Another binomial coefficient sum

Question

In my work I ran across the following binomial coefficient sum:

$$ S=\sum_{a=0}^{n-1-l} (-1)^a \binom{n}{l+1+a} \binom{l+a}{l} $$

where $n\geq 0$ and $0\leq l \leq n-1$.

I browsed the web and found (thanks to this question) the formula (5.24) in „Concrete Mathematics“ which looks like this:

$$ \sum_k \binom{l}{m+k} \binom{s+k}{n} (-1)^k = (-1)^{l+m} \binom{s-m}{n-l}$$

I went ahead and thought about the summation index. If $a<0$ then $\binom{l+a}{l}=0$ and if $a>n-1-l$ then $\binom{n}{l+1+a}$. So the summation should not be a problem. Using the formula on $S$ gives me

$$ S = (-1)^{n+l+1}\binom{l - (l+1)}{l-n}.$$

Since $l<n$ this would mean $S=0$. But numericals experiments and giving the sum to Mathematica resulted in $S=1$. What am I missing here?

Edit: Do I have to consider $a < -l$, because then $\binom{l+a}{l}\neq 0$ again?

Should we understand that $\sum\limits_k$ means $\sum\limits_{k=0}^\infty$? — ajotatxe, Jul 18 '15 at 20:47
Yes, in the book is explained that $\sum_k$ means summing over all $k\in\mathbb Z$. — Stefan Hante, Jul 18 '15 at 21:13

score 1 · Accepted Answer · answered Jul 18 '15 at 20:56

1

For $a=-\ell-1$ you would have a term

$$(-1)^{-\ell-1}\binom{n}0\binom{-1}\ell=(-1)^{\ell+1}\frac{(-1)^{\underline\ell}}{\ell!}=-1\;.$$

This term is included in the summation in $(5.24)$ from Concrete Mathematics but not in your summation, and it accounts for the difference.

answered Jul 18 '15 at 20:56

Brian M. Scott

616,228

Do you imply that $a=-l-1$ is the only term that I forgot? – Stefan Hante Jul 18 '15 at 21:05
@Wauzl: Yes. If $a<-\ell-1$, the lower number in the first binomial coefficient is negative, so the coefficient is $0$. – Brian M. Scott Jul 18 '15 at 21:07
I could have seen this, but I'm really not used to these kind of binomial coefficients. Thanks a lot. – Stefan Hante Jul 18 '15 at 21:11
@Wauzl: You’re very welcome. – Brian M. Scott Jul 18 '15 at 21:13

score 0 · Answer 2 · answered Jul 18 '15 at 23:10

Suppose we seek to evaluate $$S = \sum_{q=0}^{n-l-1} (-1)^q {n\choose l+1+q} {l+q\choose q}$$ where $l\lt n.$

Introduce the integral $${n\choose l+1+q} = {n\choose n-l-1-q} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n-l-q}} \; dz.$$

This integral is zero when $q\ge n-l$ so we may let $q$ go to infinity.

This yields for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n-l}} \sum_{q\ge 0} {l+q\choose l} (-1)^q z^q\; dz \\= \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n-l}} \frac{1}{(1+z)^{l+1}} \; dz \\= \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n-l-1}}{z^{n-l}} \; dz.$$

This is $$[z^{n-l-1}] (1+z)^{n-l-1}$$ which evaluates to one by inspection.

Thanks for your direct proof, but where did that scary integral come from? Is it in the book I mentioned, too? Is the identity you use to get rid of the sum well known? — Stefan Hante, Jul 20 '15 at 21:15

Another binomial coefficient sum

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