I came across this partial sum which I cannot find reasonable bounds on; I feel this must be known in the literature, but I do not know where to look. Here is the problem:
Let $s\in (0,1)$ and consider the binomial sum $$ \boldsymbol{(*)}\qquad \sum_{k=1}^N\, \binom{N}{k}\, \frac1{k^s} $$ are there any known good upper and lower bounds on $\boldsymbol{(*)}$?
$\newcommand{\Si}{\Sigma}$We have to lower- and upper-bound \begin{equation*} \Si_N:=\sum_{k=1}^N\,\binom Nk\, \frac1{k^s}. \end{equation*}
Note that \begin{equation*} \Si_N=2^N EX^{-s}\,1(X\ge1), \tag{10}\label{10} \end{equation*} where $X$ is a random variable with the binomial distribution with parameters $N,1/2$.
By Jensen's inequality applied to the convex function $[1,\infty)\ni x\mapsto x^{-s}$, \begin{equation*} EX^{-s}\,1(X\ge1)=P(X\ge1)\,E(X^{-s}|X\ge1) \\ \ge P(X\ge1)\,(E(X|X\ge1))^{-s}. \tag{20}\label{20} \end{equation*} Also, \begin{equation*} E(X|X\ge1)=\dfrac{EX\,1(X\ge1)}{P(X\ge1)}=\dfrac{EX}{P(X\ge1)}=\dfrac{N/2}{P(X\ge1)} \tag{30}\label{30} \end{equation*} and $P(X\ge1)=1-2^{-N}$. So, by \eqref{10}, \eqref{20}, and \eqref{30}, we get a lower bound on $\Si_N$: \begin{equation*} \Si_N\ge L_N:=M_N\,(1-2^{-N})^{1+s}, \tag{33}\label{33} \end{equation*} where \begin{equation*} M_N:=\frac{2^N}{(N/2)^s}. \tag{35}\label{35} \end{equation*}
To get an upper bound on $\Si_N$, write \begin{equation*} EX^{-s}\,1(X\ge1)=E_1+E_2, \tag{40}\label{40} \end{equation*} where \begin{equation*} E_1:=EX^{-s}\,1(X\ge\tfrac N2\,(1-h)),\quad \\ E_2:=EX^{-s}\,1(1\le X<\tfrac N2\,(1-h)), \end{equation*} and $h\in(0,1-2/N]$. Next, \begin{equation*} E_1\le(\tfrac N2\,(1-h))^{-s}. \tag{50}\label{50} \end{equation*} By Hoeffding's inequality, \begin{equation*} E_2\le P(1\le X<\tfrac N2\,(1-h)) \\ \le \exp(-\tfrac2N\,(\tfrac{Nh}2)^2)=e^{-h^2 N/2}=N^{-c^2/2} \tag{60}\label{60} \end{equation*} if $h=c\sqrt{\frac{\ln N}N}\in(0,1-2/N]$. So, in view of \eqref{10}, \eqref{40}, \eqref{50}, \eqref{35}, and \eqref{60}, we get an upper bound on $\Si_N$: \begin{equation*} \Si_N\le U_N:=M_N\,\Big[\Big(1-c\sqrt{\frac{\ln N}N}\,\Big)^{-s} +2^{-s}N^{s-c^2/2}\Big]. \tag{70}\label{70} \end{equation*}
Fixing now any real $c>\sqrt{2s}$, letting $N\to\infty$, and looking at \eqref{33} and \eqref{70}, we see that $L_N\sim M_N\sim U_N$ and hence \begin{equation*} \Si_N\sim L_N\sim U_N\sim M_N=\frac{2^N}{(N/2)^s}. \end{equation*} Thus, our lower and upper bounds $L_N$ and $U_N$ on $\Si_N$ are asymptotically exact for large $N$.
If you want a more precise value you can expand $k^{-s}$ around $k=N/2$ and sum term by term. I get $$\frac{2^N}{(N/2)^s}\Bigl(1 + \frac{s(1+s)}{2N} + \frac{s(1+s)(2+s)(3+s)}{8N^2} + O(N^{-3})\Bigr).$$
