While studying the behaviour of umbilic points on Weingarten surfaces I discovered that the following combinatorial identity must be true.

For all $l,m\in{\mathbb N}$ with $l\geq m-1\geq0$ the following holds: $ \sum_{k=m-1}^l(-1)^{k+m}\frac{k+2}{k+1}{l \choose k}{k+1 \choose m}= \left\{\begin{array}{ccl} 0&if& l>m\\ 1&if& l=m\\ -1-\frac{1}{l+1} &if& l=m-1 \end{array} \right. $

Improbable as it at first appears, it is easy to check the second and third options are true, and I have computer-checked and found it is true for all $l,m\leq 60$. Perhaps it is well-known or has an easy proof. Either would be good to know.

  • 1
    $\begingroup$ The second ($l=m$) and third ($l=m-1$) identity consist of two respectively one term in the sum, so checking is obvious. So do I understand it correctly that you are asking only for the first case ($l >m$)? (It is not clear for me from your wording.) $\endgroup$ Jul 11, 2017 at 18:25
  • $\begingroup$ A starting point for the identity in the case $l >m$ could be the binomial inversion $\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$ (see e.g. math.stackexchange.com/questions/4175/…) $\endgroup$ Jul 11, 2017 at 18:31
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    $\begingroup$ The first case is easy as the sum is the $l$th difference of a polynomial of degree $m$. $\endgroup$
    – Ira Gessel
    Jul 11, 2017 at 18:34
  • $\begingroup$ @Andreas Rüdinger yes the case $l>m$ is what needs a proof. $\endgroup$ Jul 11, 2017 at 18:59
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    $\begingroup$ @Brendan Guilfoyle If $P(k)$ is a polynomial in $k$ of degree less than $l$ then $$\sum_{k=0}^l (-1)^k \binom lk P(k) = 0.$$ In your sum, we can write $\frac{k+2}{k+1}\binom{k+1}{m}$ as $P(k)= \frac{k+2}{m}\binom{k}{m-1}$, a polynomial in $k$ of degree $m$, and the sum $\sum_{k=m-1}^l$ can be replaced with $\sum_{k=0}^l$, since the additional terms are all zero. For a brief discussion of differences of polynomials, see artofproblemsolving.com/community/c6h90529. For a slightly more detailed discussion, see arxiv.org/abs/1609.05988, pp. 13--14. $\endgroup$
    – Ira Gessel
    Jul 12, 2017 at 15:09

1 Answer 1


We obtain for $l,m\in\mathbb{N}$ with $0\leq m-1 \leq l$: \begin{align*} \color{blue}{\sum_{k=m-1}^{l}}&\color{blue}{(-1)^{k+m}\frac{k+2}{k+1}\binom{l}{k}\binom{k+1}{m}}\\ &=\frac{1}{m}\sum_{k=m-1}^l(-1)^{k+m}(k+2)\binom{l}{k}\binom{k}{m-1}\tag{1}\\ &=\frac{1}{m}\binom{l}{m-1}\sum_{k=m-1}^l(-1)^{k+m}(k+2)\binom{l-m+1}{k-m+1}\tag{2}\\ &=\frac{1}{m}\binom{l}{m-1}\sum_{k=0}^{l-m+1}(-1)^{k+1}(k+m+1)\binom{l-m+1}{k}\tag{3}\\ &=\frac{m+1}{m}\binom{l}{m-1}\sum_{k=0}^{l-m+1}(-1)^{k+1}\binom{l-m+1}{k}\\ &\qquad+\frac{l-m+1}{m}\binom{l}{m-1}\sum_{k=1}^{l-m+1}(-1)^{k+1}\binom{l-m}{k-1}\tag{4}\\ &=-\frac{m}{m+1}\binom{l}{m-1}[[l=m-1]]\\ &\qquad+\frac{l-m+1}{m}\binom{l}{m-1}\sum_{k=0}^{l-m}(-1)^k\binom{l-m}{k}\tag{5}\\ &=\left(-1-\frac{1}{m}\right)[[l=m-1]]+\frac{l-m+1}{m}\binom{l}{m-1}[[l=m]]\tag{6}\\ &\color{blue}{=\left(-1-\frac{1}{m}\right)[[l=m-1]]+1[[l=m]]} \end{align*} and the claim follows.


  • In (1) we use the binomial identity $$\binom{p+1}{q+1}=\frac{p+1}{q+1}\binom{p}{q}$$

  • In (2) we use the binomial identity $$\binom{p}{q}\binom{q}{r}=\binom{p}{r}\binom{p-r}{q-r}$$

  • In (3) we shift the index of the sum to start with $k=0$.

  • In (4) we split the sum and work similarly as in (1).

  • In (5) we do some simplifications regarding $(1-1)^{l-m+1}$ using Iverson brackets. We also shift the index of the sum again.

  • In (6) we do a similar job as in (5).

  • $\begingroup$ The context in which this arose is now in a paper on the arxiv: arxiv.org/abs/1709.00580 $\endgroup$ Sep 5, 2017 at 17:13
  • $\begingroup$ @BrendanGuilfoyle: Interesting paper (but somewhat beyond my scope) and a great pleasure to see my contribution was useful. :-) Many thanks for taking my proof into your paper and citing my name. Kind regards, $\endgroup$
    – epi163sqrt
    Sep 5, 2017 at 18:37

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