Rate of convergence of the prime zeta function P(2)

Question

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 $$c_1f(x)<\sum_{p>x}\frac{1}{p^2}<c_2f(x)?$$

Answer 1

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 $1/(x\log x)$.

Answer 2

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.)