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Thread: def of sequence

  1. #1
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    def of sequence

    Use the definition of convergent to prove :
    $\displaystyle {(2n)}^{\frac {1}{n}} \to 1 $
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  2. #2
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    Quote Originally Posted by flower3 View Post
    Use the definition of convergent to prove :
    $\displaystyle {(2n)}^{\frac {1}{n}} \to 1 $
    Yes, what have you done? What is the definition of convergence for a sequence?

    For this particular problem, you might consider taking the logarithm of every number in the sequence and looking at that new sequence.
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  3. #3
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    $\displaystyle proof:$
    $\displaystyle let \ \epsilon >0 \ want \ to \ prove \ \exists k \in N \ such \ that \ \mid (2n)^{ \frac {1}{n}} \ - 1 \mid < \epsilon ,\ \forall \ n \geq k$
    then??
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  4. #4
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    $\displaystyle (2n)^{1/n} = 2^{1/n} n^{1/n} $. Obviously $\displaystyle 2^{1/n} \rightarrow 1 $, so we need to only prove that $\displaystyle n^{1/n} \rightarrow 1 $.

    For n > 1 we have $\displaystyle n^{1/n} > 1^{1/n} = 1 $. So we can let $\displaystyle n^{1/n} = 1 + h_n $ where $\displaystyle h_n $ is some positive quantity depending on n.

    So $\displaystyle n = (1 + h_n)^n $. Now, it is clear from the binomial theorem that $\displaystyle (1 + h_n)^n \ge 1 + nh_n + \frac{n(n+1)}{2} h_n \ge \frac{n(n+1)}{2} h_n $.

    Can you show that $\displaystyle h_n $ goes to 0 and so $\displaystyle n^{1/n} = 1 + h_n $ goes to 1?
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