Is there a discrete version of de l'Hôpital's rule?

Stolz–Cesàro seems to be what you're looking for. There are two forms:

1.

Let $a_n$ and $b_n$ be two sequences approaching $0$ as $n\to\infty$, with $b_n$ decreasing. Then,

$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$

if the second limit exists.

2.

Let $a_n$ and $b_n$ be two sequences, with $b_n$ unbounded and increasing. Then,

$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$

if the second limit exists.

The discrete version of L'Hôspital's rule, in my opinion, is Abelian theorems, including the L'Hôspital's rule, Silverman-Toeplitz theorem and its sepcial case, Stolz-Cesàro theorem.

On de Bruijn's Asymptotic methods in analysis, it's said that

A theorem which derives asymptotic information about some kind of average of a function from asymptotic information about the function itself, is called an Abelian theorem. If one can find a supplementary condition under which the converse of an Abelian theorem holds, then this condition is called a Tauberian condition, and the converse theorem is called a Tauberian theorem.

The amount you gave, as far as I've tried, couldn't be easily solved with these theorems.

Let $$a_n=\frac{4^n}{\binom{2n}n\sqrt n}$$ we have $$\ln a_{n+1}-\ln a_n=\frac12\ln(n+1)-\ln(n+\frac12)+\frac12\ln n=-\frac1{4n^2}+O\left(\frac1{n^3}\right)\tag1$$ Therefore $\ln a_n$ converges as $n\to\infty$. However, the preceding equation, the asymptotic behavior of the difference, is not enough to determine the limit value, even if the result is refined. Such efforts are generally unsuccessful.

However, if $S=\lim_{n\to\infty}\ln a_n$, we could determine the asymptotic behavior of $a_n-S$ through (1) easily, since $\ln a_n=S-\sum_{k\ge n}(\ln a_k-\ln a_{k+1})$.

Remark One could determine $S$ through Stirling's formula. There's another approach, more elementary, I think:

$$a_n^2=\left(\frac{(2n-1)!!}{(2n)!!}\right)^2(2n+1)\cdot\frac n{2n+1}\to\frac1\pi$$

by Wallis product, therefore $S=1/\sqrt\pi$.