# A question on assigning finite values to divergent sums involving expression of primes

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? 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 $$c= e$$.)

• 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. – reuns Apr 20 at 0:40
• @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? – Zaza Apr 20 at 9:35
• 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 – reuns Apr 20 at 19:27

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