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Finding the limit

  1. Jun 14, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the limit of the following sequence:
    ##L_2 = \lim_{n \rightarrow +\infty} \frac {\sum_{k=0}^n (2k - 1)^p}{n^{p+1}}##
    2. Relevant equations
    3. The attempt at a solution

    Seeing that ##\lim_{n \rightarrow +\infty} n^{p+1} = + \infty ## i can apply the Stolz theorem. (Is something more necessary alongside it being divergent?).
    The upper sum when used in Stolz theorem yields just ##(2n+1)^{p}## because the rest of the sum is subtracted in form ##(n+1) - (n)##. I get

    ##\lim_{n \rightarrow +\infty} \frac{(2n+1)^p}{(n+1)^{p+1} - n^{p+1}}##
    ##\lim_{n \rightarrow +\infty} \frac{(2n+1)^p}{(n+1)^p + (n+1)^{p-1}n + (n+1)^{p-2}n^2 + ... + n^p}##
    And im stuck on how to further simplify this. Could get a hint from someone that knows?
     
  2. jcsd
  3. Jun 14, 2017 #2

    andrewkirk

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    i suggest as a next step dividing the numerator and denominator by ##n^p##.
    You should then be able to calculate separate limits for the numerator and denominator, which are neither infinity nor zero. Hence you can use the division law to calculate the limit of the ratio.
     
  4. Jun 14, 2017 #3
    I get ##\frac{2^p}{p+1}##
     
  5. Jun 14, 2017 #4

    BvU

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    Check your last step, e.g. with a simple example like p = 2. I am missing the binomial coefficients and the powers of n should be descending, not remain equal !
     
  6. Jun 14, 2017 #5

    BvU

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    Me too. (from your one-but-last step :smile: )
     
  7. Jun 14, 2017 #6
    You are right. I did mess something up. Weird that i got the right result when i divided by ##n^{p}##. So:
    ##(n+1)^{p+1} - n^{p+1} = n^{p+1} + {{p+1}\choose{1}}n^p + {{p+1}\choose{2}}n^{p-1} + ... + {{p+1}\choose{k}}n^{p-k} + ... +1 - n^{p+1}##
    ##(n+1)^{p+1} - n^{p+1} ={{p+1}\choose{1}}n^p + {{p+1}\choose{2}}n^{p-1} + ... + {{p+1}\choose{k}}n^{p-k} + ... + 1##
    If i know divide with ##n^p## i will obtain the correct result. Is this correct?
     
  8. Jun 14, 2017 #7

    BvU

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    It is. Well done !
     
  9. Jun 14, 2017 #8

    andrewkirk

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  10. Jun 14, 2017 #9
    Yeah thanks, you provided a key hint :)
     
  11. Jun 15, 2017 #10

    Ray Vickson

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    Better:
    $$(n+1)^{p+1}-n^{p+1} = n^{p+1}\left [ \left(1 +\frac{1}{n} \right)^{p+1} - 1 \right],$$
    and use ##(1+x)^{p+1} = 1 + px +O(x^2)## for small ##x = 1/n##. Thus
    $$(n+1)^{p+1}-n^{p+1} \sim (p+1)n^p$$
    for large ##n##. Here, ##\sim## denotes "asymptotic equality": ##a(n) \sim b(n)## for large ##n## if
    $$\lim_{n \to \infty} \frac{a(n)}{b(n)} = 1.$$
     
  12. Jun 15, 2017 #11
    That is a really interesting approach too. I think i like it the most :D
     
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