Chapter 8, Problem 16RE

To determine

To find: Whether the series is convergent or divergent.

Answer to Problem 16RE

The series is convergent.

Explanation of Solution

Given:

The series is ${\displaystyle \sum _{n=1}^{\infty }\frac{\cos 3n}{1+{\left(1.2\right)}^n}}$ .

Calculation:

Since $\cos 3n\le 1$ for al $n$ therefore, the equation will become as follows.

  $\frac{\cos 3n}{1+{\left(1.2\right)}^n}\le \frac{1}{1+{\left(1.2\right)}^n}\dots \left(1\right)$

Also, ${\left(1.2\right)}^n<1+{\left(1.2\right)}^n$ .

Take the reciprocal on both sides of the expression ${\left(1.2\right)}^n<1+{\left(1.2\right)}^n$ .

  $\frac{1}{{\left(1.2\right)}^n}>\frac{1}{1+{\left(1.2\right)}^n}\dots \left(2\right)$

From the equation (1) and (2),

  $\frac{\cos 3n}{1+{\left(1.2\right)}^n}\le \frac{1}{1+{\left(1.2\right)}^n}<\frac{1}{{\left(1.2\right)}^n}\dots \left(3\right)$

The comparison tests states that if $a_n$ and $b_n$ are two positive term series such that $a_n\le b_n$ for all $n$ then following condition holds true.

The series ${\displaystyle \sum b_n}$ is convergent then ${\displaystyle \sum a_n}$ is also convergent.

Also a series of the form ${\displaystyle \sum _{n=1}^{\infty }a{r}^{n-1}}=a+ar+a{r}^2+\mathrm{...}$ is convergent for $\left|r\right|<1$ .

Therefore, ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{{\left(1.2\right)}^n}}$ is convergent.

Combine the equation (3) and the comparison test.

Thus, the series is convergent.

Therefore, the series ${\displaystyle \sum _{n=1}^{\infty }\frac{\cos 3n}{1+{\left(1.2\right)}^n}}$ is convergent.