Let $$ S_k=\sum_{j=0}^{\alpha k}(-1)^j\binom{k/2}{j}\binom{\alpha k^2-j k}{k} $$ where $\alpha\in(0,1/2)$ is a constant. I'm interested in understanding the asymptotic behaviour of $S_k$.

It would be sufficient for my purposes to determine $c_\alpha=\lim_{k\to\infty} ((\log S_k)/k-\log k)$, which I believe should be something like $1+\log\alpha-h(\alpha)$ for some positive function $h$. Even an upper bound of this form would be useful.

I've no idea if this is tractable. None of the (probably very naive) ideas I've tried so far have achieved anything useful. Any pointers would be appreciated.

  • $\begingroup$ This is an alternating sum of a log-concave (hence unimodal) sequence. The first term is already way larger than what you’re hoping for, and I suspect there’s not too much cancelation here. What makes you say the sum is positive? $\endgroup$
    – Pat Devlin
    Commented Apr 22, 2020 at 22:57
  • $\begingroup$ It's positive because it's the coefficient of $z^{\alpha k^2}$ in the generating function $z^k (1-z^k)^{k/2} / (1-z)^{k+1}$, which has positive coefficients. My expectation is that there's lots of cancellation (as in the expansion for $e^{-x}$). $\endgroup$ Commented Apr 23, 2020 at 7:57

1 Answer 1


If you can find the maximum of the terms, the order of the growth of $S_k$ will be near the order of growth of the maximal term, unless you have a lot of cancellation.

To know where to start looking, let's make some rough simplifications. Looking at the simpler binomial factor in the product, as $j$ increases, that term goes up by less than a factor of $k/j$ (closer to $k/j - 1$ actually, but we are doing rough estimates now). The more complicated term increases by an amount that we can approximate by $(1 - 1/(\alpha k - j))^k$, which is not quite $e^{-1/\alpha}$, but let's pretend it is. These rates of contrary growth will cancel when $k/j$ is like $e^{1/\alpha}$, so the maximal term in the sum should have $j$ near $k/e^{1/\alpha}$, which is close enough to zero that neighboring terms will decay predictably.

Of course you will need to refine these estimates, probably to find the maximal term occurs for slightly smaller $j$, but even on the condition that the maximal term occurs at $j_0$, you should be able to estimate the decay of adjoining terms to get an asymptotic.

Gerhard "Now Go Do The Algebra" Paseman, 2020.04.22.

  • $\begingroup$ The maximum term occurs when $j$ satisfies $\log(k/2j-1)=k/(\alpha k-j)$. So $j/k$ is slightly smaller than $1/(2+2e^{1/\alpha})$. However, I'm sure there is lots of cancellation. $\endgroup$ Commented Apr 23, 2020 at 14:15
  • $\begingroup$ Then the order of growth will be a bit smaller. Have you computed S_k for many k and a fixed value of alpha? How does the size of S_k compare to the largest term? Perhaps you can bound the size of (the difference of) two consecutive terms. Gerhard "Time For A Reality Check?" Paseman, 2020.04.23. $\endgroup$ Commented Apr 23, 2020 at 14:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.