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:



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> > 
> > 1. Does this reduce to values of a known special function for arbitrary real (or complex) $s$?
> > 2. What is its asymptotic expansion for large $n$?
> > 3. Is there an efficient numerical method (avoiding cancellations) of evaluating it for large $n$?
> > 
> > 
> > 
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> 
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**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.



Alternating binomial Dirichlet series

> 
> 1. 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:



> 
> 2. 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:



> 
> 3. 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.


