Use the Integral Test to determine if the series shown below converges or diverges. Be sure to check that the conditions of the Integral Test are satisfied. \[ \sum_{n=1}^{\infty} \frac{7}{n^{2}+36} \]
Solution
Step 1 ：First, we need to check if the conditions of the Integral Test are satisfied. The function \(f(x) = \frac{7}{x^{2}+36}\) is continuous, positive, and decreasing for \(x \geq 1\).
Step 2 ：Next, we evaluate the improper integral \(\int_{1}^{\infty} \frac{7}{x^{2}+36} dx\).
Step 3 ：Let's use the substitution method to solve the integral. Let \(u = x^{2}+36\), then \(du = 2x dx\).
Step 4 ：When \(x = 1\), \(u = 37\), and when \(x = \infty\), \(u = \infty\). So, the integral becomes \(\frac{7}{2} \int_{37}^{\infty} \frac{1}{u} du\).
Step 5 ：The integral \(\int \frac{1}{u} du\) is \(\ln |u| + C\), so the integral becomes \(\frac{7}{2} [\ln |u|]_{37}^{\infty}\).
Step 6 ：Substituting the limits of integration, we get \(\frac{7}{2} (\ln |\infty| - \ln |37|)\).
Step 7 ：Since \(\ln |\infty|\) is undefined, the integral diverges.
Step 8 ：Since the integral diverges, by the Integral Test, the series \(\sum_{n=1}^{\infty} \frac{7}{n^{2}+36}\) also diverges.
Step 9 ：\(\boxed{\text{The series diverges.}}\)
