Chapter 13.4, Problem 34E

To determine

To find : If the sequence is convergent, find its limit, if it divergent explain why.

Answer to Problem 34E

The sequence is convergent with limit 3.

Explanation of Solution

Given information:

The given sequence is $a_n=\frac{12}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2$

Calculation :

A sequence $a_1$ , $a_2$ , $a_3$ …… has the limit $L$ and we write

  $\underset{n\to \infty }{\lim }{a}_n=L$

if the nth term $a_n$ of the sequence can be made arbitrarily close to $L$ by taking $n$ sufficiently large. If $\underset{n\to \infty }{\lim }{a}_n$ exists, then the sequences converges otherwise it diverges.

Find $\underset{n\to \infty }{\lim }{a}_n$.

  $\underset{n\to \infty }{\lim }\frac{12}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2=\underset{n\to \infty }{\lim }\frac{12}{n^4}\left[\frac{n^2\left(n^2+2n+1\right)}{2}\right]$

  $\underset{n\to \infty }{\lim }\frac{12}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2=\underset{n\to \infty }{\lim }\frac{3n^4+6n^3+3n^2}{n^4}$

  $\underset{n\to \infty }{\lim }\frac{12}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2=\underset{n\to \infty }{\lim }\frac{\frac{3n^4}{n^4}+\frac{6n^3}{n^4}+\frac{3n^2}{n^4}}{\frac{n^4}{n^4}}\text{ (Divide numerator and denominator by }{n}^4\text{)}$

  $\underset{n\to \infty }{\lim }\frac{12}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2=\underset{n\to \infty }{\lim }\frac{3+\frac{6}{n}+\frac{3}{n^2}}{1}$

  $\underset{n\to \infty }{\lim }\frac{12}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2=\frac{3+\frac{6}{\infty }+\frac{3}{{\infty }^2}}{1}$

  $\underset{n\to \infty }{\lim }\frac{12}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2=\frac{3+0+0}{1}\text{ (Since }\frac{a}{\infty }=0\text{)}$

  $\underset{n\to \infty }{\lim }\frac{12}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2=\frac{3}{1}$

  $\underset{n\to \infty }{\lim }\frac{12}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2=3$

Hence,

The sequence is convergent with limit 3.