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
$$c\_1f(x)<\sum\_{p>x}\frac{1}{p^2}<c\_2f(x)?$$



Rate of convergence of the prime zeta function P(2)
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](https://projecteuclid.org/euclid.ijm/1255631807) 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](https://en.wikipedia.org/wiki/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 $1/(x\log x)$.


