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Math Help - lim[a_n]=a, then lim[(a_1+a_2+...+a_n)/n]=a

  1. #1
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    Question lim[a_n]=a, then lim[(a_1+a_2+...+a_n)/n]=a

    How can I prove that if a_{n}\to a, then b_{n}=\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}\to a ?

    Can I apply the definition for Cauchy sequences here, or is this a proof for the conditions of Cauchy sequences?

    Thanks
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  2. #2
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    Quote Originally Posted by tunaaa View Post
    How can I prove that if a_{n}\to a, then b_{n}=\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}\to a ?

    Can I apply the definition for Cauchy sequences here, or is this a proof for the conditions of Cauchy sequences?

    Thanks

    I supose you must have meant b_{n}=\frac{1+\sqrt{a_2}+\sqrt[3]{a_3}+...+\sqrt[n]{a_n}}{n}\to a , and thus we must impose the condition that \forall\,n\in\mathbb{N}\,,\,\,a_n\geq 0...or something like this. Check it.

    Tonio
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    Sorry, I was thinking of another question - I actually meant:

    b_{n}=\frac{a_1+a_2+...+a_n}{n}\to a

    Thanks
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  4. #4
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    Hello,

    Look for CesÓro's theorem (or CesÓro's mean) in google.
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    Quote Originally Posted by tunaaa View Post
    Sorry, I was thinking of another question - I actually meant:

    b_{n}=\frac{a_1+a_2+...+a_n}{n}\to a

    Thanks

    This is what you need: http://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf

    The exact result you're trying to prove is in page 2.

    Tonio
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    Quote Originally Posted by tonio View Post
    This is what you need: http://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf

    The exact result you're trying to prove is in page 2.

    Tonio
    Thanks, but is there another proof that doesn't involve lim sup or lim inf?
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    Quote Originally Posted by tunaaa View Post
    Thanks, but is there another proof that doesn't involve lim sup or lim inf?

    What do you care about that?? They all are the same if the sequence converges, which is YOUR case...
    You can even modify the proof there by assuming your seq. converges and all the time lim sup and lim inf are the same.

    Tonio
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  8. #8
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    Quote Originally Posted by tonio View Post
    What do you care about that?? They all are the same if the sequence converges, which is YOUR case...
    You can even modify the proof there by assuming your seq. converges and all the time lim sup and lim inf are the same.

    Tonio
    Sorry, it's just that we've not covered lim sup and lim inf yet.
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  9. #9
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    Quote Originally Posted by tunaaa View Post
    Sorry, it's just that we've not covered lim sup and lim inf yet.
    Let \epsilon>0. Then \exists N such that \forall n>N, |a_n-a|<\epsilon, so let K=\sum_{n=1}^{N-1}a_n, and we have

    \lim_{n\to\infty}\frac{K+(n-N)(a-\epsilon)}{n}\leq\lim_{n\to\infty}\frac{a_1+a_2+..  .+a_n}{n}\leq\lim_{n\to\infty}\frac{K+(n-N)(a+\epsilon)}{n}
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  10. #10
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    Quote Originally Posted by redsoxfan325 View Post
    we have

    \lim_{n\to\infty}\frac{K+(n-N)(a-\epsilon)}{n}\leq\lim_{n\to\infty}\frac{a_1+a_2+..  .+a_n}{n}\leq\lim_{n\to\infty}\frac{K+(n-N)(a+\epsilon)}{n}
    Thanks, but can you explain how you arrived at this stage please?
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  11. #11
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by tunaaa View Post
    Thanks, but can you explain how you arrived at this stage please?
    \frac{(a_1+a_2+...+a_N)+(a_{N+1}+...+a_n)}{n}

    We know that for n>N, |a_n-a|<\epsilon\implies a-\epsilon<a_n<a+\epsilon.

    So I called the sum of the first N terms K (because the sum of a finite number of terms is just a number), and since there are n-N terms left, each of which is bounded below by a-\epsilon (and above by a+\epsilon), the inequality follows.
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