Prove that $ (\sqrt{a_{n}})_{n=1}^{\infty}\rightarrow \sqrt{a} $ when $ (a_{n})_{n=1}^{\infty}\rightarrow a $ I'm stuck on a homework question, and would be very thankful to anyone who helps. The question:
Prove or disprove the following statement: 
If:   $ (a_{n})_{n=1}^{\infty} $ is a series of positive numbers and $(a_{n})_{n=1}^{\infty}\rightarrow a $
Then: $ (\sqrt{a_{n}})_{n=1}^{\infty}\rightarrow \sqrt{a} $
Hint: Distinguish between $a = 0$ and $a > 0$
OK, so I'm pretty sure that the statement is correct, I have the part for when a = 0 proved, but I'm stuck on the second part (when a > 0) I'm having "writer's block" and can't come up with any solution that would prove it...
Any hints and help would be greatly appreciated
Thank you!
EDIT - The series is a series of positive numbers

Hint:
We'd like to show that for any $\varepsilon > 0$ there is $N$ such that $n > N \Rightarrow |\sqrt{a_n} - \sqrt{a}| < \varepsilon$. Given such an $\varepsilon$ we know there is $N$ such that $n > N \Rightarrow |a_n - a| < \varepsilon$. But what is $\sqrt{a_n} - \sqrt{a}$? We know $a_n - a = (\sqrt{a_n} - \sqrt{a})(\sqrt{a_n} + \sqrt{a})$. If $a > 0$ the you can bound the second factor by a positive constant (for you to find), call it $C$. Then take the absolute value of both sides, and let $N$ such that $n > N$ means $|a_n - a| < \varepsilon/C$.
