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.



Asymptotic approximation of a convolution of binomial coefficients
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.


