TITLE: Summing ratio of ratio of partial sums of binomial coefficients
QUESTION [5 upvotes]: I would like to approximate the following when $n \gg k$.
$\sum_{y = k + 1}^n \frac{\sum_{m = 0}^{k - 1} {y - 2 \choose m} (y - 1)}{\sum_{m = 0}^k {y - 1 \choose m}}.$
The formula can be re-written as
$\sum_{y = k + 1}^n \frac{(y - 1) + \sum_{m = 1}^{k - 1} {y - 1 \choose m + 1} (m + 1)}{\sum_{m = 0}^k {y - 1 \choose m}}.$
However since the partial sum of binomial coefficient does not closed form, I could not see any way to further simplify the formula. Any help is much appreciated. Thanks!

REPLY [3 votes]: $$ \sum_{m=0}^k \binom{z+1}{m} = \sum_{m=0}^k [  \binom{z}{m-1} + \binom{z}{m} ]
= \binom{z}{k} + 2 \sum_{m=0}^{k-1}\binom{z}{m},$$ when $z \geq k$.  So the sum
can be rewritten as
$$ \sum_{z=k}^{n-1} \frac{z}{2 + \binom{z-1}{k}/\sum_{m=0}^{k-1}\binom{z-1}{m}}. $$   Let's
call this summand $a_z$ .  Note that $a_k = k/2, a_{k+1} = (k+1)/(2+1/(2^k - 1))$, and for small
values of $j$, $a_{k+j} = (k+j)/(2 + \binom{k+j-1}{j-1}/(2^{k+j-1}-\sum_{m=0}^{j-1} \binom{k+j-1}{m}))$
.  Now as z increases, the denominator of $a_z$ eventually tends monotonically to $(z+k)/k$, which means
$a_z$ tends to $k$ from below.  Thus the entire sum has an upper bound (for $n \gg k$) of $(n-k)k$.
Further, for sufficiently large integers $l$, $a_{lk}$ is slightly larger than $lk/(l+1)$, so I imagine the actual sum differs from the upper bound by
a $O(n\log{k}) $ amount.
Gerhard "Ask Me About System Design" Paseman, 2013.05.08