How many elementary ways are there to prove that $\displaystyle\left. \frac d {ds}\,\frac 1 {\zeta(s)} \right|_{s=1} = 1 $?

Question

In a comment under this answer, a user boldly asserts that there is ONLY ONE way to prove that $$ \left. \frac d {ds}\,\frac 1 {\zeta(s)} \right|_{s=1} = 1 $$ where $\zeta$ is Riemann's zeta function.

So my question is: what examples of elementary ways of proving this best illustrate whatever diversity of such methods exists?

Answer 1

this is more an explanation of intuitively how do I think to it than an answer.

this is mainly a question on how do we define $\mu(n)$. there are only two ways to define $\mu(n) $ (there are many others but equivalent to these) :

• 1) as the Dirichlet convolution inverse of the constant sequence $1$ $$\left(\sum_{n=1}^\infty \mu(n) n^{-s} \right) \left(\sum_{m=1}^\infty m^{-s} \right) = 1$$

$$\text{or equivalently } \quad\sum_{d \mid n} \mu(d) = \delta_{n=1}$$

• 2) as the function being $\pm 1$ for square-free integers $$\mu(1) = 1, \qquad \mu(n) = \begin{cases} (-1)^k & \text{ if } n = \prod_{i=1}^k p_i, \text{ the $p_i$ being different primes}, \\[4pt] 0 & \text{otherwise} \end{cases}$$

of course these definitions are shown equivalent by the Euler product $$\zeta(s) = \sum_{n=1}^\infty n^{-s} =\prod_p \frac{1}{1-p^{-s}}$$ the question now is : if we define $\mu(n)$ with 1) or 2) how can we prove that $$\left.\frac{d \ 1/\zeta(s)}{ds}\right|_{s=1} = 1 \tag{§} $$

• 1) defines $\mu(n)$ as the Dirichlet convolution inverse of $1$ so we will have to prove that $$\zeta(s) = \frac{1}{s-1}+\mathcal{O}(1)\qquad \text{when }s\to 1 \tag{§§}$$

• now if we define $\mu(n)$ with 2), can we prove $(§)$ without relying also on 1) and hence on $(§§)$?

  I think that for proving $(§)$ only from 2) we would need something like the prime number theorem. the original (analytic) proof of the PNT clearly relies on 1) and $(§§)$.

  so our only hope is with the ''elementary'' proof of the PNT, but it also relies on the definition 1), hence you cannot prove $(§)$ without using 1) in your proof, and your proof of $(§)$ will be somewhat equivalent to proving $(§§)$.