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We know the following: $$\gamma=\lim_{n\to\infty }\left(\sum_{k=1}^n\frac{1}{k}-\ln(n)\right).$$ This could be a good candidate for renormalized sum of $\left(\sum_{k=1}^{\infty}\frac{1}{k}\right)$.

Also, we know the following: $$-\gamma=\lim_{n\to\infty }\left(\sum_{p\leq n}\frac{\ln(p)}{p-1}-\ln(n)\right).$$ I want to ask does this analogously mean that $-\gamma$ is renormalized value of $(\sum_{p}\frac{\ln(p)}{p-1})$?

Also, I wanted to ask similar question:

How to assign renormalized value of the following sums? ( Is it even possible?) I.e., is there a way we could assign finite values to the following sums? \begin{gather} \tag{1}\label{1} \sum_{p} \frac{1}{\sqrt{cp}-1} \\ \tag{2}\label{2} \sum_{p} \frac{\ln(p)}{\sqrt{cp}-1} \end{gather} Here $c$ is a constant. (I'm particularly interested in two cases, where $c=1$ and $c= e$.)

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    $\begingroup$ The word regularization is completely misleading, and thus wrong. There is no elementary function $f$ such that $\lim_{n\to\infty }\sum_{p\le n} \frac1{\sqrt{cp}-1}- f(n)$ converges (try plotting $\sum_{p\le n} \frac1{\sqrt{cp}-1}-\sum_{2\le m\le n} \frac1{\sqrt{cm}\log m}+\frac{c^{-1/2}-1/2}{m\log m}$). Telling us why you are asking those questions about $\sum_p \frac1{\sqrt{cp}-1}$ would make it easier. $\endgroup$
    – reuns
    Commented Apr 20, 2021 at 0:40
  • $\begingroup$ @renus thank you for the comment. I really apologise for my lack of knowledge. But is there any way or method that can be used to assign finite value to the sum? Also, sir, is the term 'renormalization' appropriate? $\endgroup$
    – Zaza
    Commented Apr 20, 2021 at 9:35
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    $\begingroup$ You should stop asking on mathoverflow (where answers and comments will assume a lot of prior knowledge), inventing problems about primes, and instead concentrate on the proof of $-\gamma=\lim_{n\to\infty }\left(\sum_{p\leq n}\frac{\ln(p)}{p-1}-\ln(n)\right)$, which needs (and implies) the PNT. Assuming the PNT it is an elementary exercice in complex analysis. People will be glad to help on math.stackexchange.com $\endgroup$
    – reuns
    Commented Apr 20, 2021 at 19:27

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Regarding you first question, logically, yes. At least in the theory of the divergent integrals I am currently working on, regularization operator is linear as an axiom:

$$\operatorname{reg }\int_a^b \left(f(x)+g(x)\right)dx=\operatorname{reg }\int_a^b f(x)dx+\operatorname{reg }\int_a^b g(x)dx$$

For various summation methods of divergent series this can be proven as a theorem (depending on methods in question), and for the most methods usually used and compatible with each other, this is the case in fact. So, it is natural to accept this as a postulate and check any new proposed regularization method on compliance with this rule as one of the main requirements.

Regarding your second question, I have nothing to comment.

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