# 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 $$b$$.

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.
$$\sum_{n=1}^a \frac{(-1)^{n-1}}{n^b}=2^{-b} \left(\left(2^b-2\right) \zeta (b)+(-1)^a \left(\zeta \left(b,\frac{a+2}{2}\right)-\zeta \left(b,\frac{a+1}{2}\right)\right)\right)$$ $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^b}=2^{-b} \left(2^b-2\right) \zeta (b)$$