# Sum of weighted binomial coefficients

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!

• This sum seems very similar to Stirling Numbers of the Second Kind, with the exception that you have $(k+i)$ instead of $(k-i)$: mathworld.wolfram.com/StirlingNumberoftheSecondKind.html Maybe one approach would be to expand $(k+i)^s$ as a binomial, and then re-express the sum in terms of a summation of Stirling Numbers. The other option would be to write out the sum in terms of a hypergeometric function and then use related identities to massage the result. Does your summation come from something combinatorial? – Alex R. Jun 15 '17 at 21:17
• Thanks for the information! The summation comes from some intermediate results of order statistics of Weibull distribution. It does not have a very nice combinatoric intuition - it just ended up in this format algebraically. I will look into the Stirling Numbers and hypergenmetric function that you mentioned. Can you say a bit more about "expanding $(k+i)^s$ as a binomial"? – user3026001 Jun 16 '17 at 0:54

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.
For a function $f(x)$, and a non-negative integer $m=n-k$, the expression $\Delta_mf(x)=\sum (-1)^{m-i}\binom{m}i f(x+i)$ is called $m$-th finite difference of $f$. It equals $m!$ times $f^{(m)}(\theta)$ for some $\theta$ between $x$ and $x+m$ (proof: consider the function $g(y)=f(y)-h(y)$, where $h$ is a polynomial of degree at most $m$ taking the same values as $f$ at the points $x+i$, $i=0,\dots,m$. By Rolle's theorem find $\theta$ such that $g^{(m)}(\theta)=0$.) Your conjecture follows taking $f(y)=y^s$ and $x=k$.