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

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

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

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

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

So then we have,

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

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: