I have two or three questions directly related to this wikipedia article:


I think it is exactly about my main Q (see my other Q, on hold), and I understand it quite well (sort of), but:

What actually is "the proof of Euler product formula"?

There is, of course the obvious proof under "The case s=1", but it is also obvious what it prooves:

"harmonic series" =

product of all prime numbers,

divided by the

product of (all those prime numbers minus one)

Under way, the euler product series gets inverted (1/n) and reversed (left to right). Poor euler product!

This equation is very very interesting, but what is prooved? Can you not just factor it out?

The article starts off with:

The Euler product formula for the Riemann zeta function...:

Why "...for the R. z. function "?

Then, the article has that animated 120-grid sieve (why 120 ?), and speaks of:

"There is a certain sieving property that we can use to our advantage"

I somehow see what follows, but:

Can not that "property" be explained more clearly? Why that animated sieve at all?

Can not "s" be left out, at first, in the method-prooving part ? (And afterwards expanded ?)

Why Riemann-zeta?, why any s different than 1 ?

Euler was more than 100 years before Riemann, and also Dirichlet. I don't find much/anything what HE actually did with his product formula, basically. (I know he did tons of things -- that is not the problem)

Is this not something between Eratosthenes and Euler?

Please don't blame the article -- I have been searching a bit left and right for a long time, for this "euler product series" (I don't define a name, not even "p"):

$$\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \frac{10}{11} \cdot \frac{12}{13} \cdot \frac{16}{17} \cdot \frac{\cdots}{...} \cdot \frac{p-1}{p} \cdot $$

Above is the inverse of the harmonic series.

For me, this is the basic form of the "prime product" (try to google that!). Because it looks nice and simple. This also:

$$ \prod_{n = 1} 1 - \frac {1}{p_n }$$

...should be the same. The index $n$ is for all consecutive prime numbers $p_n$.

Is it only: $$ \zeta(1)^{-1}$$

...this? (should also be the same)

Does the Euler product not exist apart from harmonic series and riemann zeta?

Well yes, euler phi totient function: another "loose end".

  • $\begingroup$ related: math.stackexchange.com/questions/1229307/euler-product-question $\endgroup$
    – user727355
    Commented Nov 23, 2019 at 11:29
  • $\begingroup$ the euler_product article itself has centrally: as is the case for the Riemann zeta-function, where a(n)=1 $\endgroup$
    – user727355
    Commented Nov 23, 2019 at 11:35

1 Answer 1


Theorema 7 of Euler's "Series Infinitas" goes like this:

The infinitely continued product out of these fractions

$\frac {2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19}{1\cdot2\cdot4\cdot6\cdot10\cdot12\cdot16\cdot18}\ \small \ {etc.}$

where the numerators are all the prime numbers, but the denominators are deficient by unity from the numerators,

This product is the same as the sum of this series

$1 + \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6} + \small{etc.}$

also infinitely.

Then he factors out the harmonic series and arrives at:

$\frac {1\cdot2\cdot4\cdot6\cdot10\cdot12\cdot16\cdot18\cdot22\ etc.} {2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23\ etc.} \cdot x = 1 $

...which he has to turn upside down because his "Observationes" are all about Goldbach and that (harmonic) series.

The next theorema has s=2, but has nothing to do with Riemann, or prime numbers, directly.

More with $\pi$ (the circle one, not the prime function one)

So this is Euler's (algebraic) proof of reciprocity between "his" prime product and the harmonic series.


Acutally, "then he factors out" is a big oversimplification. It looks like the "certain sieving method" is applied between the factoring out steps. And because the first step is one half, the other half has the same size...with the second step, the 1/3, you see that it is the other 2/3 that get substituted.

So Euler sieves out the denominators "one by one": first all even ones, then all with 3 as divisor, then with 5, etc.

$$x=1 + \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6} + \small{etc.}$$


$$\frac{1}{2}\cdot x = \frac{1}{2}+ \frac{1}{4}+ \frac{1}{6}+ \frac{1}{8}+ {etc.}$$

This is all the even denominators. The other half is the uneven denominators. (Euler only "explains": of this halved(? lat.: demta) series reamains:)

$$\frac{1}{2}\cdot x = 1 + \frac{1}{3}+ \frac{1}{5}+ \frac{1}{7} + {etc.}$$

(I keep the original "1" instead of 1/1 -- either way it is an additional possible confusion)

Then he repaeats factoring out with 1/3:

$$\frac{1}{2}\cdot\frac{1}{3}\cdot x = \frac{1}{3}+ \frac{1}{9}+ \frac{1}{15} + \frac{1}{21}+{etc.}$$

The other two thirds of that half are:

$$\frac{1}{2}\cdot\frac{2}{3}\cdot x = 1 + \frac{1}{5}+ \frac{1}{7}+ \frac{1}{11} + \frac{1}{13}+{etc.}$$

This is the series without the "nec per 2 nec per 3" divisible denominators.

He shows the same trick for "5" i.e. factor out 1/5 and then lists the other 4/5 (of the two thirds of the half of the series), and arrives at above formula.

I think this is very nice, but the subtle (?) problems with infinity etc. can be avoided by leaving out the harmonic series, and instead calculate the unsieved coprimes in the next primorial (2,6,30,210, etc.).


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