It is known that $$\frac{\partial^n\zeta(s)}{\partial s^n}=(-1)^n\sum_{k=1}^\infty{\frac{\log^nk}{k^s}}$$
Can the following alternating version of the sum be expressed in terms of well-known functions such as the Riemann zeta?
$$\sum_{k=1}^\infty{(-1)^{k+1}\frac{\log^nk}{k^s}}$$
Hint: Recall the definition of the polylogarithm function $$\mathrm{Li}_s(z)=\sum_{k\geq1}\frac{z^k}{k^s}$$ So
$$-\mathrm{Li}_{s}(-1)=\sum_{k\geq1}\frac{(-1)^{k+1}}{k^s}$$ Can you take it from here?
One may also note the formula $$-\mathrm{Li}_s(-1)=(1-2^{1-s})\zeta(s)$$