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Math Help - Testing series for convergence or divergence

  1. #1
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    Testing series for convergence or divergence

    \sum _{n=1}^{\infty }{\frac {n!}{{e}^{{n}^{2}}}}

    I tried the ratio test, from my figuring, I got that it diverges. However, the back of the book says otherwise

    I end up getting
    {\frac {n+1}{e}}

    I'm pretty sure that's wrong.
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  2. #2
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    The ratio test gives n+1 \over e^{2n+1} actually...
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  3. #3
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    Quote Originally Posted by CalculusCrazed View Post
    \sum _{n=1}^{\infty }{\frac {n!}{{e}^{{n}^{2}}}}

    I tried the ratio test, from my figuring, I got that it diverges. However, the back of the book says otherwise

    I end up getting
    {\frac {n+1}{e}}

    I'm pretty sure that's wrong.
    If you're using the ratio test, you need to work out

    \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|

     = \lim_{n \to \infty}\left|\frac{\frac{(n + 1)!}{e^{(n + 1)^2}}}{\frac{n!}{e^{n^2}}}\right|

     = \lim_{n \to \infty}\left|\frac{(n + 1)!e^{n^2}}{n!e^{(n + 1)^2}}\right|

     = \lim_{n \to \infty}\left|\frac{(n + 1)e^{n^2}}{e^{n^2 + 2n + 1}}\right|

     = \lim_{n \to \infty}\left|\frac{n + 1}{e^{2n + 1}}\right|

     = \lim_{n \to \infty}\frac{n +1}{e^{2n + 1}} since it is positive term...

     = \lim_{n \to \infty}\frac{1}{2e^{2n + 1}} by L'Hospital's Rule

     = 0.


    Therefore the series is convergent.
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  4. #4
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    Okay cool, I was just doing
    {n}^{2}+1

    Thank you so much!
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