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This is an old paper by W.Feller.

It says that

$$ \sum_{k=r}^{n-1} \binom{k}{r}\binom{2n-2-k}{n-1} = \binom{2n-1}{n-1-r}$$

But I think, by Vandermonde's Identity, it's $\binom{2n-2}{n-1-r}$

Let $$p = k-r, then \sum_{p=0}^{n-1-r} \binom{k}{k-r}\binom{2n-2-k}{n-1-k} = \sum_{p=0}^{n-1-r} \binom{k}{p}\binom{2n-2-k}{(n-1-r)-p} = \binom{2n-2}{n-1-r}$$

Is it just a Typo or not?

RobPratt
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David Lee
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  • This is because I already found a few of typos in this paper. – David Lee Aug 21 '23 at 07:16
  • I've made a sanity check for $n=4$ and $r=2$ and obtained $7$. This result is the same result as in the paper. And it is a counter example for your result. – callculus42 Aug 21 '23 at 07:28
  • It looks correct. There is a Vandermonde identity for when the top indices are constant and the bottom ones add up to a constant, and vice versa. It is the other one you use here, and all works out fine. – Michal Adamaszek Aug 21 '23 at 07:41
  • Leave a comment to let us know if you have any other questions. – callculus42 Aug 21 '23 at 08:10
  • Let $p = k-r$ then $\sum_{p=0}^{n-1-r} \binom{k}{k-r}\binom{2n-2-k}{n-1-k} = \sum_{p=0}^{n-1-r} \binom{k}{p}\binom{2n-2-k}{(n-1-r)-p} = \binom{2n-2}{n-1-r} $ – David Lee Aug 21 '23 at 08:44
  • One thing I might miss out is that top indices are not (actually) constant? So I did not apply Vandermonde to it correctly? Then, Can you show me why 2n-1? What's the well-known binomial coefficients formula the author mentioned? – David Lee Aug 21 '23 at 08:50

1 Answers1

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$$ \begin{align} \sum_{k=r}^{n-1}\binom{k}{r}\binom{2n-2-k}{n-1} &=\sum_{k=r}^{n-1}(-1)^{k-r}\binom{-r-1}{k-r}(-1)^{n-1-k}\binom{-n}{n-1-k}\tag{1a}\\ &=(-1)^{n-r-1}\binom{-n-r-1}{n-r-1}\tag{1b}\\ &=\binom{2n-1}{n-r-1}\tag{1c} \end{align} $$ Explanation:
$\text{(1a):}$ negative binomial coefficients
$\text{(1b):}$ Vandermonde
$\text{(1c):}$ negative binomial coefficients

Vandermonde in the Upper Parameters

The Binomial Theorem says that $$ \begin{align} (1+x)^{-n-1}(1+x)^{-m-1} &=\sum_{j=0}^\infty(-1)^j\binom{n+j}{n}x^j\ \sum_{k=0}^\infty(-1)^k\binom{m+k}{m}x^k\tag{2a}\\ &=\sum_{j=0}^\infty\sum_{k=j}^\infty(-1)^k\binom{n+j}{n}\binom{m+k-j}{m}x^k\tag{2b}\\ &=\sum_{k=0}^\infty\sum_{j=0}^k(-1)^k\binom{n+j}{n}\binom{m+k-j}{m}x^k\tag{2c} \end{align} $$ and that $$ (1+x)^{-(n+m+1)-1}=\sum_{k=0}^\infty(-1)^k\binom{n+m+1+k}{n+m+1}x^k\tag3 $$ Putting $(2)$ and $(3)$ together, gives $$ \sum_{j=0}^k\binom{n+j}{n}\binom{m+k-j}{m}=\binom{n+m+k+1}{n+m+1}\tag4 $$ Thus, when the summation is carried out in the upper parameter of the binomial coefficients, an extra $1$ needs to be added in the upper and lower parameters in the resultant binomial coefficient, as compared to the standard Vandermonde Identity.

robjohn
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  • WOW. Thank you very much. I think your answer is a perfect one. By the way, Exactly What was my fallacy in using standard Vandermonde in the first place? – David Lee Aug 22 '23 at 09:35
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    In the substitution of $p+r$ for $k$, the lower parameter was substituted, but the upper parameter was not. The upper parameter still contains $k$, which varies in the sum, and that $k$ was treated as a constant in your application of Vandermonde. – robjohn Aug 22 '23 at 10:06
  • That was my mistake. Thank you again. :) – David Lee Aug 25 '23 at 07:37
  • Can I ask one more thing relating to the previous one? How about this one? $$\sum_{v-1=0}^{n-1}\binom{2v}{v-1}\binom{2n-2v}{n-v}$$ My guessing is $\binom{2n}{n-1}$ and I finally got to know both uppers and lowers contains variables so that I can't apply Valdermonde, thanks to you, but I feel like it is quite reasonable because that means a path from origin to -2 passing through X-axis ONCE in symmetric randomwalk. – David Lee Aug 29 '23 at 06:40
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    I don't think that has a simple sum. Note that for $n=2$, your guess fails. – robjohn Aug 29 '23 at 09:45