# Binomial series with two binomial coefficents

My question reads: Does this formula has mathematical meaning at first place? Is it summable? $$\sum^{\infty}_{k=0}{n\choose k}{m\choose k} x^k$$

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Your "infinite" sum is nothing else than a polynomial in $\,x\,$ of degree $\,\max(n,m)\,$ – DonAntonio Oct 11 '12 at 11:21

Yes, the formula has mathematical meaning. If $a$ and $b$ are natural numbers, and $a\lt b$, the binomial coefficient $\dbinom{a}{b}$ is defined to be $0$. So the sum is effectively a finite sum.

Reamrks: The convention is useful in simplifying formulas. Without it, in the formula of the post, we would have to specify that the summation is to $\min(m,n)$. In situations with more summations, the convention can make for considerable simplifications.

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Thank you, Andre! – Moki Oct 11 '12 at 12:09

It's quite easy to verify that your sum is equal to the coefficient of $z^m$ in the product:

$$(1+xz)^n\,(1+z)^m.$$

If you set $x=1$ you can find a slight generalization of the Chu-Vandermonde identity:

$$\sum_{k=0}^{+\infty}\binom{m}{k}\binom{n}{k}=\binom{m+n}{n}.$$

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Hi Jack, thanks for answering! – Moki Oct 11 '12 at 15:00

It's summable, as $\binom pk=0$ whenever $k\geq p$. (so in the sum there are $\min\{m,n\}+1$ non-vanishing terms.

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Thank you very much for your answer! – Moki Oct 11 '12 at 11:49

It is summable and has a closed form in terms of hypergeometric function

$$F(-n,-m;\,1;\,x) \,.$$

The above hypergeometric function can be simplified to a polynomial in the following cases

1) if $n$ and $m$ are non-negative integers.

2) if $n$ or $m$ is a non-negative integer.

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Thanks a lot, Mhenni! – Moki Oct 11 '12 at 12:09
@Moki: You are welcome. – Mhenni Benghorbal Oct 11 '12 at 12:20