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Math Help - Infinite Series

  1. #1
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    Infinite Series

    I need some serious help in the attached questions...I need to determine whether the series in the picture converge, absolutely converge ot diverge...
    I realy need your guidance and help....

    Thanks everyone

    My attemp:

    I've no clue aboue 1&3...
    In 2-I know that cos(n*pi)*ln([n+1]/n ) converges by leibnitz test...But what about the first element in the sum? [sqrt(n)sin(n)tan(1/n^2)] ...
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by WannaBe View Post
    I need some serious help in the attached questions...I need to determine whether the series in the picture converge, absolutely converge ot diverge...
    I realy need your guidance and help....

    Thanks everyone

    My attemp:

    I've no clue aboue 1&3...
    In 2-I know that cos(n*pi)*ln([n+1]/n ) converges by leibnitz test...But what about the first element in the sum? [sqrt(n)sin(n)tan(1/n^2)] ...
    What have you done?

    1-n\sin\left(\frac{1}{n}\right)1-n\left(\frac{1}{n}-\frac{1}{6n^3}+\cdots\right)=\frac{1}{6n^2}+\cdots  \underset{n\to\infty}{\sim}\frac{1}{6n^2}. So this sum converges.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    The third is equally uneventful combine the fractions note the behavior of the numerator and denominator and seperately

    Spoiler:


    \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}}\underset{n\to\infty}{\sim}\fr  ac{1}{n^{\frac{3}{2}}}
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  4. #4
    MHF Contributor Drexel28's Avatar
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    For the third one. I WILL not give you the answer. Consider this though \lim_{z\to0}\frac{\tan(z)}{z}
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  5. #5
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    Tnx a lot ! ...
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  6. #6
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Drexel28 View Post
    The third is equally uneventful combine the fractions note the behavior of the numerator and denominator and seperately

    \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n+1}}\underset{n\to\infty}{\sim}\fr  ac{1}{n^{\frac{3}{2}}}
    The general term of the third series seems to be...

    \frac{1}{\sqrt{n}-1}-\frac{1}{\sqrt{n}+1}\underset{n\to\infty}{\sim}\fr  ac{2}{n}

    ... so that the series diverges...

    Kind regards

    \chi \sigma
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  7. #7
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    Quote Originally Posted by chisigma View Post
    The general term of the third series seems to be...

    \frac{1}{\sqrt{n}-1}-\frac{1}{\sqrt{n}+1}\underset{n\to\infty}{\sim}\fr  ac{2}{n}

    ... so that the series diverges...

    Kind regards

    \chi \sigma
    ITS
    <br />
\frac{1}{\sqrt{n+1}-1}-\frac{1}{\sqrt{n+1}+1}<br />
    Assuimg n > 1
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