implies that the series #\sum_{n=1}^{\infty}f(n)=\sum_{n=1}^{\infty}1/(n^2+4)# converges.

Christopher Agresta · Answer

undefined To determine whether the series $ \sum_{n=1}^{\infty} \frac{1}{n^2 + 4} $ converges or diverges, we can use the integral test.

1. Define the function $ f(x) = \frac{1}{x^2 + 4} $.
2. Verify that $ f(x) $ is positive, continuous, and decreasing for $ x \geq 1 $.
3. Integrate $ f(x) $ from $ 1 $ to $ \infty $.

$$ \int_{1}^{\infty} \frac{1}{x^2 + 4} \, dx $$

4. Evaluate the integral and determine whether the result is finite or infinite.

If the integral converges, then the series $ \sum_{n=1}^{\infty} \frac{1}{n^2 + 4} $ converges. If the integral diverges, then the series also diverges.