Is showing $\zeta(s)$ has a simple pole in the real domain applicable to the complex domain?

Can the following simple "squeeze theorem" be used to show the Riemann zeta function's singularity is a simple pole of order 1, with the derivation being done entirely in the real domain?

The method is inspired by a question: Why does $\frac{s}{s-1} > \zeta(s) > \frac{1}{s-1}$ imply $\lim_{s \to 1^{+}}(s-1)\zeta(s)=1$?

Step 1

The inequality is derived in the real domain by comparing the sum $$\sum1/n^s$$ to the integral $$\int 1/x^sdx$$.

$$\boxed{\frac{1}{s-1}<\zeta(s)<\frac{1}{s-1}+1}$$

The derivation (eg here) is for real $$s$$.

Step 2

It is then simple algebra to show

$$1<(s-1)\zeta(s)

which, again, is for real $$s$$.

As $$s\rightarrow 1^+$$,

$$1<(s-1)\zeta(s)<1$$

Step 3

We know $$\zeta(s)$$ has a singularity at $$s=1$$, so the above shows it is a removable pole of order 1, a simple pole.

Question:

This analysis was done entirely in the real domain. Is the result valid when we consider $$s$$ to be complex?

If yes, this seems an easy to way to show the Riemann Zeta function's pole is of order 1.

• What you wrote at the end of step 2 makes no sense. If you let $s\to 1+$, $s$ cannot appear in your result. Also, there is no real number strictly between $1$ and $1$.
– Gary
Commented Sep 12, 2021 at 17:00

No, that argument is not valid. If $$a \in \Bbb R$$ and $$f$$ is holomorphic in $$B_r(a) \setminus \{ a \}$$ then the existence of $$\lim_{x \to a, x \in \Bbb R} (x-a)f(x)$$ does not imply that $$f$$ has simple pole at $$a$$. A counterexample is $$f(z) = \frac 1z + e^{-1/z^2}$$ which has an essential singularity at $$z=0$$.
• thanks for this helpful comment. I am therefore puzzled by the result from this question math.stackexchange.com/questions/133870/… that $(s-1)\zeta(s)=$ by the squeeze theorem, which suggests the singularity is a removable pole of rider 1. What am I misunderstanding? Commented Sep 12, 2021 at 21:27
• @Tariq: In that question it is shown that $\lim_{s \to 1^{+}}(s-1)\zeta(s)=1$ and the notation $\lim_{s \to 1^{+}}$ implies that the limit is for real numbers $s$, approaching $s=1$ from the right. It is not claimed or shown that $\lim_{s \to 1}(s-1)\zeta(s)=1$ for complex values of $s$. Commented Sep 13, 2021 at 4:42
• @Tariq: I can only answer what you have asked, and the existence of $\lim_{s \to 1^{+}}(s-1)\zeta(s)=1$ is not sufficient to prove that a function has a simple pole, as I demonstrated with an example. Commented Sep 13, 2021 at 6:41