Asymptotic approximation of a convolution of binomial coefficients

Question

I would like to find the following limit which is somewhat similar to the usual Vandermonde's convolution for binomial coefficients. Define $L$ as follows.

$$ L \triangleq \lim_{N\to\infty} \frac{1}{2^{2N-2}\log N} \sum_{k=1}^{N} \log k \binom{2k-2}{k-1}\binom{2N-2k}{N-k}\, . $$

Note that if we simply use $\log N$ to bound $\log k$ for all summands, we can apply the Vandermonde's identity to obtain $L\le 1$. However, from numerical computations, we observe that $L$ is less than $0.8$. Show that $L=1$.

I've tried methods from Sedgewick and Flajolet's text (https://aofa.cs.princeton.edu/40asymptotic/), but failed miserably. Any help will be appreciated. :)

Edit: Following Brendan's reply, we realize that $L=1$. See replies below for the detailed derivations. Also, the summand is modified slightly to look cleaner.

Answer 1

As $k\to\infty$, $\binom{2k-2}{k-1}\sim 2^{2k-2}/\sqrt{\pi k}$. As $N-k\to\infty$, $\binom{2N-2k}{N-1}\sim 2^{2N-2k}/\sqrt{\pi (N-k)}$.

Now approximate the sum by an integral: $$\int_0^N \frac{\ln k}{\pi\sqrt{k(N-k)}\ln N}\,dk =\frac{\pi \ln N -2\pi -C}{\pi\ln N} \to 1,$$ where $C$ is a particular hypergeometric constant about 1.928.

It remains to justify the approximation, which I will leave to others.

Answer 2

Allow me to fill the details of Brendan's derivations.

We fix $0<\epsilon,\delta<1$ and show that \begin{equation} \frac{1}{2^{2N-2}}\sum_{k=1}^N \log k \binom{2k-2}{k-1}\binom{2N-2k}{N-k} \ge (1-\epsilon) \left(-2+ \log N\left(1 - \frac{2\sin^{-1}\sqrt{\delta}}{\pi}\right)\right) \end{equation} for sufficiently large $N$. Hence, choosing small $\epsilon$ and $\delta$, we have that the limit $L$ is one, as required.

To this end, we use Stirling's approximation $\binom{2N}{N} \sim \frac{2^{2N}}{\sqrt{\pi N}}$. In other words, for any $\epsilon > 0$, there exists $N_0$ such that \begin{equation} \binom{2N}{N} \geq (1-\epsilon)\frac{2^{2N}}{\sqrt{\pi N}} \text{ for $N\ge N_0$} \,. \end{equation} Given $\delta>0$, we then choose $M$ such that $\delta M\ge N_0 + 1$. Hence, for all $N\ge M$ and $\delta N\le k\le (1-\delta) N$, we have that $k-1 \ge \delta M-1\ge N_0$, and so, $\binom{2k-2}{k-1} \geq (1-\epsilon)\frac{2^{2k-2}}{\sqrt{\pi (k-1)}}$. Similarly, $N-k\ge \delta N \ge \delta M \ge N_0$ and so, $\binom{2N-2k}{N-k} \geq (1-\epsilon)\frac{2^{2N-2k}}{\sqrt{\pi (N-k)}}$. Therefore, we have that \begin{align*} \sum_{k=1}^N \log k \binom{2k-2}{k-1}\binom{2N-2k}{N-k} & \ge \sum_{k=\delta N}^{(1-\delta) N} \log k \Big(1-\epsilon\Big) \frac{2^{2N-2}}{\pi\sqrt{ (k-1)(N-k)}}\\ & = \frac{2^{2N-2}(1-\epsilon)}{\pi} \sum_{k=\delta N}^{(1-\delta) N} \frac{\log k}{\sqrt{ (k-1)(N-k)}}\,. \end{align*}

Now, we estimate the sum using an integral. That is, \begin{align*} \sum_{k=\delta N}^{(1-\delta) N} \frac{\log k}{\sqrt{ (k-1)(N-k)}} & =\sum_{k=\delta N}^{(1-\delta)N} \frac{\log k}{N \sqrt{(k-1)/N (1 - k /N)}}\\ &\sim \int_{\delta}^{1-\delta} \frac{\log(\lambda N)}{\sqrt{\lambda(1-\lambda)}} \,d\lambda \\ &= \int_{\delta}^{1-\delta} \frac{\log N}{\sqrt{\lambda(1-\lambda)}} \,d\lambda +\int_{\delta}^{1-\delta} \frac{\log \lambda}{\sqrt{\lambda(1-\lambda)}} \,d\lambda\\ &\ge \int_{\delta}^{1-\delta} \frac{\log N}{\sqrt{\lambda(1-\lambda)}} \,d\lambda +\int_{0}^{1} \frac{\log \lambda}{\sqrt{\lambda(1-\lambda)}} \,d\lambda\,. \end{align*}

Now, the last inequality follows from the fact that $\log \lambda$ is negative in the interval $(0,1)$ and we have that $\int_{0}^{1} \frac{\log \lambda}{\sqrt{\lambda(1-\lambda)}} \,d\lambda = -2\pi$. Moreover, $\int_{\delta}^{1-\delta} \frac{\log N }{\sqrt{\lambda(1-\lambda)}} \,d\lambda= (\pi-2\sin^{-1}\sqrt{\delta})\log N$. The desired inequality then follows from standard algebraic manipulations.