Show that if $\{a_n\} \to L\text{ then }\{\sqrt{a_n}\} \to \sqrt{L}$

Question

Let $\{a_n\}$ be a sequence of positive numbers. Show that if $\{a_n\} \to L$ then $\{\sqrt{a_n}\} \to \sqrt{L}$.

Here is my proof:

Consider $$|\sqrt{a_n}-\sqrt{L}|=\frac{|\sqrt{a_n}-\sqrt{L}||\sqrt{a_n}+\sqrt{L}|}{|\sqrt{a_n}+\sqrt{L}|}=\frac{|a_n-L|}{|\sqrt{a_n}+\sqrt{L}|}.$$

Let $\epsilon>0$ be given. Since $a_n$ converges we know it is bounded. Let $M=\min\{|\sqrt{a_1}|, |\sqrt{a_2}|, \cdots , |\sqrt{a_n}|, \cdots\}$. Then we know that $$\frac{|a_n-L|}{|\sqrt{a_n}+\sqrt{L}|} \leq \frac{|a_n-L|}{|M+\sqrt{L}|}.$$ Choose $N$ such that for $n>N$, $|a_n-L| < \epsilon |M+\sqrt{L}|$.

Choosing one of these $n$ gives $\frac{|a_n-L|}{|M+\sqrt{L}| } < \epsilon $ which implies $|\sqrt{a_n}-\sqrt{L}|<\epsilon$.
So it follows that $\sqrt{a_n}\to \sqrt{L}$.

Should I make any improvements? Is there anything wrong with this proof?

It's much more natural to just write that $|\sqrt{a_n}+\sqrt{L}|\geq \sqrt{L}$ for $L\neq 0$ — WoolierThanThou, Oct 23 '19 at 14:18
Thank you. I was mainly concerned about using $\min{}$ and assuming ${\sqrt{a_n}}$ is bounded — Jac Frall, Oct 23 '19 at 14:19
@Jack if the min does not exist then I was thinking of using $M=0$ but that causes problems if $L=0$. How can I resolve this problem? — Jac Frall, Oct 23 '19 at 14:29
@JacFrall It may be easiest to handle $L=0$ separately "by hand" — Hagen von Eitzen, Oct 23 '19 at 14:36
[Prove that the function $\sqrt{x}$ is continuous on domain $0,\infty )$ — , Oct 23 '19 at 15:10
It is more natural to show that if $f$ is continuous and $x_n\to x$, then $f(x_n)\to f(x)$. In fact, this is one way to define continuity on a metric space. — Math1000, Oct 23 '19 at 15:20
See also: Is this a valid proof that $(a_n) \rightarrow l$ implies $(\sqrt{a_n}) \rightarrow \sqrt{l}$? — , Oct 23 '19 at 15:37

score 3 · Answer 1 · answered Oct 23 '19 at 14:17

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Your proof fails if $L=0$ and some $a_n=0$ too, because then $M+L=0$. You will have to change it it that case. Otherwise, it is correct.

answered Oct 23 '19 at 14:17

José Carlos Santos

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I forgot to mention that one of the assumptions is that $a_n$ is a sequence of positive numbers. I will update that. Sorry about that – Jac Frall Oct 23 '19 at 14:20
But it doesn't have to have a minimum and its infimum may well still be $0$. – José Carlos Santos Oct 23 '19 at 14:21
Ok that makes sense, how can I correct this? – Jac Frall Oct 23 '19 at 14:23
In that specific situation, for each $\varepsilon>0$, you take $N\in\mathbb N$ such that $n\geqslant N\implies a_n<\varepsilon^2$. Then $n\geqslant N\implies \sqrt{a_n}<\varepsilon$. – José Carlos Santos Oct 23 '19 at 14:33

score 2 · Answer 2 · answered Oct 23 '19 at 14:20

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Some potential issues in your proof:

You're dividing by zero if $M=0$ and $a_n=0$ in $\frac{|a_n-L|}{|\sqrt{a_n}+\sqrt{L}|}$.
If $\{a_n\}$ is the always vanishing sequence, $|a_n-L| < \epsilon |M+\sqrt{L}|$ will never happens as you're using a strict inequality.

You have to correct those two cases.

answered Oct 23 '19 at 14:20

mathcounterexamples.net

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Note that I updated it so that $a_n>0$. – Jac Frall Oct 23 '19 at 14:33

score 0 · Answer 3 · answered Oct 23 '19 at 14:38

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For the $L=0$ case:

If $a_n \to 0$ then $a_n < \epsilon^2 \implies \sqrt{a_n} < \epsilon$.

If $L > 0$ then there is a $n_0$ such that $n > n_0 \implies a_n > L/2 $.

You can then use this bound.

answered Oct 23 '19 at 14:38

marty cohen

107,799

I am somewhat confused when trying to implement this. Assume $L=0$ then $\frac{|a_n-L|}{|\sqrt{a_n}+\sqrt{L}|} = \frac{|a_n|}{|\sqrt{a_n}|} = |\sqrt{a_n}|<\epsilon$. Is this correct? – Jac Frall Oct 23 '19 at 14:51
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What I am saying is that you don't need to do this when $L = 0$. – marty cohen Oct 23 '19 at 14:52
Oh I see what you mean now. Thanks – Jac Frall Oct 23 '19 at 14:53
For the case when $L>0$ it would be choose $n>N$ such that $a_n>L/2$. Then $\frac{|a_n-L|}{|\sqrt{L/2}+\sqrt{L}|}$ so choose $k>n$ such that $|a_k-L|<\epsilon |\sqrt{L/2}+\sqrt{L}|$. Then we are done – Jac Frall Oct 23 '19 at 15:05
Is this correct? – Jac Frall Oct 23 '19 at 15:15
It seems to be. – marty cohen Oct 23 '19 at 15:23

Peter Szilas · Accepted Answer · 2019-10-23T16:30:26.543

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1) Treat $L=0$, see Marty Cohen's answer.

2)$L>0$;

$|√a_n-√L| =|\dfrac{a_n-L}{√a_n+√L}| \le \dfrac{|a_n-L|}{√L}$.

Let $\epsilon >0$ be given

For $√L\epsilon >0$ there is a $n_0$ s.t. for $n \ge n_0$

$|a_n-L| <√L\epsilon$.

Then, for $n \ge n_0$:

$|√a_n-√L| \le \dfrac{|a_n-L|}{√L} \lt \epsilon$.

edited Oct 23 '19 at 16:30

answered Oct 23 '19 at 16:17

Peter Szilas

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I find your answer to be the most helpful. Why the downvotes? Is there something wrong with this? – Jac Frall Oct 27 '19 at 23:29
Jac Frall.Thanks. Perhaps something wrong with the down voter ? The answer seems OK. – Peter Szilas Oct 28 '19 at 01:55

Show that if $\{a_n\} \to L\text{ then }\{\sqrt{a_n}\} \to \sqrt{L}$

4 Answers4