$\sum_{n = 1}^\infty (-1)^n \frac{H_n}{n^s}$ 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!

The article "Generating Function Transformations Related to the Form of Dirichlet Series and Applications" was a 2012 REGS research article of mine at UIUC. The content in this article (which has apparently been moved online) has been superseded by the more recent work in this article (published in the Online Journal of Analytic Combinatorics in 2017) and this article (also to appear in OJAC in 2018). 
I think that there's another way to express these sums without having to resort to that level of abstraction and taking the limiting cases as $z \mapsto 1$ suggested by the previous post. Namely, we can perform the same trick in finding an exact expression for the alternating zeta function $\zeta^{\ast}(s)$ in terms of $\zeta(s)$ using the identity that $H_{2n}-H_n = H^{\prime}_{2n}$ where $H^{\prime}_n := \sum_{k=1}^n \frac{(-1)^{k-1}}{k}$:
$$\begin{align} 
H_1^{\ast}(s) & := -\frac{H_1}{1^s} + \frac{H_2}{2^s} - \frac{H_3}{3^s} \\ 
     & = -H_1(s) + (1+2^{-s}) \sum_{n \geq 1} \frac{H_{2n}}{n^s} \\ 
     & = -\sum_{n \geq 1} \frac{H_n}{n^s} + (2^s+1) \sum_{n \geq 1} \frac{H_{2n}}{(2n)^s}. 
\end{align}$$
This assumes that you have a known series representation for the second term on the right-hand-side of the previous equation, but I think it's definitely more in the direction you were heading with the question.
