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

I have two questions about this equality:

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?

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  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Commented Jul 19 at 17:35
  • $\begingroup$ 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 $1$. $\endgroup$
    – Will Sawin
    Commented Jul 19 at 17:41
  • $\begingroup$ @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 $\endgroup$
    – Ali Taghavi
    Commented Jul 19 at 17:42

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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 $1$.

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  • $\begingroup$ Thank you very much for your answer $\endgroup$
    – Ali Taghavi
    Commented Jul 20 at 17:57

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