Euler product proof by the fundamental theorem of arithmetic

Question

I was looking at the proof of Euler product at Mathworld https://mathworld.wolfram.com/EulerProduct.html

First we expand the product, this I understand, then “we write each term as a geometric series.” But I see that we don’t write “each term” as a geometric series. The whole term is $\frac{1}{1-\frac{1}{p_k^s}}$ but we take only this part $1/p_k^s$. Why don’t we take the whole term?

I’m not understanding this line either:

$$ 1 + \sum_{1\leq i} \frac{1}{p_i^s} + \sum_{1\leq i \leq j} \frac{1}{p_i^s p_j^s} + \sum_{1\leq i \leq j \leq k} \frac{1}{p_i^s p_j^s p_k^s} + \ldots $$

How does $\sum_{1\leq i} \frac{1}{p_i^s}$ become $\frac{1}{2^s}$?

Edit: Ioveri answer

As per Ioveri's answer, the Equation (4) below, written in general form, contains the reciprocals of all natural numbers written as prime factors. The zeta sum is the sum of the reciprocals of all natural numbers. So both sides state the same thing and they are equal.

So, even more explicitly, the first sum is the sum of the reciprocals of all primes,

$$ 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{5^s} + \frac{1}{7^s} \ldots $$

The second sum is all numbers that can be expressed as the product of two primes, including repetitions,

$$ 1 + \frac{1}{2^s\cdot 2^s} + \frac{1}{2^s\cdot 3^s} + \frac{1}{3^s\cdot 3^s} + \frac{1}{3^s\cdot 5^s} \ldots $$

The third sum is all the numbers that can be expressed as the product of three primes,

$$ 1 + \frac{1}{2^s\cdot 2^s\cdot 2^s} + \frac{1}{2^s\cdot 2^s\cdot 3^s} + \frac{1}{2^s \cdot 3^s\cdot 3^s} + \frac{1}{3^s \cdot 3^s\cdot 3^s} + \frac{1}{3^s \cdot 3^s\cdot 5^s}\ldots $$

And so on. Rearranging the equation (4) we obtain the zeta sum.

Edit: I'm adding the full proof:

$$ \begin{align} \prod_{k=1}^\infty \frac{1}{1-\frac{1}{p_k^s}} &= \frac{1}{1-\frac{1}{p_1^s}} \cdot \frac{1}{1-\frac{1}{p_2^s}} \cdot \frac{1}{1-\frac{1}{p_3^s}} \ldots \\ &= \left [ \sum_{k=0}^\infty \left ( \frac{1}{p_1^s} \right )^k \right] \left [ \sum_{k=0}^\infty \left ( \frac{1}{p_2^s} \right )^k \right] \left [ \sum_{k=0}^\infty \left ( \frac{1}{p_3^s} \right )^k \right]\\ &=\left ( 1 + \frac{1}{p_1^s} + \frac{1}{p_1^{2s}} + \frac{1}{p_1^{3s}} + \ldots \right ) \left ( 1 + \frac{1}{p_2^s} + \frac{1}{p_2^{2s}} + \frac{1}{p_2^{3s}} + \ldots \right ) \ldots\\ &= 1 + \sum_{1\leq i} \frac{1}{p_i^s} + \sum_{1\leq i \leq j} \frac{1}{p_i^s p_j^s} + \sum_{1\leq i \leq j \leq k} \frac{1}{p_i^s p_j^s p_k^s} + \ldots \text{4th equation}\\ &= 1 + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \frac{1}{5^s} + \ldots\\ &= \sum_{n=1}^\infty \frac{1}{n^s}\\ &= \zeta(s)\\ \end{align} $$

Second edit

Is this how I need to multiply the factors of the geometric series to obtain the equation 4:

$$ 1+\frac{1}{p^s_2}+\frac{1}{p^{2s}_2} + \frac{1}{p^{3s}_2} + \frac{1}{p^s_1} + \frac{1}{p^{s}_1\times p^s_2} + \frac{1}{p^{s}_1\times p^{2s}_2} + \frac{1}{p^s_1\times p^{3s}_2} + \frac{1}{p^{2s}_1} + \frac{1}{p^{2s}_1\times p^s_2} + \frac{1}{p^{2s}_1\times p^{2s}_2} + \frac{1}{p^2s_1\times p^{3s}_2} + \ldots $$

