# Weighted sum of ratio of partial sum of binomial coefficients

I would like to approximate the following sum when $n \rightarrow \infty$ and $n \gg k$,

$\sum_{x = k}^n \sum_{y > x}^n \frac{\sum_{m = 0}^{k - 1} {y - 2 \choose m}}{\sum_{m = 0}^k {y - 1 \choose m}}$.

The catch is that there is no closed form for partial sum of binomial coefficients. I wonder if there is a good way to approximate $\frac{\sum_{m = 0}^{k - 1} {y - 2 \choose m}}{\sum_{m = 0}^k {y - 1 \choose m}}$ so that the approximation converges to a value given that $n \rightarrow \infty$ and $n \gg k$. Any help is greatly appreciate.

Revision:

I re-thought about my original problem, I believe solving the following problem would make more sense. However it is still unclear to me how to approximate the formula.

Summing ratio of ratio of partial sums of binomial coefficients

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In your double sum, call the numerator of your expression s, which is short for s(y,k). Now the denominator is 2s + (y-1) choose k, which I will write as 2s + t. Note that for y > 3k, t>4s, and one can make the difference between t and s more precise. After one estimates the terms for y at most 3k, one is left with a double sum which will look like a sum of logs. Gerhard "Ask Me About System Design" Paseman, 2013.05.06 – Gerhard Paseman May 7 '13 at 5:30
It looks like the OP has posted a revised version as a new question, mathoverflow.net/questions/129931/… – Barry Cipra May 7 '13 at 12:18
(Note, the OP's new question doesn't have a sum over $x$, but the double sum here over $x$ and $y$ easily simplifies to a single sum over $y$.) – Barry Cipra May 7 '13 at 12:25
Yes I've wanted to revise this question to mathoverflow.net/questions/129931/…. Sorry about taking up the spaces. – ELW May 7 '13 at 17:02