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Math Help - Proving limit

  1. #1
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    Proving limit

    Let be x_1 \geq x_2 \geq ... \geq x_n \geq ... positive numbers and \sum _{n=1}^{\infty}x_n < \infty. Prove that \lim_{n \to \infty}nx_n =0.

    Thank you very much in advance!
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  2. #2
    Super Member girdav's Avatar
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    We have 0\leq nx_{2n}\leq x_{n+1}+\ldots +x_{2n}=s_{2n}-s_n where s_n:=\sum_{k=1}^n x_k. Hence the limit of the subsequence \left\{2nx_{2n}\right\} is 0. Now show that the limit is 0 for the subsequence \left\{(2n+1)x_{2n+1}\right\}.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by zadir View Post
    Let be x_1 \geq x_2 \geq ... \geq x_n \geq ... positive numbers and \sum _{n=1}^{\infty}x_n < \infty. Prove that \lim_{n \to \infty}nx_n =0.

    Thank you very much in advance!
    The series with positive terms \displaystyle \sum_{n=1}^{\infty} x_{n} converges and that means that there is an \varepsilon >0 and an \alpha>0 such that for n 'large enough' is...

    \displaystyle x_{n} < \frac{\alpha}{n^{1+\varepsilon}} (1)

    From (1) it follows that...

    \displaystyle \lim_{n \rightarrow \infty} n\ x_{n} =0 (2)

    Kind regards

    \chi \sigma
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  4. #4
    Super Member girdav's Avatar
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    Chisigma : are you sure that your property is true ? What about x_n :=\dfrac 1{n(\ln n)^2}?
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  5. #5
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    Thank you very much for all your answers!
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