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Which series test to use.

  1. Jan 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Does this series converge or diverge. n=1 SIGMA infinity ( (n+1)^n / ( n^(n+1) ) )

    this could also be changed to lim n-> infinity (1 + 1/n)^n , but then i ask, where the n+1 in the original equation has went?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 17, 2008 #2

    HallsofIvy

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    Since those are all positive numbers, I would be inclined to use the root test:
    [tex]^n\sqrt{\frac{(n+1)^n}{n^{n+1}}}= \frac{n+1}{n^\frac{n+1}{n}}[/itex]
    If the limit of that is less than 1, then the series converges.

    As to "where did the n+1 go", how did you get "lim (1+ 1/n)^n"?
     
  4. Jan 17, 2008 #3

    Gib Z

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    The summand can be expressed as [tex]\frac{ \left( 1 + \frac{1}{n} \right)^n }{n}[/tex], but that doesn't really help anyway.

    Halls, the root test returns 1, ie inconclusive. I haven't gone through with the calculations but I would try the ratio test next.
     
  5. Jan 17, 2008 #4
    expressing the summand as [tex]\frac{ \left( 1 + \frac{1}{n} \right)^n }{n}[/tex] does help, you just have to give up finding a test but consider finding a divergent minorante.
    [tex] \frac{1}{n} < \frac{ \left( 1 + \frac{1}{n} \right)^n }{n}[/tex] and we know that
    [tex] \sum_{n=1}^{\infty} \frac{1}{n} = \infty [/tex]
     
  6. Jan 17, 2008 #5

    Gib Z

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    Damn that is right >.< good work dalle!
     
  7. Jan 20, 2008 #6
    Okay, so this problem should be approached by the ratio test. We know it diverges, and i believe so because 1/n is a harmonic series.

    also, dalle, it looks though that may be similiar to the comparison test then?

    and "As to "where did the n+1 go", how did you get "lim (1+ 1/n)^n"?" it was a hint given by the problem and it also is equal to e.
    i'm still confused by this.

    thankyou for all the help so far.
     
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