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A conjecture on Cesaro summation and primes.

  1. Jul 8, 2007 #1
    After studying Cesaro and Borel summation i think that

    sum [tex] \sum_{p} p^{k} [/tex] extended over all primes is summable Cesaro [tex] C(n,k+1+\epsilon) [/tex]

    and the series [tex] \sum_{n=0}^{\infty} M(n) [/tex] and [tex] \sum_{n=0}^{\infty} \Psi (n)-n [/tex]

    are Cesaro-summable [tex] C(n,3/2+\epsilon) [/tex] for any positive epsilon

    hence the fact that M(0)+M(1)+M(2)+............... is Cearo summable 3/2+e is a consequence of Riemann Hypothesis.
    Last edited: Jul 8, 2007
  2. jcsd
  3. Jul 8, 2007 #2

    matt grime

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    M would be what, Jose? The Mertens function perhaps? I doubt anyone here knows the definition of Cesaro summable - if you're going to post more random speculations you could at least offer some definitions.
  4. Jul 8, 2007 #3

    Gib Z

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    It does not take much insight to see the sequence [tex]a_n = \frac{2+3+5+7+11\cdots p_n}{n}[/tex] is increasing, as the amount you are adding to the numerator with every next value of n, is larger than the amount being added to the numerator, 1. We can see The first sum you propose is not Cesaro Summable for k>1. For k=0, the Cesaro summation is equal to 1. For k<0 the Cesaro sum is 0.

    The 2nd series in your post does not make sense, M(0) is undefined...and what makes you think that [tex]\lim_{n\to\infty} \frac{ M(1) + M(2)....+ M (n)}{n}[/tex] exists? It looks to be as if it is at best oscillatory. Post a proof or any form of mathematical working...
  5. Jul 8, 2007 #4
    "CESARO, Ernesto (1859-1906). Italian geometer and analyst.

    Cesaro's summation formula. A specific method of attributing a sum to certain diĀ­ vergent series. A sequence of partial sums

    the method for k-Cesaro sum is given in webpage (for k=1 is the same as usual Cero summation)


    Using Binomial coefficient, yes M(x) is the Mertens function
  6. Jul 8, 2007 #5

    Gib Z

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    The Link you offered gives the definition for the Generalized Cesaro summation formula, my previous post thought otherwise. Either way, you are still yet to show us any mathematical working to support your claim.
  7. Jul 9, 2007 #6
    the idea i got from the fact that the sum of [tex] f(k)=k^{a} [/tex] (a being a positive integer ) from k=1,2,3,.........,n is [tex] S(n)=O(n^{a+1}) [/tex]

    the Binomial coefficient involving n+r and n for r integer are O(n^{r}) then if we put r=a+1 as n--> infinity the mean tends to a finite value.

    Also if a function is [tex] O(x^{b} [/tex] then its sum/integral is [tex] O(x^{b+1} [/tex] this is where i got the idea from.
    Last edited: Jul 9, 2007
  8. Jul 13, 2007 #7
    Since for even n we got the inequality

    [tex] 1+2^{k}+3^{k}+..........+n^{k} \le C( 1-2^{k+1}+3^{k+1}-.......+n^{k+1} [/tex]

    and due to the fact that the sum [tex] 1-2^{k+1}+3^{k+1}-....[/tex] is summable Cesaro of order (k+1) then i believe that

    [tex] 1+2^{k}+3^^{k}+......... [/tex] is r-Cesaro summable

    with [tex] k+1<r\le k+2 [/tex]
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