# Given a Cauchy sequence $a_n$, show that $\sqrt{a_n}$ is Cauchy when $a_n>0$ for all $n$.

We have a sequence $$a_n$$, that is Cauchy and every term is positive. How do I find that $$\sqrt{a_n}$$ is also Cauchy? I have seen a similar question posted but in that question $$a_n>1$$ so it is not the same. I understand well how to do it if $$a_n>1$$ but I can't understand how to alter the solution to account for the cases where $$a_n$$ and $$a_m$$ are less than 1. Thank you.

Hint:

$$|\sqrt{a_m}-\sqrt{a_n}|^2 \leqslant |\sqrt{a_m}-\sqrt{a_n}| |\sqrt{a_m}+\sqrt{a_n}|$$

implies

$$|\sqrt{a_m}-\sqrt{a_n}| \leqslant \sqrt{|a_m-a_n|}$$

If $$(a_n)_{n\in\mathbb N}$$ is a Cauchy sequence, then it converges to some $$l\geqslant0$$. Therefore, $$\left(\sqrt{a_n}\right)_{n\in\mathbb N}$$ converges to $$\sqrt l$$ and so it is a Cauchy sequence.

• Could you please provide a proof of this? – user208480 Jun 3 at 21:09
• Which assertion do you want me to prove? – José Carlos Santos Jun 3 at 21:15

Suppose $$\;a_n\xrightarrow[n\to\infty]{}L\;$$, then for any $$\;\epsilon>0\;$$ there exists $$\;N_\epsilon\in\Bbb N\;$$ s.t. $$\;n>N_\epsilon\implies |a_n-L|<\epsilon\;$$.

Observe that $$\;L\ge 0\;$$ (why?), and we shall assume now that $$\;L>0\;$$ you do the appropiratie corrections for

the case $$\;L=0\;$$ . The last inequality in line $$\;1\;$$ is equivalent with

$$-\epsilon

Thus, we get for $$\;n>N_\epsilon\;$$:

$$|\sqrt{a_n}-\sqrt L|=\frac{|a_n-L|}{a_n+L}<\frac\epsilon L\implies\sqrt{a_n}\xrightarrow[n\to\infty]{}\sqrt L$$