# Uniform upper bound for the sum over primes $\sum_{p \leq x} p^{-1+\varepsilon}$

I am reading the article D. M. Gordon and C. Pomerance, The distribution of Lucas and elliptic pseudoprime, Math. Comp. (1991) (click). In equation (27) the authors, apparently, used the following upper bound for a sum over the prime numbers: $$(\star)\quad\sum_{p \leq x} p^{-1+\varepsilon} \ll \frac{x^{\varepsilon}}{\varepsilon\log x} ,$$ for sufficiently large $x > 0$ and sufficiently small $\varepsilon > 0$, where the implicit constant in $\ll$ is absolute. (Precisely, $\varepsilon = \frac{4+\log\log\log x}{\log\log x}$, but I hope that this is not really important.) I am trying to prove ($\star$) but since now I have failed. Clearly, one way can be using partial summation and the prime number theorem, for example $$\sum_{p \leq x} p^{-1+\varepsilon} = \pi(x) x^{-1+\varepsilon} + (1-\varepsilon)\int_2^x \pi(t)t^{-2+\varepsilon}dt ,$$ but then I am now been able to prove the claim. I also throught that $$\sum_{p \leq x} p^{-1+\varepsilon} \leq \sum_{n \leq \pi(x)} n^{-1+\varepsilon} \ll \int_0^{\pi(x)} t^{-1+\varepsilon}dt = \frac{\pi(x)^\varepsilon}{\varepsilon} \ll \frac{x^\varepsilon}{\varepsilon (\log x)^\varepsilon} ,$$ but his is too weak.

Thank you in advance for any suggestion.

• Could you be more precise whether your "sufficiently large" $x$ and "sufficiently small" $\epsilon$ depend on each other? And how? – Fan Zheng Apr 23 '15 at 21:15
• @FanZheng There exists $x_0 > 0$ and $\varepsilon_0 > 0$ such that for any $x > x_0$ and $\varepsilon < \varepsilon_0$ the claim holds. – user40023 Apr 24 '15 at 7:35

The left hand side of $(\star)$ is at least $$\sum_{p\leq x}p^{-1}=\log\log x +O(1),$$ hence a necessary condition for the truth of $(\star)$ is that $$\frac{x^{\varepsilon}}{\varepsilon\log x} \gg \log\log x.$$ This condition is also sufficient, in the light of the following bound that I prove below: $$(\star\star)\quad\sum_{p \leq x} (p^{-1+\varepsilon} - p^{-1})\ll \frac{x^{\varepsilon}}{\varepsilon\log x}.$$ In particular, $(\star)$ is valid for $\varepsilon>(\log\log\log x + \log\log\log\log x)/ \log x$, since in this case $$\frac{x^{\varepsilon}}{\varepsilon\log x} > \frac{(\log\log x)(\log\log\log x)}{\log\log\log x + \log\log\log\log x} \gg \log\log x.$$

Now let me prove $(\star\star)$: $$\sum_{p \leq x} (p^{-1+\varepsilon} - p^{-1}) \ll \sum_{p\leq e^{1/\varepsilon}}\frac{\varepsilon\log p}{p}+\int_{e^{1/\varepsilon}}^x t^{-1+\varepsilon}\,d\pi(t)$$ $$\ll 1+\frac{x^\varepsilon}{\log x}+\int_{e^{1/\varepsilon}}^x \frac{t^{-1+\varepsilon}}{\log t}\,dt = 1+\frac{x^\varepsilon}{\log x}+\int_1^{\varepsilon\log x}\frac{e^t}{t}dt\,\ll\frac{e^{\varepsilon\log x}}{\varepsilon\log x}=\frac{x^\varepsilon}{\varepsilon\log x}.$$

• Are you drowning?:) – Fan Zheng Apr 23 '15 at 23:53
• Thank you very much for you proof, it was exactly what I was looking about. – user40023 Apr 24 '15 at 7:37

The main (and only) theorem in the said paper by T. Salát and S. Znám is the following one:

For any $a>0,$ let us denote the sum $\sum_{p \leq x} p^{a}$ with $S_{a}(x)$. Then, we have that

$$\lim_{x \to \infty} \frac{S_{a}(x) \cdot\log x}{x^{1+a}} = \frac{1}{1+a}.$$

Their proof depends on the Prime Number Theorem. In case you want to take a look at the paper, you can find it here: http://goo.gl/kUO6Zt

Last but not least, note that the late P. S. Bruckman may have had a different proof of this result:

https://www.siam.org/journals/categories/08-005.php

• No one reads any more :( – Igor Rivin Apr 24 '15 at 13:47

This seems to be due to Salat and Znam, 1968. The link is to a review, I have no idea how to find the actual paper.The result of Prachar cited in the review may be easier to find.

This is a direct consequence of the prime number theorem (or weaker estimates on $\pi(x)$): $$\sum_{p\le x} p^{-1+\epsilon} = \int_1^x t^{-1+\epsilon}\, d\pi(t) =\pi(x)x^{-1+\epsilon}+(1-\epsilon)\int_1^x t^{-2+\epsilon}\pi(t)\, dt ,$$ by an integration by parts. Now use that $\pi(x)\lesssim x/\log x$. The first term is already of the desired order. Split the integral into two parts, according to $t\le x^{\epsilon}/\log x$ and $t$ larger than this bound. Again, the first part is clearly $\lesssim x^{\epsilon}/\log x$. On the second interval, we have that $\log t\gtrsim \epsilon \log x$, which produces the desired bound after integrating.

• But is the estimate uniform in $\epsilon$, e.g., in the second part $\log t\ge \epsilon\log x-\log\log x$. Why is it $\gg \epsilon\log x$ with a uniform constant? – Fan Zheng Apr 23 '15 at 21:14
• @FanZheng: It sure is; $\log\log x\le (\epsilon/2)\log x$, say. – Christian Remling Apr 23 '15 at 21:17
• Maybe I'm confused, but that's true only in some sense of the two "sufficiently" above. In other words, does the "sufficient largeness" of $x$ depend on $\epsilon$? – Fan Zheng Apr 23 '15 at 21:21
• @FanZheng: Yes, I guess that's what I show: $\limsup S_{\epsilon}(x)x^{-\epsilon}\log x\le C/\epsilon$, and this $C$ doesn't depend on anything. – Christian Remling Apr 23 '15 at 23:10
• Definitely I'm in great shape with the OP's extra specification which makes $\log\log x/\epsilon$ small. – Christian Remling Apr 23 '15 at 23:21