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July 9

Riemann zeta function

From Riemann zeta function#Euler product formula we have:

The connection between the zeta function and prime numbers was discovered by Euler, who proved the identity

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^s}=\prod _{p\text{ prime}}\frac{1}{1-p^{-s}},$

where, by definition, the left hand side is Template:Math and the infinite product on the right hand side extends over all prime numbers Template:Mvar....

Both sides of the Euler product formula converge for Template:Math.... Since the harmonic series, obtained when Template:Math, diverges, Euler's formula ... implies that there are infinitely many primes.

For convergent cases, I understand this. But for the case s=1, what does this mean other than the vacuous ∞ = ∞ ? Does it mean that if you take both sides for the first k terms and take their ratio, that this ratio converges to 1 as k goes to infinity, or that their difference converges to 0, or what? Loraof (talk) 03:10, 9 July 2017 (UTC)

For s=1 both sides are infinite, but it's not exactly vacuous since it proves that there are an infinite number of primes. The idea of the ratio converging to 1 or difference converging to zero doesn't really make sense since the sum and product are taken over different sets. You are right in that infinity isn't technically a number, so saying two infinite limits are equal is a bit of a shortcut and manipulating diverging a series/product requires a bit more care to be strictly rigorous, something Euler would not have worried about. You could make it a bit more formal as follows: For a given M there is k so that $\sum _{n=1}^k\frac{1}{n}>M$. Using the argument given to prove the s>1 case, it's easy to see that $\prod _{p\text{ prime and }p\le k}\frac{1}{1-p^{-1}}>M$ since when you multiply out the product it contains every term in the sum. So the limit of this as k→∞ is infinity. This implies the number of factors goes to infinity as k→∞, in other words the number of primes is infinite. Note that something stronger is true, namely that $\sum _{p\text{ prime}}\frac{1}{p}$ diverges; this is the real improvement of Euler over Euclid.--RDBury (talk) 07:05, 9 July 2017 (UTC)

Thanks, RDBury. Sorry if I'm just being slow here, but what do you mean by "when you multiply out the product it contains every term in the sum"? Loraof (talk) 16:49, 9 July 2017 (UTC)

For example,

$\begin{array}{cc}& \frac{1}{1-2^{-1}}\frac{1}{1-3^{-1}}\frac{1}{1-5^{-1}}\\ & =(1+\frac{1}{2}+\frac{1}{4}+\dots )(1+\frac{1}{3}+\frac{1}{9}+\dots )(1+\frac{1}{5}+\frac{1}{25}+\dots )\\ & =1+\frac{1}{2}+\frac{1}{3}+\dots \end{array}$

where the sum on the right runs over natural numbers whose prime factors are restricted to 2. 3 and 5. This includes all terms from 1/1 to 1/5 as well as some others. --RDBury (talk) 18:21, 9 July 2017 (UTC)

• Template:Re Template:Tq Huh, if true that is certainly stronger, but I fail to see how it follows from what you wrote. Tigraan^{Click here to contact me} 17:01, 12 July 2017 (UTC)
  • Our article divergence of the sum of the reciprocals of the primes (which I wish was at a title like prime harmonic series) gives the (admittedly questionable) way Euler derived the result from there. Double sharp (talk) 07:49, 13 July 2017 (UTC)