Is this sequence theorem true? 
If a sequence $\{a_n\}$ of non-negative reals is convergent, then $\{\sqrt a_n \}$ is also convergent.

Is this proposition true?
I think it is true but I don't know why it does make sense.
If it is false, tell me under what condition it is true.
 A: Yes, it’s true.
Two ways of looking at it:


*

*$\sqrt{·}$ as a function $[0..∞) → [0..∞)$ is continuous, so if $a_n \overset{n → ∞}→ a$ then $\sqrt{a_n} \overset{n → ∞}→ \sqrt{a}$ (this is the convergent sequence criterion for continuity).

*More elementary: Assume the limit $a = \lim_{n → ∞} a_n$ is not zero. Let $ε > 0$. Then choose $δ = \sqrt{a}ε$, so if $|a_n - a| < δ$, you get
$$|\sqrt{a_n} - \sqrt{a}| = \frac{|a_n - a|}{|\sqrt{a_n} + \sqrt{a}|} < \frac{δ}{\sqrt a} = ε,$$
and you can proceed to show convergence by definition, using the definition of the convergence of $(a_n)$ (that’s what the $δ$ is for). And for the case $a = 0$ you may choose $δ = ε^2$ and proceed similarly.


Note that by this you essentially prove the continuity of $\sqrt{·}$ using the sequence criterion.
A: Since we are doing real analysis, the sequence $\{\sqrt{a_n}\}$ may not even be defined. What if $\{a_n\}$ is the constant sequence with value $-1$?
However, as Git Gud said in his comment, if $\{\sqrt{a_n}\}$ actually exists (i.e., if $a_n \ge 0$ for all $n$—or at least for all sufficiently large $n$ if you're willing to cheat a little), then it is true.
A: Break it down to definitions.  For $\{a_n\}$ to converge to $L$,  it means that for any $\epsilon >0,\exists N\in \mathbb N$ such that $\forall n \ge n,|a_n -L|<\epsilon $
Now,  you want to show that $\{\sqrt {a_n}\}$ converges, so first you need to know where it converges to,  the obvious (and correct) answer is $\sqrt L$.  So, to prove that, you start with an arbitrary $\epsilon >0$.  Now, we want to find a $N \in \mathbb N$  such that for any number $n\ge N$,   $|\sqrt {a_n} -\sqrt L|<\epsilon $.
From here, can you find the $N$ that will work?  Hint:  you can use ANY number greater than $0$ from the first sequence converging, so you can use numbers of the form $2\cdot \epsilon,\epsilon ^2,\sqrt \epsilon $, etc.
A: $$a_{n}>0\\|a_{n}-l|<\epsilon\\|\sqrt{a_{n}}-\sqrt{l}|<\epsilon'\\|\sqrt{a_{n}}-\sqrt{l}|<\epsilon'\frac{|\sqrt{a_{n}}+\sqrt{l}|}{|\sqrt{a_{n}}+\sqrt{l}|}\\|a_{n}-l|<\epsilon'|\sqrt{a_{n}}+\sqrt{l}|<\epsilon'(|\sqrt{a_{n}}|+|\sqrt{l}|)<\epsilon'2max{|\sqrt{a_{n}}|}\\$$if name maximum a(n) az M so we have 
$$ \epsilon'2max{|\sqrt{a_{n}}|}<\\|a_{n}-l|<\epsilon<\epsilon'2M\\\epsilon'=\frac{\epsilon}{2M+1}$$
