Question

Use any convergence tests from Sections 7.4–7.7 to determine whether the series in Exercises 36–51 converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.
$\sum _{k=2}^{\infty }\frac{1}{k^2-1}$

Short answer

The series $\sum _{k=2}^{\infty }\frac{1}{k^2-1}$converges absolutely.

Step-by-step solution

Step 1. Given Information.

The series:

$\sum _{k=2}^{\infty }\frac{1}{k^2-1}$

Step 2. Rewrite the series.

The terms of $\sum _{k=2}^{\infty }\frac{1}{k^2-1}$are positive.

$$\begin{gathered} \sum _{k=2}^{\infty }\frac{1}{k^2-1} \\ \sum _{k=2}^{\infty }{b}_k=\sum _{k=2}^{\infty }\frac{1}{k^2} \end{gathered}$$

Step 3. Solve by Limit Comparison Method.

By Limit Comparison Method,

$$\begin{gathered} \underset{k\to \infty }{lim}\frac{a_k}{b_k}=\underset{k\to \infty }{lim}\frac{\frac{1}{k^2-1}}{\frac{1}{k^2}} \\ =\underset{k\to \infty }{lim}\frac{k^2}{k^2-1} \\ =\underset{k\to \infty }{lim}\frac{k^2}{k^2(1-{\displaystyle \frac{1}{k^2}})} \\ =1 \end{gathered}$$

Step 4. Find if the series is convergent.

Since the value of $\underset{k\to \infty }{lim}\frac{a_k}{b_k}=1$, which is a positive number, and so the series is convergent.

Since the terms of the series are positive, the series is absolutely convergent.