# Show this sequence converges

• Oct 3rd 2008, 04:10 PM
Show this sequence converges
Suppose that the sequence $\{ a_n \}$ of non-negative numbers converges to a point $a$. Show that $\{ \sqrt {a_n} \} \rightarrow \sqrt {a}$
• Oct 3rd 2008, 04:53 PM
Jhevon
Quote:

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 $\epsilon$
• Oct 4th 2008, 03:22 AM
Laurent
Quote:

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)
• Oct 4th 2008, 06:37 AM
Opalg
Quote:

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.