# Prove $\lim_{n\to \infty}\frac{n}{n+1} = 1$ using Cauchy (epsilon) definition of a limit of a sequence

I have to prove $\lim_{n\to \infty}\frac{n}{n+1} = 1$ using Cauchy (epsilon) definition of a limit of a sequence.

How do I do that formally? It is clear to me that the limit is 1, after doing some manipulations with the fraction, but how do I show it formally using the Cauchy definition of a limit?

• First off, you should always show your work, give definitions, give context and motivation for the problem. Secondly: Hint? $$\frac n{n+1}=1-\frac1{n+1}$$ Jul 12, 2017 at 17:37

All these proofs start the same way: let $\epsilon>0$ be given. Now, we want to produce an $N\in\Bbb N$ such that whenever $n> N$, the inequality $$\left|\frac{n}{n+1}-1\right| < \epsilon \tag{1}$$ holds. We need to manipulate the expression $\left|\frac{n}{n+1}-1\right|$ into a form that we can say more about. Well, we can put everything over a common denominator to get $$\left|1-\frac{1}{n+1}\right|.$$ Now, pick $N$ so large that $\frac{1}{N}<\epsilon/2$. (Why is this possible?) Now conclude that if $n>N$, $(1)$ holds using what you know about absolute values in inequalities.
Take $\varepsilon>0$. Now, since$$\left|1-\frac n{n+1}\right|=\frac1{n+1}\text,$$in order to have $\left|1-\frac n{n+1}\right|<\varepsilon$, you need to have $\frac1{n+1}<\varepsilon$. So, take $p\in\mathbb N$ such that $\frac1{p+1}<\varepsilon$. Then$$n\geqslant p\Longrightarrow\frac1{n+1}\leqslant\frac1{p+1}<\varepsilon.$$
$$\left|\frac{n}{n+1}-1\right|=\left|\frac{1}{n+1}\right|$$
for a given small $\epsilon>0$, choose $$N=\left\lceil \frac{1}{\epsilon}-1\right\rceil$$ and take $n>N$