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Cesaro mean question

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Define the cesaro mean as σ=(1/n)(x1 +...+xn)
    Can it happen that xn >0 for all n, and limsup xn =∞, but limσ=0

    2. Relevant equations

    3. The attempt at a solution

    I think I am supposed to construct a piecewise sequence, with ln(n), but I can't figure this out or the life of me. The professor says there is a sequence that will work.
  2. jcsd
  3. Feb 20, 2013 #2
    So [itex]\limsup x_n = +\infty[/itex] means that there is a subsequence that diverges to infinity.

    So we know that there must be a subsequence that becomes arbitrarily large. However, if the other terms of the sequence are small and if the elements of the subsequence lie very far apart, then you can see that you can make the Cesaro mean small.

    If you don't see how to do it immediately. Can you come up with 100 positive numbers such that the largest number is greater or equal than 10, but such that the mean is smaller than 0.5 ??

    Now, can you add 1000 numbers to that sequence such that the largest number is greater than 20, but such that the mean is smaller than 0.05 ??

    Can you generalize this idea?
  4. Feb 20, 2013 #3
    I have a piecewise sequence for xn, where xn = √n for x10n , 1/n2 otherwise.

    I know that the series 1/n2 converges, so that part of the mean will go to zero when it is divided by n.
    For the other part, √n:
    at n=10, xn/n=.31
    at n=100, xn/n=.1
    at n=1000, xn/n=.031

    I think that will work, I'm just not sure how to say this in the form of a proof.
    "It looks like the mean will get very small," doesn't seem very rigourous.

    Am I on the right track at least?
  5. Feb 20, 2013 #4
    I don't see why you are calculating [itex]\frac{x_n}{n}[/itex]. That's not the Cesaro mean...
  6. Feb 20, 2013 #5
    What I have for the sequence I described is:

    But I'm not sure how to describe this formally.
  7. Feb 20, 2013 #6


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    Try defining ##x_n## such that the subsequence where ##n = 2^k## (i.e. powers of 2) increases to infinity at a suitably slow pace. Then make ##x_n## small and decreasing toward 0 for all other values of ##n##.
  8. Feb 20, 2013 #7
    I think we have found a couple that will work, the difficult part is proving it. How do I do that?
    Last edited: Feb 20, 2013
  9. Feb 20, 2013 #8


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    Well, that depends on what sequence you have chosen. But to speak in general terms, let's say we define ##a_n## to be any sequence of positive numbers such that ##\sum_{n=1}^{\infty} a_n = L## is finite. Then define ##b_n## to be zero whenever ##n## is not a power of 2. For ##n## equal to a power of 2, use something that grows slowly to infinity. Then put ##x_n = a_n + b_n##. We have
    \frac{1}{N}\sum_{n=1}^{N} x_n &= \frac{1}{N}\sum_{n=1}^{N} a_n + \frac{1}{N}\sum_{n=1}^{N} b_n \\
    &\leq \frac{1}{N}\sum_{n=1}^{\infty} a_n + \frac{1}{N}\sum_{n=1}^{N} b_n \\
    &= \frac{L}{N} + \frac{1}{N}\sum_{n=1}^{N} b_n \\
    &= \frac{L}{N} + \frac{1}{N}\sum_{n\textrm{ a power of 2}} b_n \\
    What happens next depends on how you chose ##b_n##. So what do you propose to use for ##b_n##?
  10. Feb 20, 2013 #9
    I have an=1/(10n), for n≠10k
    bn=log(n) if n=10k

    I think the argument should go as follows. Since b10 =1, b100=2, b1000= 3, etc., then,

    σn≤(1/n)[Ʃ10-n] +n*10-n

    Taking the limit of that as n→∞ would be zero. Does that make sense?
  11. Feb 20, 2013 #10


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    Yes, this will work. For the second sum, you can write
    \frac{1}{N}\sum_{n=1}^{N} b_n
    &= \frac{1}{N}\left(\sum_{n \leq N, n \textrm{ a power of 10}} \log_{10}(n)\right) \\
    &= \frac{1}{N} \sum_{k = 1}^{\lfloor \log_{10}(N) \rfloor} k \\
    &= \frac{1}{N} \frac{(\lfloor \log_{10}(N)\rfloor)(\lfloor\log_{10}(N)\rfloor+1)}{2} \\
    where ##\lfloor \cdot \rfloor## denotes the floor function. You should be able to argue pretty easily that this converges to ##0## as ##N \rightarrow \infty##.
  12. Feb 20, 2013 #11
    Thank you!!!!!!!
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