Prove that if $a_n$ is a Cauchy sequence then $\sqrt{a_n}$ is also Cauchy. If $(a_{n})_{n \in \mathbb{N}}$ is a Cauchy sequence with : $\forall n \in \mathbb{N}, \; a_{n} \geq 1$. Then $(\sqrt{a_{n}})_{n \in \mathbb{N}}$ is a Cauchy sequence as well. 
How do I show $\vert \sqrt{a_{n}} - \sqrt{a_{m}}\vert < \varepsilon $?
 A: Hint Since $a_n\ge1$ we have
$$|\sqrt{a_n}-\sqrt{a_m}|=\frac{|a_n-a_m|}{|\sqrt{a_n}+\sqrt{a_m}|}\leq\frac{1}{2}|a_n-a_m|$$
Can you take it from here?
A: Consider $f(x) = \sqrt{x}$, then
$$
f'(x) = \frac{1}{2\sqrt{x}} \leq \frac{1}{2} \quad\forall x\geq 1
$$
By the Mean-value theorem, $f$ is uniformly continuous on $[1,\infty)$.
Now show that if $(a_n)$ is cauchy, and $f$ is uniformly continuous, then $(f(a_n))$ is also Cauchy.
A: I can see two different ways of proving that $(\sqrt{a_{n}})_{n \in \mathbb{N}}$ is a Cauchy sequence :


*

*If $(a_{n})_{n \in \mathbb{N}}$ is a Cauchy sequence such that : $\forall n \in \mathbb{N}, \; a_{n} \in I$ and $f \, : \, I \, \longrightarrow \, \mathbb{R}$ is uniformly continuous, then $(f(a_{n}))_{n \in \mathbb{N}}$ is also a Cauchy sequence. To apply this here, remember that a Cauchy sequence is always bounded. In addition to this, you have : $\forall n \in \mathbb{N}, \; a_{n} \geq 1$. As a consequence, there exists $M \in \mathbb{R}^{+}$ such that : $\forall n \in \mathbb{N}, \; a_{n} \in [1,M]$. Since $x \, \longmapsto \, \sqrt{x}$ is continuous on $[1,M]$, it follows from Heine's theorem that it is uniformly continuous on $[1,M]$. It proves that $(\sqrt{a_{n}})_{n \in \mathbb{N}}$ is a Cauchy sequence.

*You can also prove it directly. Notice that : $$\forall (n,m) \in \mathbb{N}^{2}, \; \displaystyle \vert \sqrt{a_{n}} - \sqrt{a_{m}} \vert \leq \frac{\vert a_{n}-a_{m} \vert}{\vert \sqrt{a_{n}} + \sqrt{a_{m}} \vert} \leq \frac{1}{2} \vert a_{n} - a_{m} \vert$$
