Eliminating remainder in formulation of Riemann zeta function formula

Question

Summary of page 5 of Values of the Riemann zeta function at integers:

Taking the product over prime numbers $p\le N$, we get

$$\prod_{p\le N}\left(1+\frac1{p^s}+\frac1{p^{2s}}+\cdots\right) = 1+\frac1{2^s}+\frac1{3^s}+\cdots+\frac1{N^s}+\text{remainder}$$

Since $\operatorname{Re}(s)>1$, it is elementary calculus to show that the remainder can be made arbitrarily small if $N$ is sufficiently large.

Could someone explain how this is shown without relying on proof-by-intimidation? I don’t understand how $N\to\infty$ (if that is what is meant by “sufficiently large”) does anything to reduce the remainder term. The only thing I could guess is that this is maybe related to a Lagrange error, but I really have no idea.

Answer 1

$$\prod_{p\le N}\left(1+\frac1{p^s}+\frac1{p^{2s}}+\cdots\right) = \sum_{k_1=0}^{\infty} \cdots \sum_{k_j=0}^{\infty} \frac{1}{(p_{1}^{k_1} \cdots p_{j}^{k_j})^{s}}$$

where the $p_{j}$'s are the finitely many primes up to $N$. There are no repeated summands because of the fundamental theorem of arithmetic and every $m \leq N$ is realized by the summation. So it is equal to

$$ 1+\frac1{2^s}+\frac1{3^s}+\cdots+\frac1{N^s}+ \bar{\sum_{m > N}} \frac{1}{m^s}$$

where the bar indicates that the sum is taken over those $m$ with prime divisors less than or equal to $N$. Now $$ |\bar{\sum_{m > N}} \frac{1}{m^s} | \leq \bar{\sum_{m > N}} \frac{1}{|m^{s}|}= \bar{\sum_{m > N}} \frac{1}{m^{\text{Re}(s)}}$$

so this remainder term is bounded by $\frac{1}{N^{\text{Re}(s)}}$ plus the integral $$ \int_{N}^{\infty} \frac{dx}{x^{\text{Re}(s)}} $$ which go to zero as $ N \to \infty$ since $\text{Re(s)} > 1$.

Answer 2

To rephrase Parthiv Basu's answer the main point is $$\prod_{p \le N} \left(1+\sum_{k=1}^\infty p^{-sk}\right) = \sum_{n=1, Lpf(n) \le N}^\infty n^{-s}$$ where $Lpf(n)$ is the largest prime factor

thus we have $$\left|\zeta(s) - \prod_{p \le N}\left(1+\sum_{k=1}^\infty p^{-sk}\right)\right| = \left|\sum_{n=1, Lpf(n) > N}^\infty n^{-s}\right| < \sum_{n=N+1}^\infty |n^{-s}| < \int_N^\infty x^{-\Re(s)}dx = \frac{N^{1-\Re(s)}}{\Re(s)-1}$$