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Math Help - Another friggin problem with series...

  1. #1
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    Another problem with series...

    This series section is really bugging me. Am I allowed to use the word friggin? I can't get an answer that makes sense for n!/100^n with the Riemann symbol in front. I try to use the Ratio Test and then realize that the terms are positive so I don't need an absolute value.

    I do some algebraic manipulation and come up with ((n+1)/(101)^n)) I don't see how I can take a limit/simplify the denominator here. Do I divide everything by ((101)^n))? Or did I do something wrong along the way and I should be getting a different result where I can see if it is <1 (converges) >1 (diverges) or = 1 (inconclusive).
    Last edited by Undefdisfigure; January 28th 2008 at 03:54 PM.
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  2. #2
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    \sum \frac{n!}{100^n}

    Ratio test takes the infinite limit of: \frac{\frac{(n+1)!}{100^{n+1}}}{\frac{n!}{100^n}} which simplifies to: \frac{(n+1)}{100}. The series diverges because the infinite limit is greater than one.
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    As an aside, and this doesn't necessarily constitute another problem (hence no new thread). When I'm given a series n/(ln n)^n, can I use the rules of ln in the denominator and get n/((n)(ln n)) so the n's cancel out and the rest is easy?

    I don't see how my method would be breaking any rules in math. I just want to check if I'm not mistaken.
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    Quote Originally Posted by Undefdisfigure View Post
    As an aside, and this doesn't necessarily constitute another problem (hence no new thread). When I'm given a series n/(ln n)^n, can I use the rules of ln in the denominator and get n/((n)(ln n)) so the n's cancel out and the rest is easy?

    I don't see how my method would be breaking any rules in math. I just want to check if I'm not mistaken.
    The n's would not cancel out as you would have \frac{(n+1)}{n*ln|n|}
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    I haven't gotten to logarithms yet as my teacher requested that I review algebra. I don't see how you got the n+1 in the numerator. There needs to be a 1 added to the numerator when the n comes down from being an exponent to multiply the denominator?
    Last edited by Undefdisfigure; January 28th 2008 at 03:53 PM.
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    And ironically, by using your method colby I get the same answer as with my method. After an extra step is needed in your method colby, I get that
    lim n--> infinity 1/(ln n) goes to zero and so the series converges.
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    Quote Originally Posted by Undefdisfigure View Post
    I haven't gotten to logarithms yet as my teacher requested that I review algebra. I don't see how you got the n+1 in the numerator. There needs to be a 1 added to the numerator when the n comes down from being an exponent to multiply the denominator?
    You are given a_n=\frac{n}{(ln n)^n}

    Using the ratio test, you need to find \frac{a_{n+1}}{a_n}. This is equal to: \frac{\frac{n+1}{(ln(n+1))^{n+1}}}{\frac{n}{(ln n)^n}} = \frac{(n+1)(ln|n|)^n}{(n)(ln|n+1|)^{n+1}}. Note that the natural logs will not cancel each other out, but we can do the following..

    \frac{(n+1)(ln|n|)}{(n)(ln|n+1|)}\left(\frac{ln|n|  }{ln|n+1|)}\right)^n

    The limit of that obviously goes to zero because the "bottom" grows faster than the "top".
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  8. #8
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    Quote Originally Posted by Undefdisfigure View Post
    This series section is really bugging me. Am I allowed to use the word friggin? I can't get an answer that makes sense for n!/100^n with the Riemann symbol in front. I try to use the Ratio Test and then realize that the terms are positive so I don't need an absolute value.

    I do some algebraic manipulation and come up with ((n+1)/(101)^n)) I don't see how I can take a limit/simplify the denominator here. Do I divide everything by ((101)^n))? Or did I do something wrong along the way and I should be getting a different result where I can see if it is <1 (converges) >1 (diverges) or = 1 (inconclusive).
    Stirling's formula tells us that (it does tell us what k is but that is unimportant here):

    n! \sim k \frac{n^{n+1/2}}{e^n},

    so:

    \frac{n!}{100^n} \sim k \frac{n^{n+1/2}}{e^{n(1+\ln(100))}},

    and eventually n^n dominates e^{\alpha n} for any \alpha so the terms diverge, and so the series itself diverges.

    (that is for large enough n: n>e^{\alpha} and so the terms are bounded below by a multiple of a geometric series with common ratio greater than 1)

    RonL
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