Seeking explicit bounds on the error term for the reciprocal product over primes in Mertens Third Theorem

Question

The reciprocal expression of Mertens Third Theorem should give us the following limit. $$\lim_{x\to \infty}\Pi_{p \leq x} (1-\frac{1}{p})^{-1} = e^\gamma \;log(x)$$ I would be interested in learning of any known explicit upper and lower bounds on the "error term" (not sure if "error term" is the correct terminology here) as a function of $x$. Specifically, given $$y=\Pi_{p \leq x} (1-\frac{1}{p})^{-1} - e^\gamma \;log(x)$$ what are the tightest known upper and lower bounds on the range of $y$ at $x$? Additionally, is anything known concerning if and how many times $y$ changes sign? Thanks in advance.

Edit:
I appreciate the quick responses on MSE. But I feel I should clarify what it is I'm hoping for.

1) A concrete inequality for each bound. As an analogy or example, Robin gave the following upper bound governing his own famous inequality:

$$\sigma(n)-e^{\gamma}nloglogn< \frac {0.6483n}{loglogn}$$

so that given $n$ one can quickly calculate a maximum value for the difference. Someone certainly must have made a similar computation for this Mertens asymptote.

2) I think I may have read somewhere that it is not even known whether $y$ in the original question changes its sign at all or whether it remains positive for all $x$. I have been unable to find that source again and was wondering if that is correct.

2nd edit:
After running a program to get a feel for this asymptote, I would just like to include the results here in order to share how quickly and nicely the product appears to converge to $e^\gamma \;log(x)$.

The five columns below are as follows:

1) $k$

2) $k$th prime

3) $\Pi_{p \leq p_k} (1-\frac{1}{p})^{-1}$

4) $ e^\gamma \;log(p_k)$

5) $\Pi_{p \leq p_k} (1-\frac{1}{p})^{-1}- e^\gamma \;log(p_k)$

2    3         3                   1.95671             1.04329             
3    5         3.75                2.86653             0.883475            
4    7         4.375               3.46581             0.909193            
5    11        4.8125              4.27083             0.541675            
6    13        5.21354             4.56836             0.645181            
7    17        5.53939             5.04616             0.49323             
8    19        5.84713             5.24426             0.602873            
9    23        6.11291             5.58454             0.528368            
10   29        6.33123             5.9974              0.333831            
11   31        6.54227             6.11618             0.42609             
12   37        6.724               6.43131             0.292693            
13   41        6.8921              6.61414             0.277959  
....
2086 18211     17.4867             17.4719             0.014736            
2087 18217     17.4876             17.4725             0.0151092           
2088 18223     17.4886             17.4731             0.0154824           
2089 18229     17.4895             17.4737             0.0158555           
2090 18233     17.4905             17.4741             0.016424            
2091 18251     17.4915             17.4758             0.015625            
2092 18253     17.4924             17.476              0.0163881           
2093 18257     17.4934             17.4764             0.016956            
2094 18269     17.4943             17.4776             0.0167433           
2095 18287     17.4953             17.4793             0.0159461   
.....
5091 49547     19.2618             19.2546             0.00720897          
5092 49549     19.2622             19.2547             0.00752583          
5093 49559     19.2626             19.255              0.00755509          
5094 49597     19.263              19.2564             0.00657834          
5095 49603     19.2634             19.2566             0.00675124          
5096 49613     19.2638             19.257              0.00678049          
5097 49627     19.2641             19.2575             0.00666615          
5098 49633     19.2645             19.2577             0.00683897  

The main question asks: Is there a known upper bound on the 5th column strictly in terms of the 2nd column, much in the same fashion as Robin's bound given above?

3rd edit
Upon snooping around on MO, I came across this posting:
https://mathoverflow.net/questions/249147/mertens-3rd-theorem-upper-bound

With just a little reworking of the answer provided, it seems we have the following explicit lower bound from Rosser and Schoenfeld, thereby answering one part of the original question: $$-\frac{e^\gamma}{2\ln x}<\ \prod_{p\le x}\frac p{p-1}-e^\gamma\ln x\ <\quad?$$

Obviously the question mark is my own as I am still seeking an explicit upper bound on the RHS. One would suspect it approaches zero far faster than the lower bound by considering the above table of values.

I'll try to gain access to the Rosser-Schoenfeld and Diamond/Pintz material, and if I come across an explicit upper bound will share it here.

Answer 1

Mertens' theorem states that $\prod_{p \leq x} (1- 1/p) = \frac{e^{- \gamma}}{log x}(1+ O(\frac{1}{log x}))$. By taking the reciprocal on both sides,

$\prod_{p \leq x} (1- 1/p)^{-1} = e^{\gamma}(log x)(1+ O(\frac{1}{log x}))$. As $x \rightarrow \infty$, you have the result stated in your post. Now,

$\prod_{p \leq x} (1- 1/p)^{-1} -e^{\gamma}log (x) = e^{\gamma}log (x)O(\frac{1}{log (x)})=O(1)$ as $ \gamma =0.577215...$, which is the Euler - Mascheroni constant.

Answer 2

Also the error term encodes the Riemann hypothesis. The first step is $\log(\prod_{p\le x} (1-p^{-1})) = -\sum_{p\le x}p^{-1}+B+O(x^{-1/2})$. Then

$\sum_{p\le x} p^{-1}=\log \log x+M+O(x^{\sigma-1+\epsilon}) \tag{1}$ iff $\zeta(s)$ has no zero for $\Re(s)>\sigma$.

This is because $(1)$ is equivalent to $$\sum_{p\in [x,y]}p^{-1}-\sum_{n\in [x,y]} \frac1{n\log n} = O(x^{\sigma-1+\epsilon})\tag{2}$$ which implies that $$\sum_p p^{-1-s}-\sum_{n\ge 2} \frac{n^{-1-s}}{\log n}=P(s+1)+\int (\zeta(s+1)-1)$$ converges for $\Re(s) > \sigma-1$. Thus, $P(s)+\int ( \zeta(s)-1)$ is analytic for $\Re(s) >\sigma$, since $P(s)=\sum_k \frac{\mu(k)}{k}\log \zeta(ks)$ has a singularity at each non-trivial zero of $\zeta(s)$ of real part $\ge 1/2$ it means that $\zeta(s)$ has no zero for $\Re(s) >\sigma$.

The converse, that if $\zeta(s)$ has no zero for $\Re(s) >\sigma$ then $\sum_p p^{-1-s}-\sum_n \frac{n^{-1-s}}{\log n}$ converges for $\Re(s) > \sigma-1$, is a much deeper tauberian theorem, its proof follows the same lines as in the prime number theorem.

The convergence of $\sum_p p^{-1-s}-\sum_n \frac{n^{-1-s}}{\log n}$ implies $(2)$ and $(1)$.