explicit formula for a sum of sequence

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Partial sum of rows of Pascal's triangle

Working on the Euler characteristic of some topological space, I was led to the following situation. Let $a$ be a given positive integer and the sequence $u_i={1\over i!} \prod_{j=0}^{i-1} {(a-j)}$ can we simplify the expression $S_n=\sum \limits_{i=1}^{n}{u_i}$?

Edit: Working some examples I can see that $\sum \limits_{i=1}^{n} u_i={1\over n!}\sum{\alpha_ia^i}$ such that $\sum{\alpha_i}=n!$ but i don't see the general expression of $\alpha_i$ . For example $$ S_2 = {1\over 2}a+{1\over 2}a^2$$ $$ S_3 = {5\over 6}a+{1\over 6}a^3$$ $$ S_4 = {14\over 24}a+{11\over 24}a^2-{2\over 24}a^3+{1\over 24}a^4$$ $$ S_5 = \frac{47}{60} a+ \frac{1}{24} a^2+ \frac{5}{24} a^3-\frac{1}{24} a^4 + \frac{1}{120} a^5 .$$

There's a $u_i$, and then there's an $a_i$. Make up your mind! — J. M. ain't a mathematician, Nov 24 '11 at 10:48
$S_n$ is expressible in terms of the Gaussian hypergeometric function. Is that what you want? (I've gotten leery of posting closed form special function solutions after some past answers...) — J. M. ain't a mathematician, Nov 24 '11 at 11:22
$u_i$ is just the binomial coefficient a-choose-i, isn't it? So you're asking for the sum of the first $n$ terms in row $a$ of Pascal's triangle, right? Well, the first $n$ terms after $u_0=1$. — Gerry Myerson, Nov 24 '11 at 11:33
@J. M. : you mean $S_n$ is a polynomial with coefficients Pochhammer symbols wich in turn can be expressed using Gamma function ? — palio, Nov 24 '11 at 11:38
Gerry's right. $k!\binom{a}{k}=\prod_{j=0}^{k-1}(a-j)$. Alternatively you'll see it expressed as falling factorials... — J. M. ain't a mathematician, Nov 24 '11 at 11:51

explicit formula for a sum of sequence

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