Prove the limit exists I was toying around with Abel summability when I stumbled upon a limit I could not prove existed.

$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)\tag{$*$}$$

While it may not be clear that such a limit could converge, it may be helpful to note a similar example:
$$\lim_{x\to-1^+}\sum_{n=1}^\infty nx^{n-1}=\lim_{x\to-1^+}\frac1{(1-x)^2}=\frac14$$
However, a lack of closed form of $(*)$ makes it difficult for me to show it converges.  WolframAlpha returns the series as a derivative of the Lerchphi function, though it doesn't seem quite helpful.
By considering
$$f_k(x)=\sum_{n=2}^k x^n\ln(n)$$
I find that
$$f_{35}(-0.75)-0.25f'_{35}(-0.75)=0.225803586648$$
Which is a quick linear approximation of $f_{35}(x)$ centered at $x=-\frac34$.  This agrees with what I think to be the limit:
$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)\stackrel?=\eta'(0)=\frac12\ln\left(\frac\pi2\right)=0.225791352645\tag{$**$}$$
Where $\eta(s)$ is the Dirichlet eta function.
I also tried considering more elementary approaches to showing the limit exists, such as using $\ln(n+1)=\ln(n)+\mathcal O(n^{-1})$, however, I could not make use of it.
Bonus points if you can prove $(**)$.
 A: This post is to address partial proof of the conjecture, stated by @SimplyBeautifulArt in the comments section using a weaker approach via the Borel Summability, and Euler Summation. The initial attack on the problem starts from $\text{Lemma} \, (1.0)$
$\text{Lemma} \, (1.0)$
$\text{Euler Summation}$
One considers the Divigernt Series $\sum_{}a_{n}$we replace our divigrent series with the corresponding power series:$\sum_{} a_{n}x^{n}$
If such series is convergent for $|x| < 1$ and if it's limit $x \rightarrow 1^{-}$ then one defines the Euler summation of the orginal series as: 
$$\text{E}( \sum_{n} a_{n}) = \lim_{x \rightarrow 1^{-}} a_{n}x^{n}.$$
$\text{Proposition} \, (1.1)$
$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)=\lim_{N\to\infty}\frac1N\sum_{k=2}^N\sum_{n=2}^k(-1)^n\ln(n).$$
$\text{Remark}$:
We assume the RHS side converges in $\text{Proposition} \, \, (1.1).$
$\text{Lemma} \, \, (1.1)$
One can observe in our original Proposition the corresponding series on the RHS side can be rewritten as follows 
$(1)$
$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)= \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{k}(-1)^{n} + \sum_{n} \ln(n).$$
Now focusing our observations on the RHS side of our recent result, another key consideration that can be made is by applying Borel Summability as follows 
$(2.)$
$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n)= \lim_{n \rightarrow \infty} \frac{1}{N}(\lim_{t \rightarrow \infty} e^{-t} \sum_{k}\frac{t^{n}}{n!}(\sum_{n}(-1)^{n}) + \lim_{x \rightarrow 1^{-}}\sum_{n} \ln(x)^{n})$$
$\text{Remark}$
The recent development seen in $(2.)$ can't just be achieved with Borel Summability alone the series $\sum\ln(n)$ was dealt with via Euler Summation as formally discussed in $(0.0)$. So considering the formalities of Euler Summation the series $\sum\ln(n)$ can be defined as follows in $(3)$
$(3)$
$$\text{E}( \sum_{n} \ln(n)) = \lim_{x \rightarrow 1^{-}} \sum_{n} \ln(x)^{n}.$$
A: Recall the archetypal Frullani integral
$$\int_{0}^{\infty} \frac{e^{-u} - e^{-nu}}{u} \, du = \log n. $$
Since the integrand is non-negative for all $n \geq 1$, when $x \in [0, 1)$ we can apply the Tonelli's theorem to interchange the sum and integral unconditionally to get
\begin{align*}
\sum_{n=1}^{\infty} x^n \log n
&= \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} x^n \cdot \frac{e^{-u} - e^{-nu}}{u} \right) \, du \\
&= \frac{x^2}{1-x} \int_{0}^{\infty} \frac{e^{-u}(1 - e^{-u})}{u(1-xe^{-u})} \, du. \tag{1}
\end{align*}
Notice that the last integral converges absolutely. So this computation can be fed back to the Fubini's theorem, showing that exactly the same computation can be carried out to prove $\text{(1)}$ for all $|x| < 1$.
Now we would like to take limit as $x \to -1^{+}$. When $x \in (-1, 0]$, the integrand of the last integral of $\text{(1)}$ is uniformly bounded by the integrable function $u^{-1}e^{-u}(1-e^{-u})$. Therefore by the dominated convergence theorem, as $x \to -1^{+}$ we have
$$
\lim_{x\to -1^+} \sum_{n=1}^{\infty} x^n \log n
= \frac{1}{2} \int_{0}^{\infty} \frac{e^{-u}(1 - e^{-u})}{u(1+e^{-u})} \, du.
$$
This already proves that the limit exists, but it even tells more that the limit is indeed $\eta'(0)$. To this end, we first perform integration by parts to remove the pesky factor $u$ in the denominator. Then the right-hand side becomes
$$ \frac{1}{2} \int_{0}^{\infty} \frac{e^{-u}(1 - e^{-u})}{u(1+e^{-u})} \, du
= \int_{0}^{\infty} \left( \frac{e^u}{(e^u + 1)^2} - \frac{e^{-u}}{2} \right) \log u \, du$$
In order to compute this integral, it suffices to prove the following claim.

