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Thread: Proving limit

  1. #1
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    Proving limit

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

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

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

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

    From (1) it follows that...

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle x_n :=\dfrac 1{n(\ln n)^2}$?
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  5. #5
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    Thank you very much for all your answers!
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