Third edit

And these are the expansions of the sum terms of equation 4. Is this correct?:

$$\sum_{1\leq i} = \frac{1}{p^s_i} = \frac{1}{2^s} +\frac{1}{3^s} + \frac{1}{5^s} + \frac{1}{7^s} + \ldots $$

$$ \sum_{1\leq i\leq j} = \frac{1}{p^s_i\times p^s_j} = \frac{1}{2^s\times 3^s} +\frac{1}{3^s\times 5^s} + \frac{1}{5^s\times 7^s} + \frac{1}{7^s\times 11^s} + \ldots $$

$$ \sum_{1\leq i\leq j \leq k} = \frac{1}{p^s_i\times p^s_j\times p^s_k} = \frac{1}{2^s\times 3^s\times 5^s} +\frac{1}{3^s\times 5^s\times 7^s} + \frac{1}{5^s\times 7^s\times 11^s} + \frac{1}{7^s\times 11^s\times 13^s} + \ldots $$

Answer 1

How does $\sum_{1\leq i} \frac{1}{p_i^s}$ become $\frac{1}{2^s}$?

It's simply isn't. The $\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+...$ series isn't obtained by taking those sums individually but by rearranging their terms. First I'll explain those sum means:

$$\sum_{1\leq i} p^s_i \text{ is the sum of primes, each raised to the power of s}$$ $$\sum_{1\leq i \leq j} p^s_i p^s_j = \sum_{1\leq i \leq j} (p_ip_j)^s\text{ is the same sum but each number has 2 prime factor} \\ \text{(repetition also counts, e.g. } 4^{-s} = 2^{-s}2^{-s} \text{ is also a term})$$ $$\sum_{1\leq i \leq j \leq k} p^s_i p^s_j p^s_k = \sum_{1\leq i \leq j} (p_ip_jp_k)^s \text{ is the same sum but each has 3 prime factors ...}$$ And so on. This is true because each natural number larger than 1 has a unique prime factorization. And so when we add all these sums together, it's the same as "the sum of all numbers that have an arbitrary number of prime factors, each are raised to the power of $s$", or in other words, the sum of the $s^{\text{th}}$ power of those that are factorizable. And since all natural number greater than $1$ can be factorized, this mean the sum contains all $s^{\text{th}}$ powers of natural numbers except $1$. But we already have $1$ as a separate term, so the two series that you are questioning are really the same (*), it's just that one is arranged into sums of those with the same number of factors, and one is just arranged in an ascending order.

(*) Since this is an infinite rearrangement, some extra steps are actually need to prove that the sum of these two series are equal in a rigorous way. It's not too hard to do prove though

Edit: Regarding the third edit of the post: The prime factor can repeat, at least in this context since the condition for the indices is $1\leq i \leq j \leq k$ not $1 \leq i < j <k$, So the second sum, for example is: $$ \sum_{1\leq i \leq j} p^s_i p^s_j = 2^s2^s +2^s3^s + 3^s3^s + 2^s5^s + 3^s5^s + 5^s5^s + ...$$

Edit 2: Technically speaking the 4th equation mentioned by OP isn't the proper way to extend that infinite product, it's the rearranged expansion with hidden extra steps similar to what I mentioned earlier (Hint: it's related to absolute convergence)

Answer 2

On your second question: The MathWorld article explains this as follows:

Here, the rearrangement leading to equation $(1)$ follows from the fundamental theorem of arithmetic, since each product of prime powers appears in exactly one denominator and each positive integer equals exactly one product of prime powers.

The first term is $1=\frac1{1^s}$, the single sum contains the terms $\frac1{n^s}$ for all prime $n$, the double sum contains the terms $\frac1{n^s}$ for all semiprime $n$, and generally the $k$-fold sum contains the terms $\frac1{n^s}$ for all $n$ that are the product of $k$ primes. (The primes need not be distinct, which is why the summation uses $\le$.) By the fundamental theorem, that exactly covers each positive integer $n$ once. It’s not that the single sum becomes $\frac1{2^s}$; it’s just that for every positive integer $n$ (including $2$) there’s exactly one term in exactly one of the sums that yields $\frac1{n^s}$.

I’m not sure I understand your first question. The article is a bit confusing because it reuses the dummy index $k$ of the product as the dummy index of each of the geometric series, and because it calls the factors of the product “terms”, but it correctly rewrites

$$ \frac1{1-\frac1{p_1^s}} $$

as

$$ \sum_{k=0}^\infty\left(\frac1{p_1^s}\right)^k=1+\frac1{p_1^s}+\frac1{p_1^{2s}}+\frac1{p_1^{3s}}+\cdots\;, $$

which is the full geometric series.