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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

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

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.

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    $\begingroup$ 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 $1$. $\endgroup$
    – Gary
    Commented Sep 12, 2021 at 17:00

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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 $z=0$.

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  • $\begingroup$ 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? $\endgroup$
    – Penelope
    Commented Sep 12, 2021 at 21:27
  • $\begingroup$ @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 $s$. $\endgroup$
    – Martin R
    Commented Sep 13, 2021 at 4:42
  • $\begingroup$ Martin - if a pole and its order is established in the real domain, surely it applies in the complex domain too? $\endgroup$
    – Penelope
    Commented Sep 13, 2021 at 5:44
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    $\begingroup$ @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. $\endgroup$
    – Martin R
    Commented Sep 13, 2021 at 6:41

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