Tight lower and upper estimate of alternating summation of reciprocals of divisors of primorial $\sum_{d \mid p_n\#} (-1)^{\omega(d)} \dfrac{1}{d}$?

Question

Let $f(n ) = \sum_{d \mid p_n\#}(-1)^{\omega(d)} \dfrac{1}{d}$ where $p_n$ is the $n$th prime number and $p_n\# = p_n p_{n-1} \cdots p_1$.

I'm looking for a way to make a simple estimate of (upper / lower bound) $f(n)$. Can it be done?

For example, $n = 3$ gives: $$f(3) = 1 - \dfrac{1}{2} - \dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{10} + \dfrac{1}{15} - \dfrac{1}{30} = 0.2\overline{6}$$

Here is some code that will print the first few values:

from sympy import *
import matplotlib.pyplot as plt
from functools import reduce

N = 500

x_points = []
y_points = []

def powerset(s):
    x = len(s)
    masks = [1 << i for i in range(x)]
    for i in range(1 << x):
        yield [ss for mask, ss in zip(masks, s) if i & mask]

for n in range(1, N):
    P = list(primerange(2, prime(n) + 1))
    Q = powerset(P)
    
    sum = 0
    
    for q in Q:
        if q:
            prod_q = reduce(lambda x,y: x*y, q)
            sum += (-1)**(len(q)) * (1 / prod_q)
        else:
            sum += 1
    print(n, sum)

Prints:

1 0.5
2 0.33333333333333337
3 0.2666666666666667
4 0.22857142857142865
5 0.20779220779220786
6 0.19180819180819186
7 0.1805253569959452
8 0.17102402241721126
9 0.16358819535559335
10 0.15794722310195236
11 0.15285215138898603
12 0.14872101216225675
13 0.14509367040220184
14 0.14171939899750013
15 0.1387040926358517
16 0.13608703428422966

As you can see this number apparently wants to go to $0$, but I would like estimates of its actual value, not what it converges to.

Answer 1

$f(n)$ is exactly equal to $\prod_{p\le p_n} \bigl( 1-\frac1p \bigr)$, where the product runs over primes $p$. By Mertens's (third) theorem, this product is asymptotic to $1/(e^\gamma\log p_n) \sim 1/(e^\gamma\log n)$, where $\gamma$ is the Euler–Mascheroni constant. Explicit upper and lower bounds can be found in Rosser & Schoenfeld's paper.