Chapter 13.4, Problem 32E

To determine

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

Answer to Problem 32E

The sequence is convergent with limit 15.

Explanation of Solution

Given information:

The given sequence is $a_n=\frac{5}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)$

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

  $\underset{n\to \infty }{\lim }\frac{5}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)=\underset{n\to \infty }{\lim }\left(5+\frac{20}{n^2}\left[\frac{n^2+n}{2}\right]\right)$

  $\underset{n\to \infty }{\lim }\frac{5}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)=\underset{n\to \infty }{\lim }\left(5+\frac{10n^2+10n}{n^2}\right)$

  $\underset{n\to \infty }{\lim }\frac{5}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)=\underset{n\to \infty }{\lim }5+\underset{n\to \infty }{\lim }\frac{10n^2+10n}{n^2}$

  $\underset{n\to \infty }{\lim }\frac{5}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)=5+\underset{n\to \infty }{\lim }\frac{\frac{10n^2}{n^2}+\frac{10n}{n^2}}{\frac{n^2}{n^2}}\text{ (Divide numerator and denominator by }{n}^2\text{)}$

  $\underset{n\to \infty }{\lim }\frac{5}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)=5+\underset{n\to \infty }{\lim }\frac{10+\frac{10}{n}}{1}$

  $\underset{n\to \infty }{\lim }\frac{5}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)=5+\frac{10+\frac{10}{\infty }}{1}$

  $\underset{n\to \infty }{\lim }\frac{5}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)=5+\frac{10+0}{1}\text{ (Since }\frac{a}{\infty }=0\text{)}$

  $\underset{n\to \infty }{\lim }\frac{5}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)=5+10$

  $\underset{n\to \infty }{\lim }\frac{5}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)=15$

Hence,

The sequence is convergent with limit 15.