# Techniques for summing ratio of binomial coefficients

There are several identities that involve the sum of the product of binomial coefficients. However what I am searching for is an identity that involves the ratio of binomial coefficients. Specifically, I want to find a closed form expression for the sum $$\sum_{k=r-t}^n\frac{\binom{n}{k}}{\binom{k+t}{r}},$$ where $n,r \in \mathbb{N}$ are fixed and $t$ is nonpositive and fixed.

Are there any standard formulae/identities that give this or are there methods for finding this sum?

I have restricted $t$ to be nonpositive, because the case where $t$ is positive is comparatively easier (in particular, $r=1, t=1$ is straight-forward to evaluate).

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## migrated from cstheory.stackexchange.comSep 1 '12 at 7:09

This question came from our site for theoretical computer scientists and researchers in related fields.

Mods, since I haven't even received a comment over the last two days, would it be ok to re-post this question now on MathOverflow? – Ankur Aug 31 '12 at 17:46
Moderators are not checking all comments to all questions. If you have a question to moderators, it is more reliable to flag your own question, choose “it needs ♦ moderator attention,” and explains why it needs moderator attention. – Tsuyoshi Ito Sep 1 '12 at 0:42
At least for me, this question is fairly uninteresting because it looks like just a random formula, and I fail to see why it is reasonable to expect that it has a nice closed-form expression (and also because I am not good at this kind of math). If you state why you are interested in this sum, some people may care more. – Tsuyoshi Ito Sep 1 '12 at 0:43
Well, the motivation is hard to explain. It came up as an intermediate calculation. If its ok, can you help move it to math.SE? Otherwise I can merely repost it. – Ankur Sep 1 '12 at 4:11
Usually, if you have a ratio of binomial coefficients as the terms of a sum, a useful first step is to switch to hypergeometric form, and then do the manipulations on the resulting hypergeometric functions. – J. M. Sep 1 '12 at 7:54