Chapter 11.4, Problem 43E

To determine

To calculate:

The five terms and limit of the sequence, if limit does not exist then give reason.

Answer to Problem 43E

First five terms are $1,\frac{3}{5},\frac{2}{5},\frac{5}{17},\frac{3}{13}$ and limit of the sequence is $\underset{x\to \infty }{\lim }{a}_n=0$

Explanation of Solution

Given information:

  $a_n=\frac{n+1}{n^2+1}$

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $a_n=\frac{n+1}{n^2+1}$

First five terms are : $a_1,a_2,a_3,a_4,a_5$

For $n=1$ :

  $\begin{array}{l}{a}_n=\frac{1+1}{1^2+1}\\ =\frac{2}{2}\\ =1\end{array}$

For $n=2$ :

  $\begin{array}{l}{a}_n=\frac{2+1}{2^2+1}\\ =\frac{3}{5}\end{array}$

For $n=3$ :

  $\begin{array}{l}{a}_n=\frac{3+1}{3^2+1}\\ =\frac{4}{10}\end{array}$

For $n=4$ :

  $\begin{array}{l}{a}_n=\frac{4+1}{4^2+1}\\ =\frac{5}{17}\end{array}$

For $n=5$ :

  $\begin{array}{l}{a}_n=\frac{5+1}{5^2+1}\\ =\frac{6}{26}\\ =\frac{3}{13}\end{array}$

Now for finding out the limit of the sequence in case it exists:

Sequence is:

  $a_n=\frac{n+1}{n^2+1}$

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

Hence, the required limit of the sequence is $\underset{x\to \infty }{\lim }{a}_n=0$