Evaluating the limit: $\lim\limits_{x \to \infty} \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{x^{2n-1}}{(2n)! \log (2n)}$ 
How to evaluate the limit: 
  $$\lim\limits_{x \to \infty} \sum\limits_{n=1}^{\infty}  (-1)^{n-1}\frac{x^{2n-1}}{(2n)!\log (2n)}$$

I have tried to break the sum into even and odd $n$ and estimate each sum to be $\sim \frac{e^{x}}{\log (4x)}$ by estimating the summations in the range $\sum\limits_{x/c \le n \le cx} \frac{x^{4n}}{(4n+s)!\log (4n+s)}$ with the trivial bounds while $e^{-x}\sum\limits_{n > cx} \frac{x^{4n}}{(4n+s)!\log (4n+s)}$ and $e^{-x}\sum\limits_{0 \le n < x/c} \frac{x^{4n}}{(4n+s)!\log (4n+s)}$ both decay exponentially for arbitrary $c > 1$ as $x \to +\infty$ ($s=2,4$). The precise estimates I have used are similar to here.
However, I suppose stronger asymptotics will be required for calculating the limit of their difference. Any help/hint is appreciated. Thanks!
I am not much acquainted with the Laplace method or probabilistic interpretations (I'd appreciate it some references are mentioned, should the answer involve advanced tools like them.)

Machinato suggests that $f(z)=\sum_{n=2}^{\infty} \frac{(-1)^n z^{n-1}}{n! \log n}$ approaches $\log z$. This seems to be true! As evidence, here are plots of the real (red) and imaginary (blue) parts of $f(e^{2+i \theta})$, for $-\pi<\theta<\pi$. For comparison, the red and blue dashed lines are the real and imaginary parts of $\log z = 2+i \theta$. It's not clear from this image whether the limit holds for all $\theta$, but for $-\pi/2 < \theta < \pi/2$ the fit is very good. 

I'm offering a bounty for a proof of this bizarre behavior. I imagine the precise formulation is that, for $\theta$ fixed in $(-\pi,\pi)$, we have $\lim_{R \to \infty} f(R e^{i \theta}) - \log R = i \theta$ but I'll certain accept other asymptotic results of a similar flavor. (For $\theta = \pi$, it is easy to see that $f(-R)$ grows faster than any power of $R$, so it can't mimic $\log R$.)
After some more experimentation, I suspect the limit only holds for $-\pi/2<\theta<\pi/2$. Here are plots of $\mathrm{Im}(f(r e^{i \theta}))$ for $\theta = \pi/3$ (first plot) and $2 \pi/3$ (second plot), with $0 < r < 10$. In each case, the dashed line is at $\theta$. Convergence looks great in the first one, terrible in the second.


 A: EDITED. It suffices to prove the following claim.

Proposition. Assume that $f(z) = \sum_{n=2}^{\infty} a_n z^n$ has radius of convergence $R$ and define
$$ F(z)
= \sum_{n=2}^{\infty} \frac{n a_n}{n-1} z^n
= z \int_{0}^{z} \frac{f'(w)}{w} \, dw. $$
Then for all $|z| < R$, we have
$$ \sum_{n=2}^{\infty} \frac{a_n}{\log n} z^n
= \int_{0}^{1} \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} F(z e^{-t}) \, dtds. $$

Proof. The proof is a straightforward computation:
\begin{align*}
\int_{0}^{1} \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} F(z e^{-t}) \, dtds
&= \sum_{n=2}^{\infty} \frac{n a_n}{n-1} z^n \int_{0}^{1} \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} e^{-nt} \, dtds \\
&= \sum_{n=2}^{\infty} \frac{n a_n}{n-1} z^n \int_{0}^{1} \frac{ds}{n^s} \\
&= \sum_{n=2}^{\infty} \frac{a_n}{\log n} z^n.
\end{align*}

In our case, we can set $f(z) = 1 - \cos z$ and consequently $F(z) = z \operatorname{Si}(z)$, where $\operatorname{Si}(z)$ is the sine integral. Then for real $x$, we have
\begin{align*}
\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n)!\log(2n)} x^{2n-1}
&= \int_{0}^{1} \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} e^{-t} \operatorname{Si}(x e^{-t}) \, dtds \\
&\hspace{3em} \xrightarrow[x\to\infty]{} \quad
\frac{\pi}{2} \int_{0}^{1}\int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} e^{-t} \, dtds
= \frac{\pi}{2}
\end{align*}
where the limiting procedure can be easily justified by the dominated convergence theorem.
Similarly, setting $f(z) = e^{-z} - 1 + z$ gives $F(z) = z (\log z + \gamma + \Gamma(0,z))$ for $\operatorname{Re}(z) > 0$, where $\Gamma(s, z)$ is the upper incomplete gamma function. Plugging this back, we obtain
$$ \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n!\log n} z^{n-1}
= \log z + \gamma - \frac{1}{2} + \int_{0}^{1} \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} e^{-t} \Gamma(0, z e^{-t}) \, dtds. $$
Note that the last integral vanishes if we let $z \to \infty$ along the cone $|\arg(z)| \leq \frac{\pi}{2} - \delta$. So this again confirms numerical observation made by Machinato.
