I am trying 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 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.
$$ \frac{1}{(1-\frac{1}{p_{1}})}\cdot\frac{1}{(1-\frac{1}{p_{2}})}=\left(1+\frac{1}{p_{1}}+\frac{1}{p_{1}^{2}}+\ldots\right)\cdot\left(1+\frac{1}{p_{2}}+\frac{1}{p_{2}^{2}}+\ldots\right) $$
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 $p_i$.
$$ \prod_{p_{i}}\frac{1}{(1-\frac{1}{p_{i}})}=\prod_{p_{i}}\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right) $$
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 $\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right)$, we will obtain terms which are all combinations of primes, in all the combinations of powers for each prime.
$$ \prod_{p_{i}}\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right) = 1 + \frac{1}{p_1} + \frac{1}{p_2} + \ldots + \frac{1}{p_1 p_2} + \frac{1}{p_1 p_3} + \ldots + \frac{1}{p_1^2} + \frac{1}{p_2^2} + \ldots $$
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.
$$ \boxed{\prod_{p}\frac{1}{(1-\frac{1}{p})}=\sum_{n}\frac{1}{n}} $$
Discussion
If my suspicion is correct that the error is in the infinite product because it doesn't converge, how does changing $p$ for $p^s$ for $s>1$ help?
Best Answer
Your intuition is right that there is a flaw in step 3. You propose that any infinite product in which the terms tend to $1$ must converge; but this is false. The easiest example is the telescoping product $$ \prod_{j=1}^J \bigg( \frac{j+1}j \bigg) = J+1, $$ which show that $\prod_{j=1}^\infty \frac{j+1}j$ diverges to $\infty$. More generally, the convergence of the product $\prod_j (1+x_j)$ is by definition equivalent to the convergence of its logarithm $\sum_j \log(1+x_j)$, and in many circumstances this is further equivalent to the convergence of the logarithm's linear approximation $\sum_j x_j$; the example above is the case $x_j = \frac1j$. This is where we need the assumption $s>1$.
There is also a flaw (a very common one, in my experience) in step 4. You are correct that each of the terms on the right-hand side appears when we expand the left-hand side. However, there are literally uncountably many other terms as well. What about, for example, the term in the expansion where we select $\frac1{p_i^1}$ from every factor? What happens to a term like $\frac1{p_1p_2\cdots}$? And all the others where infinitely many of the factors don't equal $1$? In my opinion, it's not very easy to repair this flaw without starting over and using a more rigorous real-analysis proof.