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Math Help - Show this sequence converges

  1. #1
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    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}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tttcomrader View Post
    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
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    Quote Originally Posted by Jhevon View Post
    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)
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    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.
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