Wikipedia:Reference desk/Archives/Mathematics/2017 July 9

= July 9 =

Riemann zeta function
From Riemann zeta function we have:


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


 * $$\sum_{n=1}^\infty\frac{1}{n^s} = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}},$$


 * where, by definition, the left hand side is $ζ(s)$ and the infinite product on the right hand side extends over all prime numbers $p$....


 * Both sides of the Euler product formula converge for $Re(s) > 1$.... Since the harmonic series, obtained when $s = 1$, 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{align}&\frac{1}{1-2^{-1}}\,\frac{1}{1-3^{-1}}\,\frac{1}{1-5^{-1}} \\

& = \left(1+\frac{1}{2}+\frac{1}{4}+\dots\right)\left(1+\frac{1}{3}+\frac{1}{9}+\dots\right)\left(1+\frac{1}{5}+\frac{1}{25}+\dots\right) \\ & =1+\frac{1}{2}+\frac{1}{3}+\dots\end{align}$$
 * 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)
 * 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.|undefined 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)