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Math Help - Convergence of sequence using the definition

  1. #1
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    Convergence of sequence using the definition

    Hi, I know how to show the following sequence goes to its limit, but i need to do it using the definition, of which i am having trouble, i've started how i should though i think.

    x_{n} = \sqrt{n+1} - \sqrt{n}
    prove x_{n} \rightarrow 0

    Let \epsilon > 0 be given
    Let N=N(\epsilon ) be an integer greater than ...(answer goes here)...
    Then for all n>N we have
    |x_{n} - l| = |\sqrt{n+1}-\sqrt{n} - 0|

    |x_{n} - l| = |\sqrt{n+1} - \sqrt{n}|


    \sqrt{n+1} - \sqrt{n} = \dfrac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}}<br />
    |x_{n} - l| = \dfrac{1}{\sqrt{n+1} + \sqrt{n}}

    and that is where i get stuck.
    I'm not sure if this definition method of proof is well known so if you would like an example of a completed question just ask. Thanks.
    Last edited by Rapid_W; August 2nd 2010 at 06:43 AM. Reason: dfrac instead of frac
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  2. #2
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    Quote Originally Posted by Rapid_W View Post
    x_{n} = \sqrt{n+1} - \sqrt{n}
    prove x_{n} \rightarrow 0

    Let \epsilon > 0 be given
    Let N=N(\epsilon ) be an integer greater than ...(answer goes here)...
    Then for all n>N we have
    |x_{n} - l| = |\sqrt{n+1}-\sqrt{n} - 0|.
    The first thing to note is x_n=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}.
    Then \left|\dfrac{1}{\sqrt{n+1}+\sqrt{n}} -0\right|\le\dfrac{1}{2\sqrt{n}} .
    Can you finish?
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  3. #3
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    Yes, that is the missing link, thanks very much.

    i did think of doubling it, but i did this instead \dfrac{1}{\sqrt{2n}}, which isn't true lol

    Do i have to be able to prove that though, or state a theory?
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  4. #4
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    Prove what? That \sqrt{n+1}- \sqrt{n}= \frac{1}{\sqrt{n+1}+ \sqrt{n}}?

    Rationalize the numerator: multiply both numerator and denominator of \frac{\sqrt{n+1}- \sqrt{n}}{1} by \sqrt{n+1}+ \sqrt{n}.
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  5. #5
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    no sorry, perhaps i am being silly but, prove that \dfrac{1}{\sqrt{n+1} + \sqrt{n}} \le \dfrac{1}{2\sqrt{n}}

    also, the answer is N=N(\epsilon ) = integer > 4\epsilon ^2?
    Last edited by Rapid_W; August 2nd 2010 at 06:53 AM.
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  6. #6
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    \sqrt{n+1}+\sqrt{n}\ge\sqrt{n}+\sqrt{n}=2\sqrt{n}.

    SO?
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  7. #7
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    so 1 over then all is the same, ok got that, so if i finish of my work i get

    |x_{n}-l| \le \dfrac{1}{2\sqrt{n}}
    n > N so
    |x_{n}-l| \le \dfrac{1}{2\sqrt{N}}
    \dfrac{1}{2\sqrt{N}} < \epsilon
    If  N > 4\epsilon ^2
    |x_{n}-l| \le \epsilon

    and then i write 4\epsilon ^2 in the gap at the top.

    Is this right?
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