Claim. We have
$$ \int_{0}^{\infty} e^{-u} \log u = -\gamma, \qquad \int_{0}^{\infty} \frac{e^u \log u}{(e^u + 1)^2} \, du = -\frac{\gamma}{2} + \eta'(0). $$

Notice that the first claim is an immediate consequence of the identity $\psi(1) = -\gamma$, where $\psi$ is the digamma function. Next, term-wise integration gives
$$ \int_{0}^{\infty} \frac{u^{s-1}}{e^{\alpha u} + 1} \, du = \alpha^{-s} \Gamma(s)\eta(s) $$
where $\alpha > 0$ and $s$ is initially assumed to satisfy $\Re(s) > 1$ (so that interchanging the summation and integration works smoothly). Differentiating both sides w.r.t. $\alpha$ gives
$$ \int_{0}^{\infty} \frac{u^s e^{\alpha u}}{(e^{\alpha u} + 1)^2} \, du = \alpha^{-s-1} \Gamma(s+1)\eta(s). $$
Although we initially assumed $\Re(s) > 1$, now both sides define an analytic function for $\Re(s) > -1$, hence by the principle of analytic continuation this identity extends to this region as well. Now plugging $\alpha = 1$ and differentiating both sides w.r.t. $s$, we get
$$ \int_{0}^{\infty} \frac{u^s e^u \log u}{(e^u + 1)^2} \, du = \Gamma(s+1)\psi(s+1)\eta(s) + \Gamma(s+1)\eta'(s). $$
Plugging $s = 0$ and using known values $\psi(1) = -\gamma$ and $\eta(0) = \frac{1}{2}$, this yields
$$ \int_{0}^{\infty} \frac{e^u \log u}{(e^u + 1)^2} \, du = -\frac{\gamma}{2} + \eta'(0). $$
A: For $|z| \le 1, \Re(s)> 1$ and for $|z| < 1$ let the polylogarithm $$Li_s(z) = \sum_{n=1}^\infty n^{-s}z^n$$
For $\Re(s) > 1$ we have $Li_s(1) = \zeta(s), Li_s(-1) = -\eta(s)$.
Now for $|z| \le 1, z \ne 1$, summation by parts shows that $Li_s(z)$ is entire in $s$ and continuous in $z$ and hence $$-\eta(s) = \lim_{z \to -1, |z| \le 1} Li_s(z), \qquad -\eta'(s) = \lim_{z \to -1, |z| \le 1} \frac{\partial}{\partial s}Li_s(z)$$ $$ \lim_{z \to -1, |z| \le 1} \sum_{n=1}^\infty z^n \log n = -\eta'(0) = \frac{\log(\pi/2)}{2}$$
(see there)
