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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 $?

hrkrshnn
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Pat Green
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3 Answers3

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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?

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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.

Prahlad Vaidyanathan
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  • Isn't continuity enough? For a continuous $f$, $u_n$ Cauchy $\iff$ $u_n$ convergent $\implies$ $f(u_n)$ convergent $\implies$ $f(u_n)$ is Cauchy. – Jean-Claude Arbaut Dec 11 '13 at 13:51
  • @arbautjc He uses uniform continuity, not (ordinary) continuity. – Hanul Jeon Dec 11 '13 at 14:04
  • @tetori Yes, I know, and it's my question: isn't continuity enough? Of course, conituous on a compact $\implies$ uniformly continuous (and since $a_n$ is convergent, you can work on a compact), but it's not the point, I don't think we need this. – Jean-Claude Arbaut Dec 11 '13 at 14:10
  • @arbautjc Oh, sorry. I misunderstood your comment.
    Continuity is not enough condition. Consider $f:(0,\infty)\to\Bbb{R}$, $f(x)=1/x$, $a_n=1/n$. We know that $f$ is continuous and $a_n$ is Cauchy. But $f(a_n)$ is not Cauchy.     – Hanul Jeon Dec 11 '13 at 14:14
  • Of course, I mean, continuous, and limit of $a_n$ is in domain of continuity ;-) – Jean-Claude Arbaut Dec 11 '13 at 14:18
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I can see two different ways of proving that $(\sqrt{a_{n}})_{n \in \mathbb{N}}$ is a Cauchy sequence :

  1. 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.

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

pitchounet
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