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Math Help - determine if they are converge

  1. #1
    Senior Member
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    determine if they are converge

    \sqrt{n+1}/2\sqrt{n}

    (1+1/n)^{2n}

    2^n/n^2

    sin n / \sqrt{n}

    \sqrt {n^2+n} - n


    can anyone tell me the general methods for simplfing these sequence so it looks more obvious?
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  2. #2
    MHF Contributor
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    Hi

    \sqrt {n^2+n} - n = \frac{\left(\sqrt {n^2+n} - n\right)\left(\sqrt {n^2+n} + n\right)}{\sqrt {n^2+n} + n}

    \sqrt {n^2+n} - n = \frac{n}{\sqrt {n^2+n} +  n}

    \sqrt {n^2+n} - n = \frac{1}{\sqrt {1+\frac{1}{n}} + 1}
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  3. #3
    Super Member General's Avatar
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    Quote Originally Posted by wopashui View Post
    \sqrt{n+1}/2\sqrt{n}

    \color{red}=\frac{1}{2}\sqrt{\frac{n+1}{n}}=\frac{  1}{2}\sqrt{1+\frac{1}{n}}

    (1+1/n)^{2n}

    \color{red} \left(1+\frac{a}{n^k}\right)^{bn^k} \rightarrow e^{ab} as \color{red} n \rightarrow \infty

    2^n/n^2

    Use sandwich theorem ..

    sin n / \sqrt{n}

    Use sandwich theorem ..

    \sqrt {n^2+n} - n



    can anyone tell me the general methods for simplfing these sequence so it looks more obvious?
    ..
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  4. #4
    MHF Contributor
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    Quote Originally Posted by General View Post
    Use sandwich theorem ..
    In French we say "policemen theorem"
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