Answer to Question #158165 in Calculus for sai

Question

Use the definition of limit to prove that the sequence {n-1/n} n=1 to infinity is divergent

Answer 1

Solution

The number $L$ is the limit of the sequence $\{a_n\}$ if

$(1)\ \epsilon > 0, \ a_n \underset{e}{\simeq } L, for\ n \geq 1.$

If such an L exists, we say $\{a_n\}$converges, or is convergent; if not, $\{a_n\}$ diverges,or is divergent.

Let $L∈R$. We claim that the sequence $s_n=\begin{Bmatrix} n-\frac{1}{n} \end{Bmatrix}n$ does not converge to L. To show this, take $ε= 1$. Since $s_n=\begin{Bmatrix} n-\frac{1}{n} \end{Bmatrix}n \geq n$ , if $n >|L|+ 1$ we have that

$$|s_n−L|≥|s_n|−|L|≥n−|L|>1 =ε.$$

It follows that for any $N∈N$, if we take $n >max\{|L|+ 1,N\}$ then $|sn−L|> ε$ . Thus by the definition of the limit $(s_n)$ diverges.

The notations for the limit of a sequence in this case are:

$$\overset{lim}{n \to \infty} \begin{Bmatrix} n-\frac{1}{n} \end{Bmatrix}= \infty$$

This sequence $\begin{Bmatrix} n-\frac{1}{n} \end{Bmatrix}_{n = 1}$ is divergent