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I'm new to the hypergeometric series, and I'm trying to decipher a proof in which the author identifies a particular finite sum as a a hypergeometric series. The particular summation is:

$$ \begin{array}{l} \sum_{j=0}^{k}\frac{(-n)_j(n+\alpha+\beta+1)_j(-k)_j}{(m-k+1)_j(-m-\gamma-\delta-k)_jj!}\times \\ \sum_{l=0}^{m+n-k}\frac{(-m-n+k)_l(m-n+\gamma+\delta+k+1)_l(\gamma+\alpha+k+1)_l}{(\gamma-n+k+1)_l(\alpha+\beta+\gamma+\delta+2k+2)_ll!}= \\ _3F_2\left(\begin{array}{c} -n, & n+\alpha+\beta+1, & -k; & 1 \\ -m-\gamma-\delta-k, & m-k+1 \end{array}\right)\times \\ _3F_2\left(\begin{array}{c} -n-m+k, & m-n+\gamma+\delta+k+1, & \alpha+\gamma+k+1; & 1 \\ \gamma-n+k+1, & \alpha+\beta+\gamma+\delta+2k+2 \end{array}\right)\end{array} $$

My question is:
Why is this equality true if the upper limits on the sums are finite?

All of the definitions that I have seen for the hypergeometric functions give infinite upper limits (e.g. Abramowitz & Stegun [1], and Wolfram's function site [2]).


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Because $(-k)_i=0$ for every integers $i\geqslant k+1\geqslant1$. – Did Jan 16 '12 at 20:34
And in general, the cases where a hypergeometric function turns out to be a polynomial are the ones in which the coefficients turn out to be eventually $0$. – Robert Israel Jan 16 '12 at 20:47
Thanks, that makes sense now. – okj Jan 16 '12 at 21:03
up vote 3 down vote accepted

Let's examine the first hypergeometric series : $$_3F_2\left(\begin{array}{c} -n, & n+\alpha+\beta+1, & -k; & 1 \\ -m-\gamma-\delta-k, & m-k+1 \end{array}\right)$$

Applying the definition you get : $$\sum_{j=0}^{\infty}\frac{(-n)_j(n+\alpha+\beta+1)_j(-k)_j}{(-m-\gamma-\delta-k)_j(m-k+1)_jj!}1^j$$

$(-k)_j$ is of course the rising factorial : $(-k)_0=1$, $(-k)_j=(-k)(-k+1)\cdots(-k+j-1)$

We simply get $(-k)_j=0$ as soon as $j\gt k$ (if $k$ is a nonnegative integer) because $0$ will appear in the rising factorial and we may stop the sum at k. Note that the sum could stop earlier (for example if $n$ is a smaller nonnegative integer than $k$) but this is not excluded by interrupting the sum at $k$.

The same applies to the other hypergeometric series since the firt term is $-n-m+k$ that will become $(-n-m+k)_l$ that is $0$ for $l\gt n+m-k$.

To resume : It seems just to be a straightforward translation in both cases!

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Thanks a lot. That makes sense now. I appreciate the help! – okj Jan 16 '12 at 21:00

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