How to prove this ? $$-\frac{1}{2} = \lim_{x\to+\infty}\sum_{n=1}^{\infty}(-1)^n \frac{x^{2n-1}}{(2n)! \sqrt{\ln 2n}}$$
It reminded me of the fact that
$$-\frac\pi2 = \lim_{x\to+\infty}\sum_{n=1}^{\infty}(-1)^n \frac{x^{2n-1}}{(2n)! \ln 2n}$$
that has been proven here :
Some partial results seem to indicate that the suggested limit is not $-1/2$. From numerical simulations with $x$ increasing up to 1 million or so, I observed that the sum exhibits oscillations which increase in amplitude and decrease in frequency w.r.t. to $x$. This would suggest that the integral diverges. However, I seem to be unable to find a proof for this, as my initial plan to evaluate the integral with the stationary phase method fails. Hence, at the moment all this remains as speculations.
As the sum is difficult to numerically calculate even with modest $x$, one instead has by the Lindelöf summation formula
$$ \sum_{n=m}^{\infty} (-1)f(n) = (-1)^m \int_{-\infty}^{\infty} f\left(m - \frac{1}{2}+iy\right) \frac{dy}{\cosh(\pi y)} $$
which in this case is nicer to work with, atleast numerically. Here, with
$$S(x) = \sum_{n=1}^{\infty}(-1)^n f(n), \quad f(n) = \frac{x^{2n-1}}{\Gamma(2n+1)\sqrt{\ln(2n)}}$$
one may write
$$ S(x) = f(1) + \sum_{n=2}^{\infty}(-1)^n f(n) = -\frac{x}{2\sqrt{\ln(2)}} + \frac{x^2}{2}\int_{-\infty}^{\infty} \frac{x^{2iy}}{\Gamma(4+2iy)\sqrt{\ln(3+2iy)}}\frac{dy}{\cosh(\pi y)} $$
where I have drawn out $n=1$ to avoid issues with the natural logarithm in the denominator in cases where $y=0$. Now, the function within the integral is even w.r.t. to the real part and odd w.r.t. to the imaginary part. Hence, we have
$$ S(x) = -\frac{x}{2\sqrt{\ln(2)}} + x^2 \Re \int_{0}^{\infty} \frac{e^{2iy\ln(x)}}{\Gamma(4+2iy)\sqrt{\ln(3+2iy)}}\frac{dy}{\cosh(\pi y)} $$
As a sanity check, e.g. letting $x=10$ gives that the original sum equals $-0.5209$ whereas numerical integration gives $-0.5208$. Calculating the the value of $S(x)$ for $x$ ranging between $0$ and $1000$ gives the result below.
For the next part, my initial assumption was to use the stationary phase method, but I seem to have hoped for too much as the real part of the integral with the method evaluates each term to 0.
This is seen by Theorem 13.2 in Olver's book Asymptotics and Special Functions which gives
$$ \int_{0}^{\infty} e^{2iy\ln(x)} q(y) dy \sim -\lim_{y\to 0+} \left(q(y)\right) \cdot \frac{1}{2i\ln(x)} = \frac{i\ln^{-1}(x)}{12\sqrt{\ln(3)}} $$
where $q(y) = \left(\Gamma(4+2iy)\sqrt{\ln(3+2iy)}\cosh(\pi y)\right)^{-1}$. Hence, the real part is $0$. Using partial integrations to calculate additional terms doesn't help either, as all terms will be imaginary.
