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Convergence / Divergence of a series

  1. May 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Does the following series converge or diverge? ##∑\frac{n^5}{n^n}## (as n begins from 1 and approaches infinity)


    2. Relevant equations
    Ratio test?


    3. The attempt at a solution

    For your reference, thus far I have learned about the geometric series, the limit test, comparison test, a bit about cauchy condensation test, and the ratio test. Thanks for your time.

    I tried using the ratio test for this one and ended up getting the following:

    ##\frac{(n+1)^5}{(n+1)^{n+1}}*\frac{n^n}{n^5}=\frac{n^{n-5}}{(n+1)^{n-4}}=(\frac{n}{n+1})^{n-5}*\frac{1}{n+1}##

    (as n approaches infinity)

    I'd like to find out if my work is on the right track so far. It certainly seems that the above limit is zero, in which case the original series converges via the ratio test.

    Also, please let me know if a different strategy other than the ratio test would've worked out better. Thanks for your time.
     
  2. jcsd
  3. May 13, 2014 #2

    Dick

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    I would use the ratio test. And that looks fine. But you should show why that limit is zero. Saying "seems that it does" probably isn't good enough.
     
  4. May 13, 2014 #3

    benorin

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    I would use the comparison test: your general term [itex]\frac{n^5}{n^n}=\frac{1}{n^{n-5}}<\frac{1}{n^2}[/itex], (for sufficiently large n) a general term whose series is convergent.

    Maybe I'm giving too much help here, pm me if you think so and I'll erase it...
     
    Last edited: May 14, 2014
  5. May 14, 2014 #4
    Ok right...so clearly ##\frac{1}{n+1}## approaches zero as n approaches infinity...which leaves the other part of the equation:

    ##(\frac{n}{n+1})^{n-5}##

    I'm not sure how to find the limit of this equation as n approaches infinity...since the fraction part of it (without the exponent) approaches 1, does it mean the limit is just 1 as well?

    In which case the limit of the whole thing would be 1 * 0 = 0?
     
  6. May 14, 2014 #5
    Thanks for this! Sometimes I fail to see the forest for the trees...
     
  7. May 14, 2014 #6

    Dick

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    No, that limit isn't zero. The limit has the form ##1^\infty##. That indeterminant. You can either take the log and use l'Hopital on it, or you could write it in terms of a limit you might recognize (1+1/n)^n. What's the value of that?
     
  8. May 14, 2014 #7

    Ray Vickson

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    Even easier: let ##t_n = n^5/n^n##, so ##L_n \equiv \ln(t_n) = 5 \ln(n) - n \ln(n)##. For ##n > 7## we have ##L_n < -2 \ln(n)##, hence ##t_n < \exp(-2 \ln(n)) = 1/n^2##, so ##t_n## converges, by the comparison test.
     
  9. May 14, 2014 #8
    OK wait, the value of (1+1/n)^n is the definition of ##e##, I think? Which means...
    ##(\frac{n}{n+1})^{n-5}=\frac{1}{e}## as n goes to infinity?

    So then, continuing from the original post...
    ##Lim_{n\rightarrow\infty}[(\frac{n}{n+1})^{n-5}*\frac{1}{n+1}]## becomes ##\frac{1}{e}*0=0##?
     
  10. May 14, 2014 #9

    Dick

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    Yes, that's it.
     
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