I'll use a variable $m$ for the $m$th approximation in the sequence since the variable $n$ is in use.

On the left hand side, the $m$th approximation is $$\sum_{n=1}^{m} n^{-s}$$ with error $$\sum_{n=m+1}^\infty n^{-s}$$ which standard estimates for approximating a sum by an integral imply is well-approximated by $$\int_{m}^\infty t^{-s} dt = \frac{m^{1-s}}{s-1}.$$ The value of the error at $m$ divided by the value of the error at $m+1$ is $(1+1/m)^{s-1}$ which converges to $1$ as $m$ goes to $\infty$ so the rate of convergence is $1$.

On the other side, the analysis is similar, but more complicated. The $m$th approximation is $\prod_{ p \leq p_m} \frac{1}{1-p^{-s}} $ where $p_m$ is the $m$th prime so the error is $$ \zeta(s) \left( \prod_{p > p_m} \frac{1}{ 1-p^{-s}} -1 \right) = \zeta(s) \left(e^{ \sum_{p> p_m} \sum_{k=1}^\infty p^{-ks}/k } -1 \right) .$$

The prime number theorem implies that

$$ \sum_{p> p_m} \sum_{k=1}^\infty p^{-ks}/k \approx \int_{p_m}^\infty t^{-s} dt/\log t \approx p_m^{1-s} /\log p_m $$ and that $p_m \approx m \log m$ so the sum is $m^{1-s}$ times a logarithmic term. Exponentiating and subtracting $1$ does not much affect the size of the error, and multiplying by $\zeta(s)$ only multiplies by a constant, so again the error has the same size and the rate of convergence is $1$.