A proof of the infinitude of prime numbers using the $\zeta$ function

Question

I present some questions about a proof (from these notes) of the infinitude of prime numbers, using properties of the $\zeta$ function.

Lemma $2.4.$ $$\zeta(s) = \frac{1}{s-1} + \psi(s)$$ where $\psi(s)$ is a holomorphic function in the half plane $\{s \in \Bbb C:\operatorname{Re} s > 0\}$.

Proof of Lemma $2.4$. We have $$\psi(s) = \zeta(s) - \frac{1}{s-1} = \sum_{n\ge 1} \int_n^{n+1} (n^{-s} - t^{-s})\, dt \tag{1}$$ It suffices to show, using the Weierstrass M-Test, that the right hand side converges absolutely, and uniformly on compact subsets of $\{s \in \Bbb C:\operatorname{Re} s > 0\}$. Since $$\int_n^{n+1} (n^{-s} - t^{-s})\, dt = \int_n^{n+1} \int_t^n \frac{s}{x^{s+1}}\, dx\, dt \tag{2}$$ one can check that (see this) $$\left|\int_n^{n+1} (n^{-s} - t^{-s})\, dt \right| \le \frac{|s|}{n^{\sigma+1}} \le \frac{\sup_{s\in K}|s|}{n^{\sigma+1}} \tag{3}$$ where $K$ is a compact subset of $\{s \in \Bbb C:\operatorname{Re} s > 0\}$ and $\sigma := \inf\{\operatorname{Re} s: s\in K\} > 0$.

Now comes the proof of the main result, i.e., the infinitude of primes.

Proof of the Main Result. The presence of a pole at $s = 1$ means that the zeta function is non-zero on some neighbourhood of this point. We can therefore take the logarithm of both sides of the equation $$\prod_{p\in \mathcal P} \left( 1 - \frac{1}{p^s}\right)^{-1} = \frac{1}{s-1} + \psi(s) \tag{4}$$ for $s\ne 1$, within this neighborhood (call it $U$). $$\log\left(\prod_{p\in \mathcal P} \left( 1 - \frac{1}{p^s}\right)^{-1} \right) = \log\left(\frac{1}{s-1} + \psi(s) \right)\tag{5}$$ Taylor expanding the logarithm on the left side, we have $$\sum_{p\in \mathcal P} p^{-s} + \sum_{p\in \mathcal P} \sum_{k=2}^\infty p^{-ks} = \log\left(\frac{1}{s-1} + \psi(s) \right)\tag{6}$$ As $\psi$ is holomorphic (and thus, bounded) on $U$, the right hand side of $(6)$ will be asymptotic to $\log\left(\frac{1}{s-1} \right)$ as $s\to 1$.

1. What does asymptotic mean in a rigorous mathematical sense?

So now we only need to show that the double series on the left hand side of $(6)$ remains bounded in order to prove the theorem.

2. How does the boundedness of the series on the left help?

Note that $$\sum_{p\in \mathcal P} \sum_{k=2}^\infty p^{-ks} = \sum_{p\in \mathcal P} \frac{1}{p^s(p^s - 1)} \le \sum_{n=1}^\infty\frac{1}{n^s(n^s - 1)}$$ and the latter series converges absolutely for $\operatorname{Re}s > \frac12$. Thus once we ensure that the region $U$ on which $(6)$ is defined falls within a circle of radius $\frac12$ centered at $s = 1$, we immediately have $$\sum_{p\in \mathcal P} \frac{1}{p^s} \sim \log\left( \frac{1}{s-1}\right).$$

3. What does $\sum_{p\in \mathcal P} \frac{1}{p^s} \sim \log\left( \frac{1}{s-1}\right)$ mean, and how does it follow? I see that we can shrink $U$ appropriately to ensure the convergence of $\sum_{p\in \mathcal P} \frac{1}{p^s(p^s - 1)}$.

The infinitude of primes, follows by taking $s\to 1$ in $\sum_{p\in \mathcal P} p^{-s} \sim \log\left( \frac{1}{s-1}\right)$. Clearly, $\log\left( \frac{1}{s-1}\right)$ blows up in magnitude near $s = 1$, so $\sum_{p\in \mathcal P} \frac{1}{p^s}$ must too. However, if $\mathcal P$ were a finite set, we could swap $\lim_{s\to 1}$ with $\sum_{P\in \mathcal P}$ to get $$\infty = \lim_{s\to 1} \log\left( \frac{1}{s-1}\right) = \lim_{s\to 1}\sum_{p\in \mathcal P} p^{-s} = \sum_{p\in \mathcal P} \lim_{s\to 1} p^{-s} = \sum_{p\in \mathcal P} \frac{1}{p} < \infty$$ which is a contradiction.

Thanks a lot!