Can a Finite Zeta Sum Be Expressed as a Finite Prime Product?

Question

The Riemann zeta function has the following representations for $$\text{Re}(s) > 1$$:

• As a Dirichlet series: $$ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} $$

• And as an Euler product: $$ \zeta(s) = \prod_{p \text{ prime}} \left(1 - \frac{1}{p^s} \right)^{-1} $$

These are equal only in the limit as $$N \to \infty$$.

However, I am not asking about approximations.

My question is this:

Can the finite sum $ \sum_{n=1}^N \frac{1}{n^s} $ be expressed as an exact finite product involving primes, possibly using the prime factorizations of integers up to N?

To clarify:

• I am not asking for a truncated version of the Euler product as an approximation.
• I'm asking whether there exists an exact expression of the finite sum $\sum_{n=1}^N \frac{1}{n^s}$ as a finite product over primes — possibly involving multiplicities, prime factorizations, or some clever rearrangement using prime powers.
• In other words, is there a way to express the finite sum using a function similar in spirit to the Euler product, but in a way that is exact for finite $N$ and can be implemented programmatically to compute the sum precisely?
• If such an expression exists, what is its form?

I understand that for the full zeta function, the infinite sum equals the infinite product. But I'm curious whether there is a meaningful finite product identity that holds exactly for the finite sum.

• J.G.: Hint: if $p$ is the greatest prime $\le N,ドル we need a 1ドル/(1-p^s)$ factor for the 1ドル/p^s$ term, but since 2ドルp>N$ there's a problem if $p\ge3$.
• paul garrett: A reasonable question, but I'd think not, for slightly complicated (and not so interesting) reasons.
• Riemann's Last Theorem 0bq.com: Paul Garrett, if I understand correctly, you’re saying that no answer to this question is known at present?

Answer 1

The answer is negative for $N \ge 3$ as follows.

Assume $$\sum_{n=1}^N \frac{1}{n^s}=\zeta_N(s)=\Pi_{p \le N}(\sum_{k \ge 0}a_k(p)p^{-ks})$$ where each term in the product converges somewhere. Then since there are finitely many terms in the product, they all converge absolutely for some $\Re s >A$ so we can switch sum and product and get that $a_0(p) \ne 0$ (free term is 1ドル=\Pi_P a_0(p)$) and $a_k(p)=0, p^k >N$ but $a_k(p) \ne 0$ if $p^k<N$ (just by looking at the $p^{-ks}$) terms on both sides.

In particular $a_1(p) \ne 0$ for all $p \le N$

But then picking $p\le N$ odd prime st 2ドルp>N$ which is possible when $N \ge 3$ we get that RHS contains a nonzero $(2p)^{-s}$ term and LHS doesn't and that is a contradiction!