Chapter 12.4, Problem 52E

To determine

To find the first five terms of the sequence ${\mathrm{a}}_n=\frac{\left(3n-1\right)!}{\left(3n+1\right)!}$ and its limit.

Answer to Problem 52E

The first five terms of given sequence are $\frac{1}{12},\frac{1}{42},\frac{1}{90},\frac{1}{156},\frac{1}{240}$ and its limit is $0$ .

Explanation of Solution

Given information:

Given ${\mathrm{a}}_n=\frac{\left(3n-1\right)!}{\left(3n+1\right)!}$

Since we have ${\mathrm{a}}_n=\frac{\left(3n-1\right)!}{\left(3n+1\right)!}$

  $\begin{array}{l}=\frac{\left(3n-1\right)!}{\left(3n+1\right)\left(3n\right)\left(3n-1\right)!}\\ \therefore {\mathrm{a}}_n=\frac{1}{\left(3n+1\right)\left(3n\right)}\end{array}$

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

  $\begin{array}{l}\text{Thus},forn=1,{\mathrm{a}}_1=\frac{1}{\left(3\left(1\right)+1\right)\left(3\right)}\\ \Rightarrow {\mathrm{a}}_1=\frac{1}{4\times 3}\\ \Rightarrow {\mathrm{a}}_1=\frac{1}{12}\\ forn=2,{\mathrm{a}}_2=\frac{1}{\left(3(2)+1\right)\left(3(2)\right)}\\ \Rightarrow {\mathrm{a}}_2=\frac{1}{7\times 6}\\ \Rightarrow {\mathrm{a}}_2=\frac{1}{42}\end{array}$

  $\begin{array}{l}forn=3,{\mathrm{a}}_3=\frac{1}{\left(3(3)+1\right)\left(3(3)\right)}\\ \Rightarrow {\mathrm{a}}_3=\frac{1}{10\times 9}\\ \Rightarrow {\mathrm{a}}_3=\frac{1}{90}\end{array}$

  $\begin{array}{l}forn=4,{\mathrm{a}}_4=\frac{1}{\left(3(4)+1\right)\left(3(4)\right)}\\ \Rightarrow {\mathrm{a}}_3=\frac{1}{13\times 12}\\ \Rightarrow a_3=\frac{1}{156}\\ forn=5,{\mathrm{a}}_5=\frac{1}{\left(3(5)+1\right)\left(3(5)\right)}\\ \Rightarrow {\mathrm{a}}_3=\frac{1}{16\times 15}\\ \Rightarrow a_3=\frac{1}{240}\end{array}$

Thus, the first five terms are $\frac{1}{12},\frac{1}{42},\frac{1}{90},\frac{1}{156},\frac{1}{240}$

The limit of sequence is limit of ${\mathrm{a}}_n=\frac{1}{\left(3n+1\right)\left(3n\right)}$ as $n\to \infty$

  $\begin{array}{l}i.e.\underset{n\to \infty }{\lim }{\mathrm{a}}_n=\underset{n\to \infty }{\lim }\frac{1}{\left(3n+1\right)\left(3n\right)}\\ =\underset{n\to \infty }{\lim }\frac{1}{n^2\left(3\right)\left(3+\frac{1}{n}\right)}\\ \therefore \underset{n\to \infty }{\lim }{\mathrm{a}}_n=0\left(\because \underset{n\to \infty }{\lim }\frac{1}{n}=0\right)\end{array}$

Thus, the limit of ${\mathrm{a}}_n$ is $0$

Hence, the first five terms of given sequence are $\frac{1}{12},\frac{1}{42},\frac{1}{90},\frac{1}{156},\frac{1}{240}$ and its limit is $0$