Question

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.

14.$\mathbf{\sum }_{\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{1}}{\sqrt{{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{9}}}$

Short answer

The series$\sum _1^{\infty }\frac{1}{\sqrt{n^2+9}}$ is divergent.

Step-by-step solution

Definition of convergent and divergent

If the partial sum${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series tend to a limit S, then the series is called convergent.

If the partial sum${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series do not approach to a limit, the series is called divergent.

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

Apply the integral test

The given series is $\sum _1^{\infty }\frac{1}{\sqrt{n^2+9}}$.

Use the integral in the given series as:

$\underset{}{\overset{\infty }{\int }}\frac{1}{\sqrt{n^2+9}} dn$

Now, solve the integral is as follows:

Use the integral formula $\int \frac{1}{\sqrt{x^2+a^2}} dx=\log \left|x+\sqrt{x^2+a^2}\right|+c$, where c is a constant.

$\begin{array}{rcl}\underset{}{\overset{\infty }{\int }}\frac{1}{\sqrt{n^2+9}} dn& =& {\left[\log \left|n+\sqrt{n^2+9}\right|\right]}^{\infty }\\ & =& \left[\log \left|\infty +\sqrt{\infty +9}\right|\right]\\ & =& \infty \end{array}$

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