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?

Answer 1

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 $n \ge n_0$:

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

Answer 2

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.

Answer 3

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.

Answer 4

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.