Alternating binomial Dirichlet series I have come across the following deceptively simple expression:
$$ H_n^s=\sum_{j=1}^n(-1)^{j-1}\left(\begin{array}{c}n\\j\end{array}\right)j^{-s} $$
We have (using eg mathematica, though probably not difficult to prove): $H_n^0=1$, $H_n^1=H_n$ (the harmonic numbers), expressions involving hypergeometric series with unit argument for integer $s<0$ and involving polygamma functions for integer $s>1$.  For fixed $n$ the sum is of course finite.  My (closely related) questions are:


    
*
    
*Does this reduce to values of a known special function for arbitrary real (or complex) $s$?
    
*What is its asymptotic expansion for large $n$?
    
*Is there an efficient numerical method (avoiding cancellations) of evaluating it for large $n$?     


Edit: Using Noam's approach I found two more terms that check numerically:
$$ H_n^s=\frac{(\ln n)^s}{\Gamma(s+1)}+\frac{\gamma(\ln n)^{s-1}}{\Gamma(s)}+\frac{6\gamma^2+\pi^2}{12\Gamma(s-1)}(\ln n)^{s-2}+\ldots $$
where $\gamma=0.577\ldots$ is the Euler constant.  Further asymptotics very welcome.
 A: 
  
*
  
*Does this reduce to values of a known special function for arbitrary real (or complex) $s$?
  

Answered by Johannes Trost in a comment: it's also known as a
"Roman harmonic number".  But this and the associated references
do not yield answers to the next two questions, so I continue:


  
*What is its asymptotic expansion for large $n$?
  

I don't have a full asymptotic expansion, but for starters
$$
H_n^s = \frac{(\log n)^s}{\Gamma(s+1)} + O((\log n)^{\sigma-1})
$$
holds for each $s$ of positive real part $\sigma$.  This follows from
an integral formula that is also relevant to the final question:


  
*Is there an efficient numerical method (avoiding cancellations) of evaluating it for large $n$?
  

One standard approach to alternating sums such as
$\sum_{j=0}^n (-1)^j {n \choose j} f(j)$ is to write them as
weighted averages over $X$ of
$\sum_{j=0}^n (-1)^j {n \choose j} X^j = (1-X)^n$,
which in turn requires writing $f$ as a Laplace or Mellin transform.
Here we're dealing with $\sum_{j=1}^n$, not $\sum_{j=0}^n$, but
the same method applies: taking $X = e^{-x}$, we find
$$
H_n^s = \frac1{\Gamma(s)} \int_0^\infty (1 - (1-e^{-x})^n) \, x^s \, \frac{dx}{x}.
$$
The integrand is smooth, and positive for $s \in {\bf R}$
(whence $H_n^s > 0$ for all $n$ and $s>0$),
so this formula can be used to evaluate
$H_n^s$ with numerical integration techniques,
and to estimate it asymptotically.
For large $n$ the factor $1 - (1-e^{-x})^n$
behaves like the characteristic function of the interval $[0, \log n]$,
which makes $H_n^s$ asymptotic to
$$
\frac1{\Gamma(s)} \int_0^{\log n} x^s \, \frac{dx}{x}
= \frac{(\log n)^s}{\Gamma(s+1)}.
$$
A bit more care with the difference between $1 - (1-e^{-x})^n$
and $\chi_{[0,\log n]}$ yields the error estimate $O((\log n)^{\sigma - 1})$,
and with further work it may be possible to derive more precise
asymptotic estimates.
A: Not an answer, but this may help with asymptotics:
According to Maple the o.g.f. for $H^s_n$ is
$$ \sum_{j=1}^\infty j^{-s} (-1)^{j-1} \sum_{n=j}^\infty {n \choose j} x^n
    = {\frac {1}{-1+x}{\it polylog} \left( s,{\frac {x}{-1+x}} \right) }$$
In particular this should be analytic for $|x|<1$.
EDIT: For positive integer values of $s$,  Dilcher's formula says
$$ H_n^s = \sum_{1 \le i_1 \le i_2 \le \ldots \le i_s \le n} \dfrac{1}{i_1 i_2 \ldots i_s} $$
