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.

10.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{e}}^{\mathbf{n}}}{{\mathbf{e}}^{\mathbf{2}\mathbf{n}}\mathbf{+}\mathbf{9}}$

Short answer

The series $\sum _{n=1}^{\infty }\frac{e^n}{e^{2n}+9}$is convergent.

Step-by-step solution

Definition of convergent and divergent

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

If the partial sum${\mathit{S}}_{\mathbf{n}}$ 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 $\sum _{n=1}^{\infty }\frac{e^n}{e^{2n}+9}$.

Use the integral in the given series as:

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

Solve the integral as follows:

Let, $t=e^n$, so $dt=e^n dn$.

Solve the integral

Substitute the value of t and dt into the integral and change the variable from n into t.

$\underset{}{\overset{\infty }{\int }}\frac{1}{t^2+9} dt=\underset{}{\overset{\infty }{\int }}\frac{1}{t^2+3^2} dt$

Use the integral formula $\int \frac{1}{x^2+a^2} dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+c$.

$\begin{array}{rcl}\underset{}{\overset{\infty }{\int }}\frac{1}{t^2+9} dt& =& \frac{1}{3}{\left[{\tan }^{-1}\left(\frac{t}{3}\right)\right]}^{\infty }\\ & =& \frac{1}{3}\left[{\tan }^{-1}\left(\frac{\infty }{3}\right)\right]\\ & =& \frac{1}{3}\left[{\tan }^{-1}\left(\infty \right)\right]\\ & =& \frac{1}{3}\left[\frac{\pi }{2}\right]\\ & =& \frac{\pi }{6}\end{array}$

Here, the series approaches to a finite value, therefore the given series converges.