We can prove that $$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\zeta(n)=\gamma$$
In fact, If we let $f(z)=\sum_{m=2}^\infty\frac{(-1)^m}m z^m$, then by the method which used in this question,
$$\sum_{n=2}^{\infty}\frac{(-1)^n}{n}\zeta(n)=\sum_{n=1}^nf\left(\frac1 n\right)=\sum_{n=1}^\infty\sum_{m=2}^\infty\frac{(-1)^m}{mn^m}=\sum_{n=1}^\infty\left(\frac1 n+\log\left(1-\frac1 n\right)\right)=\gamma$$
Is there any known value for $\displaystyle \sum_{n=2}^{\infty}\frac{(-1)^n}{n^k}\zeta(n)$ for every natural number $k\ge2$? What is the best result?
I only get $$ \sum _{n=2}^{\infty }{\frac { \left( -1 \right) ^{n}\zeta \left( n \right) }{{n}^{2}}} = \gamma+\int _{0}^{1}\!{\frac {\ln \left( \Gamma \left( s+1 \right) \right) }{s}}{ds} $$ Probably not much help.