As shown in my previous answer, the value of the sum that you see is 
$$\lim_{t\uparrow1}\sum_{n=2}^\infty (-t)^n \ln n. $$

Here is a "manual" way to show this. Writing 
\begin{equation}
	\ln n=\int_0^\infty dz\,\frac{e^{-z}-e^{-nz}}z,
\end{equation}
for $t\uparrow1$ we have 
\begin{equation}
\begin{aligned}
&\sum_{n=2}^\infty (-t)^n \ln n \\ 
	&=\sum_{n=2}^\infty (-t)^n \int_0^\infty dz\,\frac{e^{-z}-e^{-nz}}z \\ 
	&=\int_0^\infty \frac{dz}z\sum_{n=2}^\infty (-t)^n (e^{-z}-e^{-nz}) \\ 
	&=\frac{t^2}{1+t}\int_0^\infty \frac{dz}z\,e^{-z}\frac{1-e^{-z}}{1+t e^{-z}} \\ 
	&\to\frac12\int_0^\infty \frac{dz}z\,e^{-z}\frac{1-e^{-z}}{1+e^{-z}} \\ 
	&=-\int_0^1 \frac{dx}{\ln x}\,\frac{1-x}{1+x} 
	= \frac12\,\ln\frac\pi2, 
\end{aligned}	
\end{equation} 
by (say) Formula 4.267.1 on p. 545 in I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, Seventh Edition, 2007.