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Math Help - prove that a sequence converges

  1. #1
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    prove that a sequence converges

    Prove that the sequence x_{n}=n^{1/n} converges.
    I know it converges to 1. We need to use the definition of sequence convergence: x_{n} \rightarrow L \Leftrightarrow(\forall\epsilon>0)(\exists N\in \mathbb{N})(\forall n\in \mathbb{N})(n>N \Rightarrow |x_{n}-L|<\epsilon)
    Another thing, obviously |x_{n}-L|=|n^{1/n}-1|=n^{1/n}-1 since n^{1/n}-1\geq0
    Last edited by santiagos11; April 30th 2010 at 03:12 PM.
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  2. #2
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    Quote Originally Posted by santiagos11 View Post
    Prove that the sequence x(n)=n^(1/n) converges.
    Let y_n=\sqrt[n]{n} - 1 so y_n\ge 0 and
    by the binominal theorem n=(y_n+1)^n>\frac{n(n-1)}{2}y_n^2.
    Therefore 0\le y_n<\sqrt{\frac{2}{n-1}}.
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    Well, you are certanly right. But how does that fit the standard definition?
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  4. #4
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    Quote Originally Posted by santiagos11 View Post
    Well, you are certanly right. But how does that fit the standard definition?
    Having taught this 'stuf' for thirty+ years I have never seen a \epsilon\-\delta proof given for this limit.
    So I can't imagine what you mean be 'standard'.
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  5. #5
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    For example, here is a partial proof:
    Let \epsilon>0. Clearly 1\leq n^{1/n}\leq 2 \ \forall n\in \mathbb{N}
    case 1: \epsilon>1
    Let N=1. Suppose n>N. Then |x_{n}-1|=|n^{1/n}-1|=n^{1/n}-1. On the other hand \epsilon + 1 > 2 \geq n^{1/n}, so n^{1/n}-1<\epsilon. Therefore, |x_{n}-1|<\epsilon.
    Do you see how this follows the definition that I gave in the top?
    Now it remains to show the same, in the case 0<\epsilon\leq1
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  6. #6
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    Quote Originally Posted by santiagos11 View Post
    For example, here is a partial proof:
    Let \epsilon>0. Clearly 1\leq n^{1/n}\leq 2 \ \forall n\in \mathbb{N}
    case 1: \epsilon>1
    Let N=1. Suppose n>N. Then |x_{n}-1|=|n^{1/n}-1|=n^{1/n}-1. On the other hand \epsilon + 1 > 2 \geq n^{1/n}, so n^{1/n}-1<\epsilon. Therefore, |x_{n}-1|<\epsilon.
    Do you see how this follows the definition that I gave in the top?
    Now it remains to show the same, in the case 0<\epsilon\leq1
    That is a waste of effort. There is absolutely no reason to reinvent the wheel.
    The object of proof at this level is to deepen understanding, not to show off.
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  7. #7
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    Well yeach I gues you are right. However I do have another problem:
    Prove that that the following sequence,defined recursively, converges or diverges:
    x_{1}=1
    x_{n}=\sqrt{1+x_{n-1}}
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