The power series $\sum_{n=1}^\infty \ln(n)z^n$ has radius of convergence $1$ and $z=1$ is a singular point. Is $z=1$ an isolated singularity? If yes, what kind of isolated singularity?
I am only able to deduce that $z=1$ cannot be a pole.
Such type of questions appear naturally when one tries to relate the singularities of the power series and those of the Dirichlet series associated to the same sequence.
Let $f(z)$ be your function. Then $g(z)=f(z)(1-z)$ is equal to $$ (1-z)\sum_n \ln(n)z^n=\sum_{n\geq 2} (\ln(n)-\ln(n-1))z^n $$
Now, $\ln(n)-\ln(n-1)=\frac{1}{n}+O\left(\frac{1}{n^2}\right)$, which gives us $$ g(z)=\sum_{n≥1} \left(\frac{1}{n}+g_n\right)z^n, $$
where $g_n=\ln(n)-\ln(n-1)-\frac{1}{n}=O\left(\frac{1}{n^2}\right)$ for $n>1$ and $g_1=-1$, so that $$ f(z)=\frac{\ln(1-z)}{z-1}+\frac{h(z)}{z-1}. $$
Here $h(z)$ is holomorphic in the unit disc and continuous on its boundary. Hope this answers your question on the type of singularity in $z=1$.
Edit:
Let me also present a slightly less elementary way to study properties of this series, based on my favorite method of lots of contour integration. We will use the derivative of Riemann zeta-function, so this is more in the spirit of the question.
Let $x\in \mathbb C$ be a number with positive real part. Using the formula
$$ e^{-nx}=\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty} \Gamma(s)(nx)^{-s}ds, $$
we obtain
$$ f(e^{-x})=\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty}\Gamma(s)\zeta'(s)x^{-s}ds. $$
From this we easily get
$$ f(e^{-x})=\mathrm{Res}_{s=1}\,\Gamma(s)\zeta'(s)x^{-s}+\sum_{n\geq 0}\mathrm{Res}_{s=-n}\,\Gamma(s)\zeta'(s)x^{-s}. $$
The first summand is actually a bit different from all the other, because we get double pole. From expansions
$$ \Gamma(s)=1-\gamma(s-1)+O((s-1)^2), \zeta'(s)=\frac{1}{(s-1)^2}+O(1) $$
and
$$ x^{-s}=\frac{1}{x}-\frac{(s-1)\ln x}{x}+O((s-1)^2) $$
(here $\gamma$ is the Euler-Mascheroni constant) we get
$$ \mathrm{Res}_{s=1}\,\Gamma(s)\zeta'(s)x^{-s}=-\frac{\ln x+\gamma}{x}, $$
which corresponds to the first part of my answer and also gives $h(1)=-\gamma$. The rest is way easier to compute and we obtain
$$ f(e^{-x})=-\frac{\ln x+\gamma}{x}+\sum_{n\geq 0}\frac{(-1)^n\zeta'(-n)x^n}{n!}. $$
Now, from this answer about derivative of zeta we see that this series has a nonzero radius of convergence (namely, $2\pi$) and we can even see singularities at $x=2\pi i n$ for $n\in \mathbb Z$, which is of course what one should expect because of singularity of $f$ at $z=1$.
