# Thread: Show this sequence converges

1. ## Show this sequence converges

Suppose that the sequence $\{ a_n \}$ of non-negative numbers converges to a point $a$. Show that $\{ \sqrt {a_n} \} \rightarrow \sqrt {a}$

Suppose that the sequence $\{ a_n \}$ of non-negative numbers converges to a point $a$. Show that $\{ \sqrt {a_n} \} \rightarrow \sqrt {a}$
hint: note that $\sqrt{a_n} - \sqrt{a} = \frac {(\sqrt{a_n} - \sqrt{a})(\sqrt{a_n} + \sqrt{a})}{\sqrt{a_n} + \sqrt{a}} < \frac {a_n - a}{\sqrt{a}} \le \frac {|a_n - a|}{\sqrt{a}}$

of course, we want the last expression to be less than $\epsilon$

3. Originally Posted by Jhevon
hint: note that $\sqrt{a_n} - \sqrt{a} = \frac {(\sqrt{a_n} - \sqrt{a})(\sqrt{a_n} + \sqrt{a})}{\sqrt{a_n} + \sqrt{a}} < \frac {a_n - a}{\sqrt{a}} \le \frac {|a_n - a|}{\sqrt{a}}$
(just add absolute values from the beginning to make it quite rigorous)

4. Originally Posted by Jhevon
hint: note that $\sqrt{a_n} - \sqrt{a} = \frac {(\sqrt{a_n} - \sqrt{a})(\sqrt{a_n} + \sqrt{a})}{\sqrt{a_n} + \sqrt{a}} < \frac {a_n - a}{\sqrt{a}} \le \frac {|a_n - a|}{\sqrt{a}}$
Also note that you may have to consider the case a=0 separately.