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

Question

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!

Answer 1

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 $k \in \mathbb N.$