Monday Math 59

Recall the Riemann zeta function

Now, consider the product . We see:
and we have the zeta function series with the even terms removed. Now, let us multiply the above by . Then we see:

where the remaining terms are those where k is not divisible by 2 or 3. Multiplication of the above by will eliminate those remaining terms divisible by 5; we may continue this procedure through the primes, and (in analogy with the Sieve of Eratosthenes) will, in the infinite product, eliminate every term, k>1, giving
, where pn is the nth prime; thus the Riemann zeta function may be expressed as the product
Thus we see a connection between the primes and the Riemann zeta function. In fact, through this identity (first proved by Euler), the divergence of at s=1 (divergence of the harmonic series) implies that there are infinitely many primes.


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2 Responses to “Monday Math 59”

  1. Monday Math 60 « Twisted One 151’s Weblog Says:

    […] and as , and for k>0, we have Now, via the geometric series, so and thus . Now, we showed previously that , and so . To find this last product, note that . Thus , and again using , we see that , and […]

  2. Monday Math 88 « Twisted One 151's Weblog Says:

    […] that Thus, the series inside the prime product has only two non-zero terms, and , using the Euler product for the Riemann zeta function . We can similarly see: .   Now, for a more challenging example, consider Euler’s […]

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