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Math Help - Limit of a sequence

  1. #1
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    Limit of a sequence

    Prove that the limit of {Sqrt(n+1)-Sqrt(n)} = 0.
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  2. #2
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    Quote Originally Posted by tarheelborn View Post
    Prove that the limit of {Sqrt(n+1)-Sqrt(n)} = 0.

    Multiply by conjugate: \sqrt{n+1}-\sqrt{n}=\left(\sqrt{n+1}-\sqrt{n}\right)\,\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{  n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}=...

    Tonio
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    Quote Originally Posted by tarheelborn View Post
    Prove that the limit of {Sqrt(n+1)-Sqrt(n)} = 0.
    Dear tarheelborn,

    You can show this using the limit definition for a sequence.

    i.e:~\lim_{n\rightarrow\infty}\{a_n\}=a\Longleftri  ghtarrow~\forall~\epsilon>0~~\exists~n_{o}\in{Z^+}  ~such~that~n>n_{o}\Rightarrow{\mid~a_{n}-a~\mid}<\epsilon

    Of course you will have to multiply by the conjugate as tonio had pointed out.

    Hope you can continue.
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  4. #4
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    Quote Originally Posted by tonio View Post
    Multiply by conjugate: \sqrt{n+1}-\sqrt{n}=\left(\sqrt{n+1}-\sqrt{n}\right)\,\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{  n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}=...

    Tonio
    Actually, I got this far on my own. I am not sure how to solve in terms of Epsilon.
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  5. #5
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    Would it be acceptable to prove this using 1/sqrt(n+1) instead of the whole denominator 1/(sqrt(n+1)-sqrt(n))?
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  6. #6
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    Quote Originally Posted by tarheelborn View Post
    not sure how to solve in terms of Epsilon.
    Well \frac{1}{{\sqrt {n + 1}  + \sqrt n }} \leqslant \frac{1}<br />
{{2\sqrt n }} < \varepsilon .
    So how large must n be?
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  7. #7
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    So n would be < 1/4*epsilon^2, right?
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  8. #8
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    Quote Originally Posted by tarheelborn View Post
    So n would be < 1/4*epsilon^2, right?
    Try greater than: n \geqslant \left\lfloor {\frac{1}{{4\varepsilon^2 }} + 1} \right\rfloor .
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