Let $H_{n,k}=\sum_{j=1}^{n}\frac{1}{j^k}$ and $\displaystyle\lim_{n \to \infty}H_{n,k}=H_{\infty,k}$ for $k >1$. I need to prove that
\begin{equation} \displaystyle\lim_{n \to \infty} \sum_{k=2}^{\infty} \frac{(-1)^k H_{n,k} }{k} = \sum_{k=2}^{\infty} \frac{(-1)^k H_{\infty,k} }{k}. \end{equation}
I tried to prove that for all $\epsilon >0$, there is $N$ natural number such that for all $n>N$: \begin{equation} \left|\sum_{k=2}^{\infty} \frac{(-1)^k (H_{\infty,k}-H_{n,k}) }{k}\right|<\epsilon, \end{equation}
for this, I tried to use the fact that $\displaystyle\lim_{n \to \infty} (H_{\infty,k}-H_{n,k})=0$, but with this I could not complete the test.