91Ó°ÊÓ

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.

The given series is $\underset{n=1}{\overset{∞}{∑}}\frac{e^n}{e^{2n}+9}$.

Use the integral in the given series as:

$\underset{}{\overset{∞}{∫}}\frac{e^n}{e^{2n}+9} dn$

Solve the integral as follows:

Let, $t=e^n$, so $dt=e^n dn$.

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

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

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

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

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