Alternating binomial Dirichlet series

Question

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:

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 $n$?

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.

Answer 1

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.

Answer 2

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 $$ H_n^s = \sum_{1 \le i_1 \le i_2 \le \ldots \le i_s \le n} \dfrac{1}{i_1 i_2 \ldots i_s} $$