Asymptotic expansion of $\sum_{\substack{p \in \mathbb{P}\\ p\le x}}{\frac{\log(p)}{p-1}}$

I am trying to determine the asymptotic expansion of the following series:

$$\sum_{\substack{p \in \mathbb{P}\\ p\le x}}{\frac{\log(p)}{p-1}}$$

First I did that :

$$\frac{\log(p)}{p-1} = \sum_{k=1}^{\infty}{\frac{\log(p)}{p^k}} = \frac{\log(p)}{p} + \sum_{k=2}^{\infty}{\frac{\log(p)}{p^k}} \label{eq:eq1} \tag{1}$$

$$\sum_{\substack{p \in \mathbb{P}\\ p\le x}}{\frac{\log(p)}{p-1}} = \sum_{\substack{p \in \mathbb{P}\\ p\le x}}{\frac{\log(p)}{p}} + \sum_{k=2}^{\infty}{\sum_{\substack{p \in \mathbb{P}\\ p\le x}}{\frac{\log(p)}{p^k}}} \tag{2}$$

The second sum of the RHS is clearly converging. Also, its limit can can be expressed in term of the derivative of the prime zeta function ($$P(s)$$ which is absolutely convergent for $$\Re(s)\gt 1$$).

The problem is coming from the first sum as it is divergent. However, from the Prime Number theorem we have (Approximations for the nth prime number : Wikipedia FR):

$$\frac{n}{p_n} = \frac{1}{\log(p_n)}+\frac{1}{\left(\log(p_n)\right)^2} + \frac{2}{\left(\log(p_n)\right)^3} + \frac{6}{\left(\log(p_n)\right)^4} + o\left(\frac{1}{\left(\log(p_n)\right)^4}\right) \tag{3}$$

Giving :

$$\frac{\log(p_n)}{p_n} = \frac{1}{n}+\frac{1}{n\log(p_n)} + \frac{2}{n\left(\log(p_n)\right)^2} + \frac{6}{n\left(\log(p_n)\right)^3} + o\left(\frac{1}{n\left(\log(p_n)\right)^3}\right)$$

And

$$\frac{1}{n\log(p_n)} = \frac{1}{p_n}-\frac{1}{n\left(\log(p_n)\right)^2} - \frac{2}{n\left(\log(p_n)\right)^3} - \frac{6}{n\left(\log(p_n)\right)^4} + o\left(\frac{1}{n\left(\log(p_n)\right)^4}\right)$$

Finally:

$$\frac{\log(p_n)}{p_n} = \frac{1}{n}+\frac{1}{p_n} + \frac{1}{n\left(\log(p_n)\right)^2} + \frac{4}{n\left(\log(p_n)\right)^3} + o\left(\frac{1}{n\left(\log(p_n)\right)^3}\right) \tag{4}$$

Summing that, we have :

$$\sum_{\substack{p_n \in \mathbb{P}\\ p_n\le x}}{\frac{\log(p)}{p}} = \sum_{n=1}^{\pi(x)}{\frac{1}{n}} + \sum_{\substack{p_n \in \mathbb{P}\\ p_n\le x}}{\frac{1}{p_n}} + O(1)$$

Which gives, using appropriate asymptotic expansions:

$$\sum_{\substack{p_n \in \mathbb{P}\\ p_n\le x}}{\frac{\log(p)}{p}} = \log(\pi(x)) + \log(\log(x)) + O(1)$$

The problem here is the $$\log(\pi(x))$$ as the asymptotic expansion of $$\pi(x)$$ is quite bad for my use.

Do you see a way to improve this bound ? Or another way to express it ?

• The Mertens theorem gives $\log x + O(1)$, you need the PNT and a summation by parts to improve it to $\log x +C+o(1)$ where $C=0$ (I think) is found from $\sum_{p^k \le x} \frac{\log p}{p^k} = \log x+o(1)$ Dec 7, 2022 at 18:42
• Thanks for your reply, I didn't know the Mertens theorem. Dec 8, 2022 at 10:07
• However, regarding the C, I don't know how to go that far, I have: $\sum_{2\le n \le x}{\frac{1-\alpha(n)\log(n)}{n}} = \frac{\lfloor x\rfloor - \vartheta(x)}{x} - \int_{2}^{x} \frac{\lfloor x\rfloor - \vartheta(x)}{t^2}dt$ But then we only have $\vartheta(x) = x + O\left(\frac{x}{\log(x)}\right)$ which doesn't seems enough. Dec 8, 2022 at 10:19

The PNT in the form $$\sum_{n=1}^N \Lambda(n) = N+O(\frac{N}{\log^2 N})$$ and a summation by parts gives that $$\sum_{p^k\le x} \frac{\log p}{p^k} = \frac{\sum_{n\le x}\Lambda(n)}{x} + \sum_{n\le x-1} (\sum_{m\le n} \Lambda(m))(\frac1n-\frac1{(n+1)})$$ $$= 1+o(1)+\sum_{n\le x-1} (n+O(\frac{n}{\log^2 n}))(\frac1{n^2}+O(\frac1{n^3}))= \log x + C+ o(1)$$

$$\int_1^\infty x^{-s-1}dx = 1/s$$, differentiating wrt $$s$$ gives $$\int_1^\infty \log(x) x^{-s-1}dx = 1/s^2$$.

Then write $$\int_1^\infty (C+o(1))x^{-s-1}dx= \int_1^\infty (\sum_{p^k\le x} \frac{\log p}{p^k}-\log x)x^{-s-1}dx = \frac1s \frac{-\zeta'(s+1)}{\zeta(s+1)}-\frac1{s^2}$$ From that $$\zeta(s)=\frac1{s-1}-\gamma+O(s-1)$$ at $$s=1$$ we know that $$\frac{-\zeta'(s)}{\zeta(s)}= \frac{1}{s-1}-\gamma+O(s-1)$$ as $$s\to 1^+$$ so that $$\frac1s \frac{-\zeta'(s+1)}{\zeta(s+1)}-\frac1{s^2}- \frac{\gamma}s$$ is bounded as $$s\to 0^+$$.

This implies that $$C=-\gamma$$.

Finally $$\sum_{p\le x} \frac{\log p}{p-1}=o(1)+\sum_{p^k\le x} \frac{\log p}{p^k}= \log x-\gamma+o(1)$$

It can be shown that improving $$\log x+O(1)$$ (the Mertens theorem) to $$\log x + C+o(1)$$ is equivalent to the PNT, so there is no simpler way.

• Ok, I understand what you meaned now. Thank you very much ^^ Dec 8, 2022 at 12:20
• The expansion of $-\zeta’(s+1)/\zeta(s+1)$ near origin should be $\frac1s-\gamma+O(|s|)$ instead of what you had. Dec 10, 2022 at 9:14
• I see, that's why I had $-\gamma$ here : Dec 12, 2022 at 9:50
• As a constant term appearing here when summing by part : $\sum_{n\le x}{\frac{\left(\pi(n)-\pi(n-1)\right)\log(n)}{n}} - \log(x) = \sum_{n\le x}{\frac{\left(\pi(n)-\pi(n-1) -\frac{1}{\log(n)}\right)\log(n)}{n}} - \gamma + o(1)$ Dec 12, 2022 at 9:57