number theory – Convergence of infinite products flaw in this derivation of Euler product formula?

I am writing to write a derivation of the Euler product formula in a way that is accessible to younger students and those not mathematically trained at university.

Because of this I am starting from the simplest starting point that I can, and use the simplest possible steps in the derivation.

I arrive at the following which I know is wrong as it is missing the exponent $s>1$ to ensure convergence.

$$ prod_{p}frac{1}{(1-frac{1}{p})}=sum_{n}frac{1}{n} $$

My question is to ask where I am introducing the error is in the following simple derivation.

Step 1

Start with a familiar series, valid for $|x| < 1$.

$$ frac{1}{(1-x)}=1+x+x^{2}+x^{3}+ldots $$

We’re interested in primes, and might be tempted to set $x$ to a prime $p$, but we can’t because $|x|$ needs to be $<1$. Let’s try $frac{1}{p}$ which is always $<1$.

$$ frac{1}{(1-frac{1}{p})}=1+frac{1}{p}+frac{1}{p^{2}}+frac{1}{p^{3}}+ldots $$

This series converges to a finite value.

Step 2

Let’s now pick two different primes $p_1$ and $p_2$, and multiply the analogous series for each.

$$ frac{1}{(1-frac{1}{p_{1}})}cdotfrac{1}{(1-frac{1}{p_{2}})}=left(1+frac{1}{p_{1}}+frac{1}{p_{1}^{2}}+ldotsright)cdotleft(1+frac{1}{p_{2}}+frac{1}{p_{2}^{2}}+ldotsright) $$

I don’t think there is a problem here.

We are multiplying two infinite series, each one known to converge to a finite value. Am I wrong?

Step 3

Now we extend from two primes, $p_1$ and $p_2$, to all primes $p_i$.

$$ prod_{p_{i}}frac{1}{(1-frac{1}{p_{i}})}=prod_{p_{i}}left(1+frac{1}{p_{i}}+frac{1}{p_{i}^{2}}+ldotsright) $$

I think this is the flaw in my derivation. We have an infinite product of finite values.

Each factor $frac{1}{(1-frac{1}{p_{i}})}$ is $>1$ so the infinite product looks like it might not converge.

However each factor tends to $rightarrow 1$ as $i rightarrow infty$. This suggests the infinite product is ok. Am I wrong?

Step 4

If we multiply out those brackets, each of the form $left(1+frac{1}{p_{i}}+frac{1}{p_{i}^{2}}+ldotsright)$, we will obtain terms which are all combinations of primes, in all the combinations of powers for each prime.

$$ prod_{p_{i}}left(1+frac{1}{p_{i}}+frac{1}{p_{i}^{2}}+ldotsright) = 1 + frac{1}{p_1} + frac{1}{p_2} + ldots + frac{1}{p_1 p_2} + frac{1}{p_1 p_3} + ldots + frac{1}{p_1^2} + frac{1}{p_2^2} + ldots $$

Importantly, each combination occurs only once.

Step 5

The fundamental theorem of arithmetic tells us that each integer can be written as a unique product of primes. This means that each integer is represented in the terms we arrived at in the previous step.

Because each combination of primes occurs only once, each integer is represented only once.

That leads us to the Euler product formula.

$$ boxed{prod_{p}frac{1}{(1-frac{1}{p})}=sum_{n}frac{1}{n}} $$