Binomial Transform of $a_i = 1/(i+1)^s$ I'm trying to find a nice asymptotic expression for the expression
$$b_n = \sum_{i=0}^{n-1} \binom{n-1}{i}(-1)^i \frac{1}{(1+i)^{1+\frac{1}{\beta}}}$$
to show that $b_n \approx An^c$ for some constants $A,c$ with $c < 1$ (this is related to a previous question I asked on the expected maximum of Weibull distributions (https://stats.stackexchange.com/questions/286224/expectation-of-maximum-of-i-i-d-weibull-random-variables#comment547532_286227))  
I know from a related question (A finite sum involving the binomial coefficients and the harmonic numbers) that there is an easy formula for this sum if $\beta = -1$. When $\beta= -1$, the binomial transformation becomes $H_n$, the $n^{th}$ harmonic number and I can obtain $\sum_{i=0}^{n-1} \binom{n-1}{i}(-1)^i \frac{1}{(1+i)} = H_n = \Theta(\log n)$ . 
Is there a similar asymptotic approximation to $\sum_{i=0}^{n-1} \binom{n-1}{i}(-1)^i \frac{1}{(1+i)^{s}}$ for arbitrary $s > 1$ when $n$ is large?
 A: We have 
$$ \frac{1}{(1+i)^{1+\frac{1}{\beta}}} = \frac{1}{\Gamma\left(1+\frac{1}{\beta}\right)}\int_{0}^{1} x^i\left(-\log x\right)^{\frac{1}{\beta}}\,dx\tag{1}$$
hence the whole sum can be written as
$$ \frac{1}{\Gamma\left(1+\frac{1}{\beta}\right)}\int_{0}^{1}(1-x)^{n-1}\left(-\log x\right)^{\frac{1}{\beta}}\,dx = \frac{1}{\Gamma\left(1+\frac{1}{\beta}\right)}\int_{0}^{+\infty} t^{\frac{1}{\beta}}e^{-t}(1-e^{-t})^{n-1}\,dt\tag{2}$$
and approximations can be recovered from the asymptotic expansion for the (incomplete) $\Gamma$ function. The sum is actually a fractional derivative of Euler's beta function, and the integral appearing in the LHS of $(2)$ can be approximated by applying Holder's inequality.
As an alternative, from
$$ \int_{0}^{1}(1-x)^{n-1}(-\log x)^{\frac{1}{\beta}}\,dx \approx \int_{0}^{+\infty}e^{-(n-1)x}(-\log x)^{\frac{1}{\beta}}\,dx \tag{3} $$
it follows that the expected behaviour for $n\to +\infty$, for $\beta\leq 1$, is
$$ b_n \approx \frac{\log(n-1)^{\frac{1}{\beta}}}{(n-1)\,\Gamma\left(1+\frac{1}{\beta}\right)}.\tag{4}$$
