Chapter 11, Problem 93CR

To determine

To evaluate: the first five terms of the sequence and find the limit of the sequence.

Answer to Problem 93CR

The first five terms of sequence is $-\frac{1}{9},\frac{1}{14},\frac{3}{19},\frac{5}{24}\text{ and }\frac{7}{29}$ .

The limit of sequence is $\frac{2}{5}$ .

Explanation of Solution

Given:

  $a_n=\frac{2n}{n^2+1}$

In this, finding the first term as n = 1,

  $\begin{array}{l}{a}_n=\frac{2n}{n^2+1}\\ a_1=\frac{2\left(1\right)}{{\left(1\right)}^2+1}\\ a_1=1\end{array}$

In this, finding the first term as n = 2,

  $\begin{array}{l}{a}_n=\frac{2n}{n^2+1}\\ a_2=\frac{2\left(2\right)}{{\left(2\right)}^2+1}\\ a_2=\frac{4}{5}\end{array}$

In this, finding the first term as n = 3,

  $\begin{array}{l}{a}_n=\frac{2n}{n^2+1}\\ a_3=\frac{2\left(3\right)}{{\left(3\right)}^2+1}\\ a_3=\frac{3}{5}\end{array}$

In this, finding the first term as n = 4,

  $\begin{array}{l}{a}_n=\frac{2n}{n^2+1}\\ a_4=\frac{2\left(4\right)}{{\left(4\right)}^2+1}\\ a_4=\frac{8}{9}\end{array}$

In this, finding the first term as n = 5,

  $\begin{array}{l}{a}_n=\frac{2n}{n^2+1}\\ a_5=\frac{2\left(5\right)}{{\left(5\right)}^2+1}\\ a_5=\frac{5}{13}\end{array}$

Finding the limit of sequence,

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{2n}{n^2+1}\\ \underset{n\to \infty }{\lim }\frac{2n}{n^2+1}=\underset{n\to \infty }{\lim }\left(\frac{n\left(2\right)}{n\left(n+\frac{1}{n}\right)}\right)\\ \underset{n\to \infty }{\lim }\left(\frac{\left(2\right)}{\left(n+\frac{1}{n}\right)}\right)=0\end{array}$

Hence, the first five terms of sequence is $1,\frac{4}{5},\frac{3}{5},\frac{8}{9}\text{ and }\frac{5}{13}$ .

The limit of sequence is $0$ .