Is there a further simplification for alternating harmonics with order?

Question

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 $b$.

Answer 1

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.

Answer 2

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