$\sum_{n = 1}^\infty (-1)^n \frac{H_n}{n^s}$ in terms of $\sum_{n = 1}^\infty \frac{H_n}{n^s}$ I have been looking for an equation that relates the above sums to no avail. Perhaps, I am missing some important Harmonic identities.
In the sums, $H_n$ represents the $n^{th}$ harmonic number.
Thanks in advance!

This might not be an answer, but it's too long for a comment.
First, calculate the ordinary generating function (OGF) of <$H_n$>, which is
$$
H(z)=\sum_{n=0}^\infty H_n z^n.
$$
Note that each term in  is the convolution of <$a_n$> $=$ < $1,1,1$, $\ldots$ > and the reciprocals $=$<$0,1,1/2,1/3$, $\ldots$>, i.e.,
$$
H_n=\sum_{k=0}^n a_k b_{n-k}.
$$
The OGFs of < $a_n$ > and < $b_n$ > are
$$
A(z)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}, 
$$
and
$$
B(z)=\sum_{n=1}^\infty \frac{z^{n+1}}{n}=-z \ln(1-z)
$$
respectively.
Thus, 
$$
H(z)=A(z)B(z)=\frac{-z \ln(1-z)}{1-z}.
$$
Let $G^{(j)}(z)$ denote the $j$th-order derivative of $G(z)$, and let
$$
{k \brace j}_* = -\frac{1}{j} {k \brace j-1}_* + \frac{1}{j} {k-1 \brace j}_* 
+[k=j=1]_\delta +[k=j=0]_\delta. \tag{1}
$$
By equation (3.2) in Generating Function Transformations Related to the Form of Dirichlet Series and Applications, 
$$
\sum_{n=1}^\infty \frac{g_n z^n}{n^k}= \sum_{j=1}^\infty {k+2 \brace j}_* z^j G^{(j)}(z). \tag{2}
$$
The OP's series is
$$
\sum_{n = 1}^\infty (-1)^n \frac{H_n}{n^k} = \sum_{j=1}^\infty {k+2 \brace j}_* (-1)^j H^{(j)}(-1),
$$
for $k \in \mathbb N.$
