Is there a further simplification for alternating harmonics with order? I know that the harmonic number $H_a ^{(b)}$ is $$\sum_{n=1}^a \frac{1}{n^b}$$
I was wondering if, for the generalized alternating harmonic number $\bar H_a^{(b)}$, there was a closed formula. For example, when $b$ is 1, $$\sum_{n=1}^a \frac{(-1)^{n-1}}{n^b}$$
is conditionally convergent to $\ln(2)$ because of the taylor series representation of $\ln(1+x)$. I'm asking if similar techniques are possible to find the value with a variable $b$.
 A: For $x > 1$, 
$$  \sum_{n=1}^\infty \frac{1}{n^x} = \zeta(x)  $$
and
$$  \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^x} = (1 - 2^{1-x})\zeta(x)  \text{,}  $$
where $\zeta(x)$ is the Riemann zeta function.  The fact you cite for $x = 1$ follows from the "coincidence", for $|x|<1$, 
$$  \ln(1+x) = \sum_{n = 1}^\infty (-1)^{n+1} \frac{x^k}{k}  \text{,}  $$
which we may evaluate at $x = 1$ to get the alternating harmonic series.
The corresponding coincidence for $x = 2$ is the dilogarithm:
$$  \mathrm{Li}_2(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}  \text{,}  $$
which we may evaluate ate $x = -1$ to get the alternating series with $b=2$ about which you ask.  More generally, the polylogarithms are the functions for the series you ask about.
$$  \mathrm{Li}_b(x) = \sum_{n=1}^\infty \frac{x^n}{n^b}  \text{,}  $$
which we evaluate at $x = -1$ to get the alternating series.
A: $$\sum_{n=1}^a \frac{(-1)^{n-1}}{n^b}=2^{-b} \left(\left(2^b-2\right) \zeta (b)+(-1)^a \left(\zeta \left(b,\frac{a+2}{2}\right)-\zeta
   \left(b,\frac{a+1}{2}\right)\right)\right)$$
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^b}=2^{-b} \left(2^b-2\right) \zeta (b)$$
