# Rate of convergence of the Riemann zeta function and the Euler product formula

We consider the Euler product formula $$\sum_{n=1}^\infty \frac{1}{n^s}=\prod_p \frac{1}{1-p^{-s}}$$

1)Does the rate of convergence of each side depend on $$s$$?

2)For given $$s$$ is the rate of convergence of the left hand side equal to the rate of convergence of the right hand side of the equality?

• How do you define the rate of convergence? It's not clear if "rate of convergence" can be defined in such a way that the rate of convergence of a sum can be compared to a product, nor two sums/products whose index sets are different. Commented Jul 19 at 17:35
• If you interpret both sides as sequences with partial products and partial sums, right-hand side is still indexed by the primes rather than the natural numbers, so you still have to decide whether the $n$th member of the sequence is the $n$th prime or the product over primes up to $n$. But I guess this doesn't matter since in all cases the rate of convergence will be $1$. Commented Jul 19 at 17:41
• @WillSawin the rate of convergence has a standard definition in numerical analysis: Let $a_n$ positive goes to zero then the rate is unique q for which the limit $\frac{a_{n+1}}{a_n^q}$ is finite and non zero. In case of non convergence of the later limit you may replace by limsup or lim inf appropriately. now if $a_n$ goes to $\ell$ you consider $|a_n-\ell$ instead of $a_n$. An interesting example is that rate of convergence of secant algorithm is the golden number Commented Jul 19 at 17:42

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 $$\zeta(s) \left( \prod_{p > p_m} \frac{1}{ 1-p^{-s}} -1 \right) = \zeta(s) \left(e^{ \sum_{p> p_m} \sum_{k=1}^\infty p^{-ks}/k } -1 \right) .$$
$$\sum_{p> p_m} \sum_{k=1}^\infty p^{-ks}/k \approx \int_{p_m}^\infty t^{-s} dt/\log t \approx p_m^{1-s} /\log p_m$$ 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 $$1$$.