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Math Help - D'Alembert's ratio question

  1. #1
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    D'Alembert's ratio question

    hi everyone,
    Use D'Alembert's ratio test to test the convergent of the following series:


    i)1+ \frac{2^2}{2!}+\frac{3^3}{3!}+\frac{4^4}{4!}

    im stuck, i found the general term is
    \frac{n^n}{n!}
    i cant make out the solution,,,please help me...
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  2. #2
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    If \lim_{n\to \infty}\frac{a_{n+1}}{a_n} exists and is less than 1, the serie converges, if is great than 1, the serie diverges.

    \frac{a_{n+1}}{a_n}=\dfrac{\frac{(n+1)^{n+1}}{(n+1  )!}}{\frac{n^n}{n!}}= \frac{(n+1)^{n+1}n!}{(n+1)!n^n}=(\frac{n+1}{n})^n
    Appliying the limit, the result is e, then, the serie diverges.
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  3. #3
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    thank you felper for helping to clarify, am just wondering, how you get the result e? need some advise on this.

    thank you so much
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  4. #4
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    Quote Originally Posted by anderson View Post
    thank you felper for helping to clarify, am just wondering, how you get the result e? need some advise on this.

    thank you so much
    It's a standard limit - see

    http://en.wikipedia.org/wiki/E_(mathematical_constant)
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  5. #5
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    \lim_{n\to\infty}\left(\dfrac{n+1}{n} \right)^n=\lim_{n\to\infty}\left(1+\dfrac{1}{n} \right)^n=e

    http://en.wikipedia.org/wiki/E_(mathematical_constant)

    this article could help you =)
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  6. #6
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    hi felper
    thank you so much,for the article as well, now i understand.

    thank you again.
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