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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.

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Hint:

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

implies

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

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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.

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

$$-\epsilon<a_n-L<\epsilon\implies -\epsilon-2L<-\epsilon+2L<a_n+L<\epsilon+2L \implies \color{red}{|a_n+L|=a_n+L<\epsilon+2L}$$

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

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

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