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Math Help - Limit of a sequence...

  1. #1
    MHF Contributor chisigma's Avatar
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    Limit of a sequence...

    Does the sequence...

    a_{n} = \frac{1}{\sqrt{n}}\cdot \sum_{i=1}^{n-1} \frac{1}{\sqrt{i+\alpha}} , 0<\alpha<1 (1)

    ... have finite limit or not?...

    I will appreciate very much an answer from you!...

    Kind regards

    \chi \sigma
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  2. #2
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    Quote Originally Posted by chisigma View Post
    Does the sequence...

    a_{n} = \frac{1}{\sqrt{n}}\cdot \sum_{i=1}^{n-1} \frac{1}{\sqrt{i+\alpha}} , 0<\alpha<1 (1)

    ... have finite limit or not?...

    I will appreciate very much an answer from you!...

    Kind regards

    \chi \sigma
    a_n=\frac{1}{n} \sum_{i=0}^{n-1}\frac{1}{\sqrt{\frac{i + \alpha}{n}}} - \frac{1}{\sqrt{\alpha n}}. therefore: \lim_{n\to\infty} a_n = \lim_{n\to\infty}\frac{1}{n} \sum_{i=0}^{n-1}\frac{1}{\sqrt{\frac{i + \alpha}{n}}} = \int_0^1 \frac{1}{\sqrt{x}} \ dx= 2.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Excellent!...

    Kind regards

    \chi \sigma
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  4. #4
    MHF Contributor chisigma's Avatar
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    Instead of opening a new thread I can use this and ask you for another limit...

    \lim_{x \rightarrow \infty} x\cdot e^{- c\cdot \sqrt{\ln x}}

    ... where c>0 is a constant... Does it exist?... if yes, what is its value?...

    Thank you very much!...

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    Instead of opening a new thread I can use this and ask you for another limit...

    \lim_{x \rightarrow \infty} x\cdot e^{- c\cdot \sqrt{\ln x}}

    ... where c>0 is a constant... Does it exist?... if yes, what is its value?...

    Thank you very much!...

    Kind regards

    \chi \sigma
    let \ln x = t^2. then \lim_{x\to\infty}xe^{-c\sqrt{\ln x}}=\lim_{t\to\infty}e^{t^2-ct}=\infty.
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