It's prime time

Solution:

I'm unsure as to whether there are better solutions, but here is the solution I came up with:

Take the Euler product form of the Riemann Zeta function for $\Re (s)>1$ $\zeta (s)=\prod_{n=1}^{\infty} \frac1{1-P_{n}^{-s}}$ Taking the natural log of this and differentiating with respect to $s$ brings us to $\frac{\zeta '(s)}{\zeta (s)}=-\sum_{n=1}^{\infty} \frac{P_{n}^{-s}\ln P_n}{1-P_{n}^{-s}}$ which is easily seen to be equivalent to the following equation by multiplying each term in the series by $\frac{P_n^s}{P_n^s}$ $-\frac{\zeta '(s)}{\zeta (s)}=\sum_{n=1}^{\infty} \frac{\ln P_n}{P_n^s-1}$ I eluded to this solution in the answer choices with $\color{#D61F06} \frac{\zeta '(2)}{\zeta (2)}$ in order to make the solution less abstract for some observant people, but this answer is actually incorrect due to the sign being incorrect (it should be $\color{#3D99F6} -\frac{\zeta '(2)}{\zeta (2)}$ ). From this equation you can get to an easy solution for the answer by putting in $s=2$ , however none of the answers in the choices fit the form. In order to correct this, the value for $\zeta '(2)=\zeta (2) \times \left[ \ln{2\pi}+\gamma - 12\ln{A} \right]$ must be substituted in (for which I don't expect anyone to be able to remember $\zeta '(2)$ ) of the top of their heads, I had to do some searching for this result). This brings the result of: $\boxed{L= \color{#20A900} 12\ln{A}-\gamma-\ln{2\pi}}$ END OF SOLUTION

Extra/Irrelevant information:

In decimal notation this means $L\approx 0.56996099\dots$ . By looking at the partial sums of the equation you can see that this series does not converge too quickly, for example if we define a sequence of partial sums by $L_x=\sum_{n=1}^{x} \frac{\ln P_n}{P_n^2-1}, \quad x \in \mathbb{N}$ and then look at a reasonably large value, for example $L_{1000} \approx 0.56983533\dots$ , which is only correct to 3dp.

Obviously this is a side effect of the general result $-\frac{\zeta '(s)}{\zeta (s)}=\sum_{n=1}^{\infty} \frac{\ln P_n}{P_n^s-1}, \quad \Re (s)>1$ which means I technically found infinity results to extremely similar series, it just happens that the only value I could find for $\zeta '(s)$ with $\Re (s)>1$ was $\zeta '(2)$ and so the problem for $\sum_{n=1}^{\infty} \frac{\ln P_n}{P_{n}^{2}-1}$ is the one I settled with.

I don’t know if this problem is anywhere else on this website or the internet, but if it does and someone has a more elegant solution than mine I would love to see it.