Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Product of a specific sequence of prime number squares

  1. Dec 26, 2011 #1
    Here is a good question for maths enthusiasts. I really find this sum very tough. Any help, advice or guidance for this question will be greatly appreciated.

    find the product upto n terms

    (1+(1/2^2)) . (1+(1/3^2)) . (1+(1/5^2)) . (1+(1/7^2)) . (1+(1/11^2))......upto n terms

    Where the nth term is (1+(1/P^2)) or (1+(p^-2)) where p is a prime number.

    I feel this question is really tough and couldn't even get the logic how to solve it.
     
    Last edited: Dec 26, 2011
  2. jcsd
  3. Dec 26, 2011 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    You should read up on the Riemann-zeta function.
     
  4. Jan 15, 2012 #3
    In the infinite products below, the factors run over all primes [itex] p_{k} \in \mathbb{P}[/itex]


    Theorem: [itex]\prod \frac{p_{k}^{s}+1}{p_{k}^{s}}= \frac{\zeta(s)}{\zeta(2 s)}, with \zeta(s) [/itex] the Riemann zeta function


    Proof: With the Euler product of the Riemann zeta function we have:

    [itex] \zeta(2 s) = \prod (1 - p_{k}^{-2 s})^{-1} = \prod \frac{p_{k}^{2 s}}{p_{k}^{2 s} - 1} = \prod \frac{p_{k}^{s}}{p_{k}^{s}-1} \frac{p_{k}^{s}}{p_{k}^{s}+1} = \zeta(s) \prod \frac{p_{k}^{s}}{p_{k}^{s}+1}[/itex] Q.E.D.


    Now, with Im(s) = 0 and Re(s) = n we have for the Euler zeta function:

    [itex]\prod \frac{p_{k}^{n}+1}{p_{k}^{n}}= \frac{\zeta(n)}{\zeta(2 n)}[/itex]

    and for the special case n = 2

    [itex]\prod \frac{p_{k}^{2}+1}{p_{k}^{2}}= \frac{\zeta(2)}{\zeta(4)} = \frac{15}{π^{2}}[/itex]

    and that answers the posted question
     
    Last edited: Jan 15, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook