Given positive integers $n$ and $k$, ($1\leqslant k\leqslant n-1$), and a real constant $s\in(0,1)$, I'm considering the following summation:
$$\sum\_{i=0}^{n-k}(-1)^i\binom{n-k}{i}(k+i)^s$$
My goal is to show that this summation, for any choice of $n$ and $k$, is **negative**. Can anyone give me some possible directions to think about? Thanks!



Sum of weighted binomial coefficients
Here is a possible direction of approach.


**Lemma.** *If we denote the function*
$$f(n,k):=\sum\_{i=0}^{n-k}(-1)^i\binom{n-k}i(k+i)^s,$$
*then $f(n+1,k+1)-f(n,k)=-f(n+1,k)$.*


**Proof.** Consider the difference $f(n+1,k)-f(n,k)$ instead:
\begin{align} f(n+1,k)-f(n,k)
&=\sum\_{i=0}^{n+1-k}(-1)^i\binom{n+1-k}i(k+i)^s-
\sum\_{i=0}^{n-k}(-1)^i\binom{n-k}i(k+i)^s \\
&=(-1)^{n+1-k}(n+1)^s+\sum\_{i=0}^{n-k}(-1)^i\binom{n-k}i
\frac{i\,(k+i)^s}{n+1-k-i} \\
&=(-1)^{n+1-k}(n+1)^s+\sum\_{i=0}^{n-k}(-1)^i\binom{n-k}{i-1}(k+i)^s \\
&=-(-1)^{n-k}(n+1)^s-\sum\_{j=0}^{n-k-1}(-1)^j\binom{n-k}j(k+1+j)^s \\
&=-\sum\_{j=0}^{n-k}(-1)^j\binom{n-k}j(k+1+j)^s \\
&=-f(n+1,k+1). \end{align}
The proof follows. $\square$


Now, one may proceed with some induction on $n$ and for all $1\leq k<n$. You see, if $f(n+1,k)<0$ then $f(n+1,k+1)>f(n,k)$. The following fact can be brought to bear: whenever $d<m$, 
$$\sum\_{i=0}^m(-1)^i\binom{m}ii^d=0.$$


Next, try to convince yourself that for fixed $k$:
$$\lim\_{n\rightarrow\infty}f(n,k)=0.$$
This statement and the above lemma (and remark) would allow the conclusion you desire.


