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For an application in statistical group theory, we need explicit upper and lower bounds that an expert in number theory (I am not one) may know how to prove.

Question 1: What are "good" bounds $f_1(x)<\displaystyle\sum_{p>x}\frac{1}{p^2}<f_2(x)$ where $p>x$ is prime?

For our application, sharper bounds than the following $f_1(x),f_2(x)$ are desirable: $$ \frac{1}{12\left(\frac{x}{\log(x)-4}+1\right)^4}<\sum_{p>x}\frac{1}{p^2}<\frac{1}{x-1}.$$ The lower bound can hold for $x>x_1$ and the upper bound for $x>x_2$ provided $x_1,x_2$ are "small".

Question 2: Is there a function $f(x)$ and explicit positive constants $c_1,c_2$ such that $$c_1f(x)<\sum_{p>x}\frac{1}{p^2}<c_2f(x)?$$

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Indeed, this is a type of question that is standard to those in the know (pretty much exactly what MathOverflow is for!). The famous paper of Rosser and Schoenfeld contains many estimates for $\pi(x)$, the number of primes up to $x$: for example, their Corollary 1 tells us that $$ \frac x{\log x} < \pi(x) < 1.25506 \frac x{\log x} \text{ for } x\ge17. $$ Bounds for your sum (for $x\ge17$ in this case) can then be found using partial summation. For example, for the lower bound, \begin{equation*} \begin{split} \sum_{p>x} \frac1{p^2} &= \int_x^\infty \frac1{t^2} \,d\pi(t) \\ &= \frac{\pi(t)}{t^2}\bigg|_x^\infty + \int_x^\infty \pi(t) \frac2{t^3} \,dt \\ &> \frac1{t\log t}\bigg|_x^\infty + \int_x^\infty \frac2{t^2\log t} \, dt \\ &= -\frac1{x\log x} - 2\mathop{\rm li}\bigg( \frac1x \bigg), \end{split} \end{equation*} where $\mathop{\rm li}(t) = \int_0^t dt/\log t$ is the logarithmic integral function. [The second expression on the top line is a Riemann–Stieltjes integral; if you're not familiar with those, one can verify by hand that the sum equals the expression on the second line.) This last expression can be shown to be asymptotic to $1/(x\log x)$ for large $x$.

The calculation of the upper bound is exactly the same except with an extra factor of $1.25506$. Indeed, a more careful calculation (using sharper bounds from Rosser and Schoenfeld from earlier on the same page) will give both upper and lower bounds that are asymptotic to $1/(x\log x)$.

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  • $\begingroup$ The last displayed line has sign errors. $\endgroup$ – Glasby May 19 '19 at 11:45
  • $\begingroup$ @Glasby Two errors fixed, thank you! $\endgroup$ – Greg Martin May 19 '19 at 17:51
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An answer to Question 2 follows from the lemma below by letting $y\to\infty$.

Lemma. Suppose $x,y$ are real numbers with $12\leqslant x\leqslant y$ and $p$ denotes a prime. Then $$\sum_{x<p\leqslant y}\frac{1}{p^2}\leqslant\frac{2.22}{\lfloor x\rfloor\log\lfloor x\rfloor}-\frac{1.61}{\lfloor y\rfloor\log\lfloor y\rfloor}.$$

For an elementary proof using forward differences (and avoiding Riemann-Stieltjes integrals) see Lemma 6 of the preprint arXiv:1911.12613. A lower bound with a similar form (but with different constants) can be obtained using the same methods. As the lower bound for $x$ increases from 12, the constant 2.22 decreases to 1 as predicted by Greg Martin's answer. (For our application we needed explicit constants for an upper bound.)

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