I am trying to determine the asymptotic behavior (as $n \to \infty$) of the quantity \begin{equation*} \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(1+k)^r}, \end{equation*} for any $r > 0$.

Notice that in the case where $r$ is a positive integer we have \begin{equation*} \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(1+k)^r} \sim \frac{1}{\Gamma(r)} \frac{(\ln n)^{r-1}}{n+1}. \end{equation*} This formula follows by differentiating $r$ times with respect to $z$ the equation \begin{equation*} \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{z+k} = n!\prod_{k=0}^n (z+k)^{-1}. \end{equation*}

I am trying to determine the asymptotic behavior (as $n \to \infty$) of the quantity \begin{equation*} \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(1+k)^r}, \end{equation*} for any $r > 0$.

Notice that in the case where $r$ is a positive integer we have \begin{equation*} \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(1+k)^r} \sim \frac{1}{\Gamma(r)} \frac{(\ln n)^{r-1}}{n+1}. \end{equation*} This formula follows by differentiating $r-1$ times with respect to $z$ the equation \begin{equation*} \sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{z+k} = n!\prod_{k=0}^n (z+k)^{-1}. \end{equation*}