# What is the reason to use hypergeometric functions?

I would be grateful if anyone could explain the purpose of using hypergeometric functions. If a function exists in closed form, e.g. $\sum\limits_{k \geq 0}z^k = {}_2 F_1 \bigg[{{1\; 1}\atop{1}} \vert z \bigg] = \frac{1}{1-z}$, but in case it doesn't, why bother rewriting it, e.g. $\sum\limits_{k \leq m}\binom{n}{k} = \sum\limits_{k \geq 0} \binom{n}{m-k} = \binom{n}{m} {}_2 F_1 \bigg[{{-m\; 1}\atop{n-m+1}} \vert 1 \bigg]$

since it doesn't yield a closed form or approximation of it?

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Closed form is in the eye of the beholder. – André Nicolas Feb 8 '12 at 22:12
Let's take your second sum as an example; it turns out that knowing the $-m$ numerator parameter is a negative integer is crucial, since it is a known property of hypergeometric functions that they degenerate to polynomials whenever one or both of their numerator parameters are negative. – J. M. Feb 8 '12 at 22:37