Chapter 12, Problem 92RE

To determine

To write: the first five terms of the sequence and the limit of the sequence.

Answer to Problem 92RE

The first five terms of the sequence are

  $-\frac{1}{2},-\frac{9}{8},-\frac{7}{6},-\frac{37}{32},-\frac{57}{50}$

And

  $\underset{n\to \infty }{\lim }{a}_n=-1$

Explanation of Solution

Given information:

The sequence is

  $a_n-\frac{1}{2n^2}\left[3-2n\left(n+1\right)\right]$

Calculation:

Consider the following sequence

  $a_n-\frac{1}{2n^2}\left[3-2n\left(n+1\right)\right]$

Substitute $n=1$ in the expression $a_n=\frac{1}{2n^2}\left[3-2n\left(n+1\right)\right]$

  $\begin{array}{c}{a}_1=\frac{1}{2\times 1^2}\left[3-2\times 1\left(1+1\right)\right]\\ =-\frac{1}{2}\end{array}$

Substitute $n=2$ in the expression $a_n=\frac{1}{2n^2}\left[3-2n\left(n+1\right)\right]$

  $\begin{array}{c}{a}_2=\frac{1}{2\times 2^2}\left[3-2\times 2\left(2+1\right)\right]\\ =-\frac{9}{8}\end{array}$

Substitute $n=3$ in the expression $a_n=\frac{1}{2n^2}\left[3-2n\left(n+1\right)\right]$

  $\begin{array}{c}{a}_3=\frac{1}{2\times 3^2}\left[3-2\times 3\left(3+1\right)\right]\\ =-\frac{7}{6}\end{array}$

Substitute $n=4$ in the expression $a_n=\frac{1}{2n^2}\left[3-2n\left(n+1\right)\right]$

  $\begin{array}{c}{a}_4=\frac{1}{2\times 4^2}\left[3-2\times 4\left(4+1\right)\right]\\ =-\frac{37}{32}\end{array}$

Substitute $n=5$ in the expression $a_n=\frac{1}{2n^2}\left[3-2n\left(n+1\right)\right]$

  $\begin{array}{c}{a}_5=\frac{1}{2\times 5^2}\left[3-2\times 5\left(5+1\right)\right]\\ =-\frac{57}{50}\end{array}$

Therefore the first five terms of the sequence are

  $-\frac{1}{2},-\frac{9}{8},-\frac{7}{6},-\frac{37}{32},-\frac{57}{50}$

Limits of the sequence can be found as

  $\begin{array}{c}\underset{n\to \infty }{\lim }{a}_n=\frac{1}{2n^2}\left[3-2n\left(n+1\right)\right]\\ =\frac{1}{2n^2}\left(3-2n^2-2n\right)\\ =\frac{-2}{2}\\ =-1\end{array}$

The value of the limit is found to be $-1$ .