Chapter 8, Problem 5RE

To determine

To find: Whether the sequence is convergent or divergent. If it is convergent then find its limit.

Answer to Problem 5RE

The given sequence is convergent, and limit is 0.

Explanation of Solution

Given:

The series is $a_n=\frac{n\sin n}{n^2+1}$ .

Calculation:

To check the convergence of the series find the limit $n\to \infty$ , and if the limit exists then $a_n$ is convergent otherwise divergent.

Since $-1\le \sin n\le 1$ for all value of $n$ .

So, $\sin n$ tends to a finite value between $-1$ and $1$ , but $\sin n$ does not tend to a fixed value.

Find the limit of the series $a_n=\frac{n\sin n}{n^2+1}$ at $n\to \infty$ .

  $\begin{array}{c}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left(\frac{n}{n^2+1}\right)\left(\underset{n\to \infty }{\lim }\sin n\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{1/n}{\frac{1}{n^2}+1}\right)\left(\text{a}\text{finite}\text{value}\text{between}\text{-1}\text{and}\text{1}\right)\\ =\frac{0}{0+1}\left(\text{a}\text{finite}\text{value}\text{between}\text{-1}\text{and}\text{1}\right)\\ =0\end{array}$

The limit of the series exists.

Therefore, the given sequence is convergent, and limit is 0.