Find the sum $\sum_{n=1}^{\infty} \frac{\ln n}{n(n- \ln n)}$ Find the following sum : $\sum_{n=1}^{\infty} \frac{\ln n }{n(n- \ln n)}$
I couldn't find any way to solve. 
I just want a hint, how to initiate?
 A: I checked the actual problem (Problem 3.1.5.(b) of $\text{[1]}$, see also this image). Its exact formulation is as follows:
$$ \sum_{n=1}^{\infty} \frac{1}{n} \left[ \frac{\ln n}{n - \ln n} \right]. $$
Now, in this textbook, the bracket $[\cdot]$ is exclusively used for the greatest integer function. In other words, $[x] = \lfloor x \rfloor$. Now the hint perfectly makes sense, because
$$ 0 \leq \frac{\ln n}{n - \ln n} < 1 \quad \Rightarrow \quad \left\lfloor \frac{\ln n}{n - \ln n} \right\rfloor = 0 $$
and the sum is zero.

$\text{[1]}$ Kaczor, Wiesława J., and Maria T. Nowak. Problems in Mathematical Analysis: Real numbers, sequences, and series. Vol. 1. American Mathematical Soc., 2000., p64
A: $$S=\sum_{n=1}^{\infty} \frac{\ln n}{n(n-\ln n)}= \sum_{n=1}^{\infty} \left( \frac{1}
{n-\ln n}- \frac{1}{n} \right)\sim\sum_{n=1}^{\infty} \left( \frac{1}{n} (1+\frac{\ln n}{n}+\frac{\ln^2 n}{n^2}+ \frac{\ln^3 n}{n^3}...) -\frac{1}{n}\right).$$ $$\implies S= \sum_{n=1}^{\infty} \left(\frac{\ln n}{n^2}+\frac{\ln^2 n}{n^3}+ \frac{\ln^3 n}{n^4}+...\right)=-\zeta'(2)+....$$ As per  the suggestion of @Toby Mak above.
Further, using $$\sum_{n=0}^{\infty} \frac{1}{n^z}= \zeta(z) \implies 
\sum_{n=1}^{\infty}(-1)^k \frac{\ln^{k} n}{n^z} =\zeta^{(k)}(z)$$
It comes the same result as predicted by @Sangchui Lee above here in his comment as $$S=\sum_{k=1}^{\infty} (-1)^k \zeta^{(k)} (k+1)$$
However, numeric strongly suggests a value of 1.28495... for S which is very close to $1+\gamma/2~~!$, where $\gamma$ is the Eular number equal to 0.5772...$
