Why do you need to prove the error term goes to zero for the complete derivation of the Euler Product Formula?

I am doing a project on the Riemann-Zeta Function which begins by examining the Euler Product Formula. I understand the proof up until the point where it is made 'rigorous'. In other words, I understand the use of the fundamental theorem of arithmetic to show that $$\prod_{p\text{ prime}} \frac{1}{1-\frac{1}{p^s}}$$

is equal to a sum of terms of the form $$\frac{1}{(p_{1}^{n_{1}}p_{2}^{n_{2}}p_{3}^{n_{3}} \cdots p_{r}^{n_{r}})^s}$$
for distinct primes $p_i$ and natural numbers $n_i$. This, to me, completes the proof that $$\sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p\text{ prime}} \frac{1}{1-\frac{1}{p^s}}$$

However, in rigorous proofs, there is talk of an 'error term' using the Big O function or something similar. I don't understand why this is necessery, what it shows, and how it is showing it. Any help is greatly appreciated.

• Probably in the proof you read the author first consider the finite product up to a prime $P$ and then shows that $$\left|\zeta\left(s\right)-\prod_{p\leq P}\left(1-\frac{1}{p^{s}}\right)^{-1}\right|\leq\frac{1}{\left(P+1\right)^{ \sigma }}+ \frac{1}{\left(P+2\right)^{\sigma}}+\dots$$ so if you take $P \rightarrow \infty$ the RHS goes to zero. – Marco Cantarini Dec 5 '15 at 8:34
• Possible duplicate of Euler Product formula for Riemann zeta function proof – Klangen Dec 17 '18 at 12:04
• @Klangen This is earlier. – GNUSupporter 8964民主女神 地下教會 Dec 17 '18 at 13:43