I'm trying to show that the sequence $\sqrt{n+1} – \sqrt{n}$ converge toward $L = 0$.
We have :
$$
\begin{align}
|u_n – L| = \left| \left( \sqrt{n+1)} – \sqrt{n} \right) \cdot \left( \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} \right) \right| = \left| \frac{1}{\sqrt{n+1} + \sqrt{n}} \right| \leq \left| \frac{1}{\sqrt{n+
1}} \right|
\end{align}
$$
For every $\epsilon > 0$, by the Archimedean property, we can take
$$
N \geq \frac{1}{\epsilon^2} – 1
$$
Such that if $n \in \mathbb{N}$ and $n \geq N$,
$$
|u_n – L| \leq \left| \frac{1}{\sqrt{n+1}} \right| \leq \frac{1}{\sqrt{N+1}} \leq \epsilon
$$
Eventually, we can write :
$$
\lim_{n \rightarrow \infty} \sqrt{n+1} – \sqrt{n} = 0
$$
Is it a correct way to prove that the sequence converge ?
Best Answer
This is very good. You are missing some parentheses, as was mentioned. I would rephrase your application of the Archimedean property. Generally it refers to your $N$ as being an integer. So, you would need $$N\ge \frac{1}{\epsilon^2} - 1.$$ Also you are using $N$ for two different things, just use $\mathbb{N}$ (
\mathbb{N}) or $\mathbb{Z}^+$ (\mathbb{Z}^+) for the set of natural numbers. I don't know your prefference but I like to end with a statement such as "since for any $\epsilon >0$ there exists an $N\in \mathbb{N}$ such that $n\ge N$ implies $|u_n|<\epsilon$, by definition, $$\lim_{n\rightarrow \infty} \sqrt{n+1} -\sqrt{n} = 0."$$ Very good work.