$ C_{MRB}=\sum _{x=1}^{\infty } (-1)^x\left(e^{\frac{\log x}{x}}-1\right)$ Is its absolutely convergent neighbor prime number theorem related?
$ C_{MRB}=\sum _{x=1}^{\infty } (-1)^x\left(e^{\frac{\log x}{x}}-1\right)=\sum _{x=1}^{\infty } (-1)^x \left(e^{\frac{\log (x)}{x}}-\frac{\log (x)}{x}-1\right)+\frac{1}{2} \left(2 \gamma \log (2)-\log ^2(2)\right).$
where $ C_{MRB}$ is the MRB constant and $\gamma$ is Euler's Gamma constant. Is that equivalence of the two series prime number theorem related?
I've been studying the series,
$C_{MRB}=\sum _{x=1}^{\infty } (-1)^x\left(e^{\frac{\log x}{x}}-1\right)$
It's all well and good, but I never liked the fact that it is only conditionally convergent. To my relieve the Late Richard Crandall discovered a way of converting it to an absolutely convergent one: $$ C_{MRB}=\sum _{x=1}^{\infty } (-1)^x \left(e^{\frac{\log (x)}{x}}-\frac{\log (x)}{x}-1\right)+\frac{1}{2} \left(2 \gamma \log (2)-\log ^2(2)\right) $$
I can tell that the new series is absolutely convergent because the series of its sequences' absolute values converges. So that is not my question.
I might be overthinking it, but I wonder if he had the prime counting function in mind when he derived the absolutely convergent series? Or would that be useful in further study about the MRB constant? An GPT-4 search produced the phrase "Abscissa of absolute convergence."
The reason for my madness is that the prime counting function is as x-> infinity, $\pi( x)=\frac{x}{log(x)}$. And the term he added to the continually convergent series is its inverse, which coincidental (or not) is also the exponent on $e.$