How many terms are required to get $D$ digits of Riemann zeta prime function $\zeta_p(s) = \sum_p \frac{1}{p^s}$?
Sebah & Gourdon mentions that finding $\zeta_p(2)$ to 20 digits by using $\sum_p \frac{1}{p^2}$ directly would require computing all the primes upto $10^{20}$. So how did they get that?
From the prime number theorem, the density of primes around $n$ is asymptotically $1/\ln n$. So the convergence properties of the given series should be similar to those for $$\sum_n\frac1{n^s\ln n},$$ (summed over all natural numbers) and in particular, a reasonable approximation to the tail of the series will be the integral $$\int_N^\infty\frac{dx}{x^s\ln x}.$$ Computing reasonable asymptotics for this integral is left as an exercise. ;-)
Edit: An upper bound is clearly $$\frac1{\ln N}\int_N^\infty\frac {dx}{x^s}=\frac{s-1}{N^{s-1}\ln N},$$ and this is probably not a bad estimate for $s=2$ since $x^2$ grows a lot faster than $\ln x$ does.
Now, to get accuracy $\varepsilon=10^{-20}$ for $s=2$, you need to solve $N\ln N=1/\varepsilon=10^{20}$.