Chapter 12.4, Problem 51E

To determine

To find the first five terms of the sequence ${\mathrm{a}}_n=\frac{\left(n+1\right)!}{n!}$ and its limit.

Answer to Problem 51E

The first five terms of given sequence are $2,3,4,5,6$ and its limit doesn’t exist.

Explanation of Solution

Given information:

Given ${\mathrm{a}}_n=\frac{\left(n+1\right)!}{n!}$

Since we have ${\mathrm{a}}_n=\frac{\left(n+1\right)!}{n!}$

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

The first five terms of the sequence ${\mathrm{a}}_n=n+1$ are obtained by substituting $n=1,2,3,4,5$ in general form of ${\mathrm{a}}_n$ .

  $\begin{array}{l}\text{Thus},forn=1,{\mathrm{a}}_1=1+1\\ \Rightarrow {\mathrm{a}}_1=2\\ forn=2,{\mathrm{a}}_2=2+1\\ \Rightarrow {\mathrm{a}}_2=3\end{array}$

  $\begin{array}{l}forn=3,{\mathrm{a}}_3=3+1\\ \Rightarrow {\mathrm{a}}_3=4\end{array}$

  $\begin{array}{l}forn=4,{\mathrm{a}}_4=4+1\\ \Rightarrow {\mathrm{a}}_4=5\\ forn=5,{\mathrm{a}}_5=5+1\\ \Rightarrow {\mathrm{a}}_5=6\end{array}$

Thus, the first five terms are $\frac{1}{5},\frac{4}{7},1,\frac{16}{11},\frac{25}{13}$

The limit of sequence is limit of ${\mathrm{a}}_n=n+1$ as $n\to \infty$

  $\begin{array}{l}i.e.\underset{n\to \infty }{\lim }{\mathrm{a}}_n=\underset{n\to \infty }{\lim }\left(n+1\right)\\ \therefore \underset{n\to \infty }{\lim }{\mathrm{a}}_n=\infty \end{array}$

Thus, the limit doesn’t exist as limit of ${\mathrm{a}}_n$ not approaches to a unique value as $n\to \infty$

Hence, the first five terms of given sequence are $2,3,4,5,6$ and its limit doesn’t exist.