Question

Use the integral test to find whether the following series converge or diverge. Hint and warning: Do not use lower limits on your integrals.

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{3}}^{\mathbf{\infty }}\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{2}}\mathbf{-}\mathbf{4}}$

Short answer

The series$\sum _{n=3}^{\infty }\frac{1}{n^2-4}$ is divergent.

Step-by-step solution

Definition of convergent and divergent.

If the partial sums${\mathit{S}}_{\mathbf{n}}$ of an infinite series tend to a limit S, the series is called convergent. If the partial sums${\mathit{S}}_{\mathbf{n}}$ of an infinite series don't approach a limit, the series is called divergent.

The limiting value S is called the sum of the series.

Integral test.

The given series $\sum _{n=3}^{\infty }\frac{1}{n^2-4}$.

Use the integral in the given series, $\underset{}{\overset{\infty }{\int }}\frac{1}{n^2-4} dn$.

The integral is also written as follows:

$\underset{}{\overset{\infty }{\int }}\frac{1}{n^2-4} dn=\underset{}{\overset{\infty }{\int }}\left[\frac{1}{4\left(n-2\right)}-\frac{1}{4\left(n+2\right)}\right] dn$

Solve integral.

Solve the integral is as follows:

Use the integral formula, $\int \frac{1}{x} dx=\ln x+c$, where is a constant.

$\begin{array}{rcl}\underset{}{\overset{\infty }{\int }}\left[\frac{1}{4\left(n-2\right)}-\frac{1}{4\left(n+2\right)}\right] dn& =& {\left[\frac{1}{4}\ln \left(n-2\right)-\frac{1}{4}\ln \left(n+2\right)\right]}^{\infty }\\ & =& \frac{1}{4}\ln \left(\infty -2\right)-\frac{1}{4}\ln \left(\infty +2\right)\\ & =& \infty \end{array}$

Hence, the series approaches to infinite therefore the given series diverges.