Chapter 9.5, Problem 14E

To determine

whether the given series converges or diverges.

Answer to Problem 14E

The series ${\displaystyle \sum _{n=1}^{\infty }\frac{e^n}{1+e^{2n}}}$ converges.

Explanation of Solution

Given information: The series ${\displaystyle \sum _{n=1}^{\infty }\frac{e^n}{1+e^{2n}}}$ is given.

Concept used :

If $a_n=f\left(n\right)$ is a sequence of non-negative terms, where $f\left(n\right)$ is a continuous, positive and decreasing function of $x$ for all $x\ge N$ , then the series ${\displaystyle \sum _{n=N}^{\infty }{a}_n}$ either both converge or both diverge.

Calculation:

Since the function $f\left(x\right)=\frac{e^x}{1+e^{2x}}$ is a continuous, positive and decreasing function of $x$ on the interval $\left[1,\infty \right)$ , Integral test can be applied.

Therefore,

  ${\displaystyle \underset{1}{\overset{\infty }{\int }}f\left(x\right)}={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{e^x}{1+e^{2x}}}dx$

Now, consider $e^x=u$ such that $e^{x}dx=du$ such that the limit varies from $e$ to $\infty$ . Use the mentioned substitutions and limits to the right side of the obtained equation and simplify.

  $\begin{array}{c}{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{e^x}{1+e^{2x}}}dx={\displaystyle \underset{e}{\overset{\infty }{\int }}\frac{du}{1+u^2}}\\ ={\left[{\tan }^{-1}u\right]}_e^{\infty }\\ =\left[{\tan }^{-1}\infty -{\tan }^{-1}e\right]\\ =\frac{\pi }{2}-{\tan }^{-1}e\\ \approx 0.35251\end{array}$

This implies that the integral is convergent. Therefore, by Integral Test, the given series is a convergent series.