Chapter 12.4, Problem 53E

To determine

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

Answer to Problem 53E

The first five terms of given sequence are $-1,\frac{1}{2},-\frac{1}{3},\frac{1}{4},\frac{1}{5}$ and its limit is $0$ .

Explanation of Solution

Given information:

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

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

  $\begin{array}{l}\text{Thus},forn=1,{\mathrm{a}}_1=\frac{{\left(-1\right)}^1}{1}\\ \Rightarrow {\mathrm{a}}_1=-1\\ forn=2,{\mathrm{a}}_2=\frac{{\left(-1\right)}^2}{2}\\ \Rightarrow {\mathrm{a}}_2=\frac{1}{2}\end{array}$

  $\begin{array}{l}forn=3,{\mathrm{a}}_3=\frac{{\left(-1\right)}^3}{3}\\ \Rightarrow {\mathrm{a}}_3=\frac{-1}{3}\end{array}$

  $\begin{array}{l}forn=4,{\mathrm{a}}_4=\frac{{\left(-1\right)}^4}{4}\\ \Rightarrow {\mathrm{a}}_4=\frac{1}{4}\\ forn=5,{\mathrm{a}}_5=\frac{{\left(-1\right)}^5}{5}\\ \Rightarrow {\mathrm{a}}_5=-\frac{1}{5}\end{array}$

Thus, the first five terms are $-1,\frac{1}{2},-\frac{1}{3},\frac{1}{4},\frac{1}{5}$

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

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

Hence, the first five terms of given sequence are $-1,\frac{1}{2},-\frac{1}{3},\frac{1}{4},\frac{1}{5}$ and its limit is $0$ .