Chapter 11.4, Problem 49E

To determine

To calculate:

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

Answer to Problem 49E

First five terms are $2,3,4,5,6$ and limit of the sequence does not exist.

Explanation of Solution

Given information:

  $a_n=\frac{\left(n+1\right)!}{n!}$

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $a_n=\frac{\left(n+1\right)!}{n!}$

Now Simply factorial the given term:

  $\begin{array}{l}{a}_n=\frac{\left(n+1\right)!}{n!}\\ =\frac{\left(n+1\right)!.n!}{n!}\\ =n+1\end{array}$

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

For $n=1$ :

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

For $n=2$ :

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

For $n=3$ :

  $\begin{array}{l}{a}_n=3+1\\ =4\end{array}$

For $n=4$ :

  $\begin{array}{l}{a}_n=4+1\\ =5\end{array}$

For $n=5$ :

  $\begin{array}{l}{a}_n=5+1\\ =6\end{array}$

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

For the rational function $f\left(x\right)=\frac{N\left(x\right)}{D\left(x\right)}$

Where,

  $N\left(x\right)=a_n{x}^n+\mathrm{...}+a_0$

  $D\left(x\right)=b_m{x}^m+\mathrm{...}+b_0$

The limit $x$ approaches positive or negative infinity as as below:

  $\underset{x\to \infty }{\lim }f\left(x\right)=\{\begin{array}{l}0,ifn<m\\ \frac{a_n}{b_m},ifn=m\end{array}$

The limit does not exist if $n>m$

Look at the $n^{th}$ term of the sequence,

  $\begin{array}{l}{a}_n=n+1\\ =\frac{n+1}{n^0}\end{array}$

As the value of the $n^0=1$

Hence, the degree of the numerator is grater then the degree of the denominator. So the limit of the function does not exist.