Approximating $\sum_{p\in\Bbb P} \frac1{p\ln p}$ It's known that the sum
$$\sum_{p\in\Bbb P} \frac1{p\ln p} \approx 1.6366$$
converges approximately to the indicated value, see here for example.
How is this approximation calculated?  The series converges painfully slow.  For example, when naively summing up to primes below $10^7$, the sum is $0.06$ off the real value (assuming all digits of "$1.6366$" are correct), whereas the summands are of size around $10^{-8}$:
sum(10000000) = 1.5745747442808875 , delta = 6.20e-09
So what's the technique / transformation used to speed up the computation and to keep rounding errors at bay?

FYI, here is the ad-hoc Sage script which produced the line above:
#!/usr/bin/env sage

def f(x):
    return float(1 / (p * log(p)))

bound = 1e7

s = 0
for p in Primes():
    if p > bound:
        break
    s += f(p)

print ("sum(%.0f) =" % bound, s, ", delta = %.2e" % f(bound))


 A: This is just a rough idea, as I'm no expert on your series. We want$$\sum_{p\in\Bbb P}\frac{1}{p\ln p}=\sum_{p\in\Bbb p,\,p\le n}\frac{1}{p\ln p}+\sum_{p\in\Bbb p,\,p>n}\frac{1}{p\ln p}.$$The last sum can be estimated with an integral by the prime number theorem. The crudest estimate is$$\sum_{p\in\Bbb p,\,p>n}\frac{1}{p\ln p}\approx\int_n^\infty\frac{\pi(p)dp}{p^2\ln p}\approx\int_n^\infty\frac{dp}{p\ln^2p}=\frac{1}{\ln n}.$$You already get the given decimal places right with $n=10^4$, as this Python shows:
from math import log
def primes_up_to(n):
    statuses = [False]*2+[True]*(n-1)
    k = 2
    while k*k<=n:
        for i in range(2*k, n+1, k): statuses[i] = False
        k += 1
        while not statuses[k]: k += 1
    return [p for p, b in enumerate(statuses) if b]
#All of the k below get 1.6366
for k in range(4, 8):
    n = 10**k
    print(sum(1/(p*log(p)) for p in primes_up_to(n))+1/log(n))

This first stab can be further improved with (i) estimating the sum as an integral plus other stuff and (ii) estimating $\pi(p)$ more precisely than as $\frac{p}{\ln p}$.
