Expansion about $x=1$ for $\sum_{n=2}^\infty \frac{x^n}{n\log n}$ $$\sum_{n=2}^N \frac{1}{n\log n}$$
diverges as $N\rightarrow \infty$, because the integral $$\int_2^N \frac{{\rm d}t}{t\log t}=\log(\log N) - \log(\log 2)$$
diverges. The singularity is double logarithmic and I therefore expect the series
$$f(x)=\sum_{n=2}^\infty \frac{x^n}{n\log n}$$
to have a double logarithmic singularity at $x=1$ i.e.
$$f(x) \sim \log \left(-\log\left(1-x\right)\right)$$
as $x\rightarrow 1$. Is there a simple way to derive the expansion about $x=1$?
Since $\frac{x^n}{n\log n}$ is monotonic, it may be useful/effective to deploy Euler-Maclaurin Expansion i.e. calculate the integral
$$\int_2^\infty \frac{e^{n\log x}}{n\log n} \, {\rm d}n \, ,$$
but no anti-derivative seems to exist.

Ultimately this question is related to the existence of these types of integrals $$\int_2^\infty \frac{e^{-tx}}{\log x} \, {\rm d}x$$ for $t>0$ and $t\rightarrow 0$ in terms of (semi)-elementary functions. I found the following asymptotic expansion for any $a>0$
\begin{align}
\int_a^\infty \frac{e^{-xt}}{\log x} \, {\rm d}x &\stackrel{u=xt}{=} \frac{-1}{t\log t} \int_{at}^\infty \frac{e^{-u}}{1 - \frac{\log u}{\log t}} \, {\rm d}u \\
&= \frac{-1}{t\log t} \sum_{n=0}^\infty \frac{1}{\log^n t} \int_{at}^\infty e^{-u} \log^n u \, {\rm d}u \\
&\stackrel{t\rightarrow 0}{=} \frac{-1}{t\log t} \sum_{n=0}^\infty \frac{\Gamma^{(n)}(1)}{\log^n t} \, . \tag{1}
\end{align}
The limiting integral for $t\rightarrow 0$ converges and the error by extending the range to $0$ is ${\cal O}\left(t\log^n(t)\right)$. Therefore the asymptotic expansion follows.

Integrating the asymptotic expansion (1) with respect to $t$ then yields
$$\int_a^\infty \frac{e^{-xt}}{x\log x} \, {\rm d}x = \log(-\log(t)) - \sum_{n=1}^\infty \frac{\Gamma^{(n)}(1)}{n\log^n t} + C$$
and it can be shown $C=0$.
 A: This should really be a comment but its a bit too long.  I think i can get the double logarithmic singularity but its not the right form, so i'm doing something wrong.  But perhaps it will give you some ideas. Note we have
$$f\left( x \right)=\sum\limits_{n=2}^{\infty}{\frac{{{x}^{n}}}{n\log \left( n \right)}}=\sum\limits_{n=2}^{\infty }{\frac{{{x}^{n}}}{n}}\int\limits_{0}^{\infty}{{{e}^{-\log \left( n \right)t}}dt}=\sum\limits_{n=2}^{\infty}{\frac{{{x}^{n}}}{n}}\int\limits_{0}^{\infty }{{{n}^{-t}}dt}=\int\limits_{0}^{\infty }{-x+L{{i}_{1+t}}\left( x \right)dt}$$
Where $Li$ is the polylogarithm.  Break the integral into two
$$f\left( x \right)=\int\limits_{0}^{a}{-x+L{{i}_{1+t}}\left( x \right)dt}+\int\limits_{a}^{\infty }{-x+L{{i}_{1+t}}\left( x \right)dt}$$
The second can bounded for $a>0$ and so we’re left with something like
$$\lim f\left( x \right)=-1+\lim \int\limits_{0}^{a}{L{{i}_{1+t}}\left( x \right)dt}+C$$
for some constant $C$.  Using, for $x$ near $1$,
$$L{{i}_{1+t}}\left( x \right)=\Gamma \left( -t \right){{\left( -\log \left( x \right) \right)}^{t}}+\sum\limits_{k=0}^{\infty }{\frac{\zeta \left( t+1-k \right)}{k!}{{\log }^{k}}\left( x \right)}$$
And assume $a$ is very small and so expanding about $t=0$ we find
$$L{{i}_{1+t}}\left( x \right)=-\log \left( -\log \left( x \right) \right)+\sum\limits_{k=1}^{\infty }{\frac{\zeta \left( t+1-k \right)}{k!}{{\log }^{k}}\left( x \right)}+O\left( t \right)$$
And so we get something like
$$f\left( x \right)\sim-x-\log \left( -\log \left( x \right) \right)$$
for $x$ near $1$, which potentially shows it blows up like a double log, but this isnt asymptotic.  Maybe the right choice of scaling with the 'a' terminal would drag it out kicking and screaming.
