Proof of Zeta Function product formula

Question

In my Complex Analysis classes I was studying the Riemann's zeta function. At some point my teacher was able to demonstrate the product formula in this way:

"Suppose M and N are positive integers with M>N. Observe now that, by the fundamental theorem of arithmetc, any positive integer $ n \le N$ can be written uniquely as a product of primes, and that each prime that occurs in the product must be less than or equal to N and repeated less than M times. Therefore:"

$$\zeta(s)\le \prod_{p \le N} (1+p^{-s}+p^{-2s}+...+p^{-Ms}) \mbox{ (1)}$$, p primes.

Letting N tend to infinity, we have:

$$\zeta(s)\le \prod_{p}\frac{1}{1-p^{-s}}$$ And then, using a similar argument with the fundamental theorem of arithmetic:

$$\prod_{p \le N} (1+p^{-s}+p^{-2s}+...+p^{-Ms}) \le \zeta(s) \mbox{ (2)}$$ And conclude that:

$$\zeta(s)=\prod_{s} (\frac{1}{1-p^{-s}})$$

Assuming $s$ real

But I'm not understanding how to get the inequalities (1) and (2). Can someone explain it to me?

Answer 1

I think the argument goes as follows: Let $s>0$.

Fix $N>0$. Then consider $\sum^N_{n=1}n^{-s}$. Next, consider the product \begin{align} \prod_{p-\text{prime}\leq N}\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\ldots \right) \end{align} which when expanded is a sum of reciprocal of integers raised to the $s$ power. Therefore, we have the inequalities \begin{align} \sum^N_{n=1}\frac{1}{n^s} \leq&\ \prod_{p-\text{prime}\leq N}\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\ldots \right) \leq \prod_{p-\text{prime}}\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\ldots \right) \\ =&\ \prod_{p-\text{prime}}\left(1-\frac{1}{p^s}\right)^{-1}. \end{align} But since this holds for all $N$, then it follows \begin{align} \zeta(s) \leq \prod_{p-\text{prime}}\left(1-\frac{1}{p^s} \right)^{-1}. \end{align}

The reverse inequality is trivial, since \begin{align} \prod_{p-\text{prime} \leq N}\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\ldots \right), \end{align} when expanded, is a sum of reciprocal of integers raised to the $s$ power. However, the key observation here is that non of the integers are similar. Hence we have that \begin{align} \prod_{p-\text{prime} \leq N}\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\ldots \right) \leq \zeta(s) \end{align} for all $N$. Take the limit, we get that \begin{align} \prod_{p-\text{prime}}\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\ldots \right) \leq \zeta(s). \end{align} Hence we have the desired equality.

Answer 2

I always have trouble understanding infinite product proofs. Either the proof isn't written carefully enough to my liking or I don't spend enough time trying to understand why the proof as written is correct. Anyway, here is an alternative proof that you might like better.

For $\operatorname{Re}(s) > 1$, divide both sides of the equation $\zeta(s) = \sum\limits_{n \geq 1} \frac{1}{n^s} $ by $2^s$ to get $\frac{\zeta(s)}{2^s} = \sum\limits_{2 \mid n} \frac{1}{n^s}$. Subtract the second equation from the first. The left hand side is $\zeta(s) - \frac{\zeta(s)}{2^s} = \zeta(s)(1 - \frac{1}{2^s})$. The right hand side is the sum over all odd integers:

$$\zeta(s)(1 - \frac{1}{2^s}) = \sum\limits_{2 \nmid n} \frac{1}{n^s}$$

We can do this kind of thing because the sum defining $\zeta(s)$ is absolutely convergent, which means we can rearrange terms at will. Similarly, if we multiply both sides of this equation by $(1 - \frac{1}{3^s})$, then we get

$$\zeta(s)(1 - \frac{1}{2^s})(1 - \frac{1}{3^s}) = \sum\limits_{2, 3 \nmid n} \frac{1}{n^s}$$

If $p_1 < p_2 < \cdots$ are the prime numbers, then inductively we get

$$\zeta(s) \prod\limits_{i=1}^N (1 - \frac{1}{p_i^s}) = \sum\limits_{p_1, ... , p_N \nmid n} \frac{1}{n^s}$$

The limit as $N$ goes to infinity of the right hand side exists, and is equal to the sum of $\frac{1}{n^s}$ over all positive integers which are not divisible by any prime number. There is only one such positive integer:

$$\lim\limits_{N \to \infty} \sum\limits_{p_1, ... , p_N \nmid n} \frac{1}{n^s} = 1$$

Hence

$$\lim\limits_{N \to \infty} \prod\limits_{i=1}^N (1 - \frac{1}{p_i^s}) = \prod\limits_{i=1}^{\infty} (1 - \frac{1}{p_i^s})$$

exists as a limit. We have made use of the simple fact that if $w, \alpha_n, \beta_n : n \in \mathbb N$ are complex numbers with $w \alpha_n = \beta_n$ for each $n$, and if $\beta = \lim\limits_n \beta_n$ exists and is not zero, then $\alpha = \lim\limits_n \alpha_n$ also exists, $w \alpha = \beta$, and $w$ and $\alpha$ are nonzero. Thus

$$\zeta(s) \prod\limits_{i=1}^{\infty} (1 - \frac{1}{p_i^s}) = 1$$

and in particular, $\zeta(s) \neq 0$ for $\operatorname{Re}(s) > 1$. The same proof works with any reordering of the prime numbers, so

$$\prod\limits_p (1 - \frac{1}{p^s})$$

is well defined as a limit, independent of the order in which the product is taken, with

$$\zeta(s) \prod\limits_p (1 - \frac{1}{p^s}) = 1$$

or

$$ \prod\limits_p (1 - \frac{1}{p^s}) = \zeta(s)^{-1}$$

In any convergent infinite product of nonzero terms whose limit is nonzero, we can commute taking inverses with taking limits, because $z \mapsto z^{-1}$ is continuous on $\mathbb{C} - 0$:

$$\zeta(s) = \prod\limits_p (1 - \frac{1}{p^s})^{-1}$$