Find an N such that the convergent sequence is within ε of its limit

by RoRi

February 28, 2016

Consider the convergent sequence $\{ a_n \}$ with terms defined by

$a_n = \frac{1}{n}.$

Let $L = \lim_{n \to \infty} a_n$ . Find the value of $L$ and values of $N$ such that $|a_n - L| < \varepsilon$ for all $n \geq N$ for each of the following values of $\varepsilon$ :

$\varepsilon = 1$ ,
$\varepsilon = 0.1$ ,
$\varepsilon = 0.01$ ,
$\varepsilon = 0.001$ ,
$\varepsilon = 0.0001$ .

First, we know

$\lim_{n \to \infty} \frac{1}{n} = 0 \quad \implies \quad L= 0.$

So then we have,

$\begin{align*} |a_n - L| < \varepsilon && \implies && \left| \frac{1}{n} \right| &< \varepsilon \\ && \implies && \frac{1}{n} &< \varepsilon \\ && \implies && n &> \frac{1}{\varepsilon}. \end{align*}$

Thus, if $N > \frac{1}{\varepsilon}$ then for every $n \geq N$ we have $|a_n| < \varepsilon$ . We compute for the given values of $\varepsilon$ as follows:

$\varepsilon = 1$ implies $N = \frac{1}{1} = 1$ .
$\varepsilon = 0.1$ implies $N = \frac{1}{.1} = 10$ .
$\varepsilon = 0.01$ implies $N = \frac{1}{.01} = 100$ .
$\varepsilon = 0.001$ implies $N = \frac{1}{.001} = 1000$ .
$\varepsilon = 0.0001$ implies $N = \frac{1}{.0001} = 10000$ .

Related

3 comments

August 19, 2018 at 7:47 pm

Roberto Furini Filho says:

N>1 , N>10, N>100…..N>10000

Reply
- July 16, 2020 at 9:32 am
  
  Anonymous says:
  
  Greetings!
  
  I think you mean n>1, n>10, $\dots$ as \textbf{N refers to the natural number n has to be larger than} for $| f - L | < \epsilon$ to be true.
  
  Cheers!
  
  Reply
- July 16, 2020 at 9:45 am
  
  Anonymous says:
  
  Roberto Furini Filho is right since $n\geN$.
  
  Reply

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