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Limit with parameter

  1. Dec 10, 2004 #1
    Find limit n-->infinity of sequence a_n:

    a_n = (1^k+2^k+...+n^k)/(n^(k+1)), where k is parameter.

    IThanks from advance for any help.

    I tried to compute this limit using Stolz Theorem, but I don't know if I can do it in this way.
     
  2. jcsd
  3. Dec 11, 2004 #2

    learningphysics

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    What value can k take? The analysis below is for k being an integer and k>=0. For k<0 the denominator is not strictly increasing, so Stolz theorem doesn't hold.

    For k>=0 and k being an integer, I think Stolz theorem is a good way to solve the problem:

    By applying it... try to find the limit as n->infinity of:

    [tex]\frac{(n+1)^k}{(n+1)^{(k+1)}-n^{(k+1)}}[/tex]

    You can factor the denominator:

    [tex](n+1)^{(k+1)}-n^{(k+1)}=[(n+1)-n][(n+1)^k +(n+1)^{(k-1)}n+(n+1)^{(k-2)}n^2+...][/tex]

    Then if you divide both the numerator and denominator by [tex](n+1)^k[/tex] it should be easy to see the limit. Hint: the answer depends on k.
     
    Last edited: Dec 11, 2004
  4. Dec 12, 2004 #3
    One more question: What if k=-1. The limit doesn't exist or limit equals to infinity?
     
  5. Dec 12, 2004 #4

    learningphysics

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    Yes, the limit does not exist (or another way to say it is the limit is infinity).

    If you plug in k=-1 into your sequence you get:

    [tex]a_n=\begin{array}{c}n\\\sum\\m=1\end{array}1/m[/tex]



    Since [tex]\sum1/m[/tex] is a divergent series, as n->infinity a_n->infinity.
     
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