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Summation involving von Mangoldt function

  1. Aug 3, 2008 #1
    Please help me in solving the problem,
    find the sum
    Sum{r=2 to infinity} (von Mangoldt(r)-1)/r



    Your help is appreciated.
     
  2. jcsd
  3. Aug 11, 2008 #2
    do you mean [tex] \sum _{n=2}^{\infty} \frac{ \Lambda (n) -1}{n} [/tex] ??


    i think is divergent
     
  4. Aug 12, 2008 #3
    hi mhill,
    can you prove that the series is divergent?



    -Ng
     
  5. Aug 12, 2008 #4

    CRGreathouse

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    [tex]\frac1n \sum_{k=1}^n\Lambda(k)=1+o(1/\log n)[/tex]

    so your series seems to be something like

    [tex]\sum\frac{1}{n\log n}\approx\log\log n[/tex]

    Obviously this is very heuristic here.
     
  6. Aug 12, 2008 #5

    CRGreathouse

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    OK, it diverges.

    [tex]\sum_{n=2}^{\infty} \frac{\Lambda(n) -1}{n}=\sum_p\sum_{k=1}^\infty\(\frac{\log p-1}{p}+\frac{\log p-1}{p^2}+\cdots\)=\sum_p\frac{\log p-1}{p-1}[/tex]
    and we all know that
    [tex]\sum_p\frac1p=+\infty[/tex]
     
  7. Aug 13, 2008 #6
    when n is a prime or prime power, the summation is okay.

    but suppose when n=6, then the sum is (von Mangoldt(6) -1)/6 , which is = -1/6,

    as n runs from 2 to infinity,can we settle the problem of convergency or divergency?



    -Ng
     
  8. Aug 13, 2008 #7

    CRGreathouse

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    My post addressed the case where n runs from 2 to infinity, which diverges.
     
  9. Aug 15, 2008 #8
    Hi CRGreatHouse,

    In your post 1673, the summation on LHS runs from n=2 to infinity, (n=2,3,4,5,6,7,8,...)

    But the summation on RHS runs over all primes.(p=2,3,5,7,...)


    From the definition of von Mangoldt function,when n=6,10,12,14,15,18,... , the summand

    became (-1/n) whenever n is not equal to any prime or prime power.

    Is something missing ?


    -Ng
     
  10. Aug 15, 2008 #9

    CRGreathouse

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    Right, right... yeah, I calculated it for numerator [itex]\Lambda[/itex] first, forgetting about the -1 term, and when I added it back in forgot that part.

    But wouldn't that also suggest divergence (in the other direction), since the prime powers are density 0, the reciprocal primes vary as log log n, and the reciprocal integers vary as log n?

    Numerical experimentation would be nice here.
     
  11. Aug 16, 2008 #10
    I have tried numerical calculation and the sum seems to converge to ~ -1.16

    Can we approach the problem from Zeta function?



    -Ng
     
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