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Math Help - limit

  1. #1
    Super Member dhiab's Avatar
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    limit

    Calculate this limit ( n ــــ > + infinity ) :

     <br />
\frac{n!e^{n}}{n^{n+\frac{1}{2 }}}<br />
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  2. #2
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    Hello, dhiab!

    \lim_{x\to\infty} \frac{n!e^{n}}{n^{n+\frac{1}{2 }}}<br />

    Ratio Test: . \left|\frac{a_{n+1}}{a_n}\right| \;\;=\;\;\left|\frac{(n+1)!e^{n+1}}{n^{n+\frac{3}{  2}}} \cdot \frac{n^{n+\frac{1}{2}}}{n!e^n}\right| \;\;=\;\left|\frac{(n+1)!}{n!}\cdot \frac{e^{n+1}}{e^n} \cdot\frac{n^{n+\frac{1}{2}}}{n^{n+\frac{3}{2}}} \right|

    . . . . . . . . =\;\; \left|\frac{n+1}{1}\cdot\frac{e}{1} \cdot\frac{1}{n}\right| \;\;=\;\;\left|\frac{n+1}{n}\cdot e\right|

    Divide top and bottom by n\!:\quad \left|\frac{1 + \frac{1}{n}}{1}\cdot e\right|


    Then: . R \;=\;\lim_{x\to\infty}\left|\frac{1 + \frac{1}{n}}{1}\cdot e\right| \;=\; \left|\frac{1+0}{1}\cdot e\right|  \;=\;e


    Since R > 1, the series diverges.

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  3. #3
    Senior Member
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    Anchorage, AK
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    Possible error

    Soroban,

    I think you made an error in your ratio test: since a_n=\frac{n!e^n}{n^{n+\frac{1}{2}}},
    then
    a_{n+1}=\frac{(n+1)!e^{n+1}}{\left(\color{red}n+1\  color{black}\right)^{n+\frac{3}{2}}},
    which prevents the cancelling you did.

    Dhiab, the key to this limit is Stirling's approximation:
    n!\approx\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n} for large n.

    --Kevin C.
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