It is known that the first order error term in the Shannon entropy formula for a binomial distribution is $1/n$ (for example, see the Wikipedia page Binomial distribution), where in the limit $n \to \infty$ the entropy of a binomial approaches the Gaussian one. There are quite technical papers which have calculated the entropy of a binomial distribution exactly; however, I would like to see this error correction $1/n$ to the Gaussian in a simpler way. So, I do the following.

Here, in Equation (4), the binomial distribution in the limit $n \to \infty$ has been written as $$f(x) = \frac{1}{\sqrt{2\pi npq}} e^{- (x - np)^2/2npq} \bigg[1+\mathcal{O}\Big(\frac{1}{\sqrt{n}}\Big)\bigg].$$.

In order the limit $n \to \infty$ to be well-defined and $f(x)$ doesn't vanish, we standardize $f(x)$ by defining $\mu \equiv np$, $\sigma \equiv \sqrt{npq}$, and $g(x) \equiv \sigma f(\sigma x+\mu)$, then we have: $$g(x) = \frac{1}{\sqrt{2\pi}} e^{- x^2/2} \bigg[1+\mathcal{O}\Big(\frac{1}{\sqrt{n}}\Big)\bigg],$$ where the term outside the bracket is a Gaussian with mean $0$ and standard deviation $1$.

Now, we can calculate the differential Shannon entropy of $g(x)$ as $S = -\int dx \, g(x) \ln g(x)$, which reads $$S = \frac{1}{2} \ln 2 \pi e + \mathcal{O}\Big(\frac{1}{\sqrt{n}}\Big).$$

The first term is correctly the entropy of the standard Gaussian but the error is of the order $1 / \sqrt{n}$. If we assume the result quoted in Wikipedia is correct, what is wrong with my approach?

**Edit 1 (Comment on answer 1)**

By the method of moment-generating functions, one can show that the standardized binomial distribution, $B(n, p)$, approaches the moment-generating function of the standard Gaussian, $\mathcal{N}(0,1)$, when $n \to \infty$. Thus, one expects the entropy of the former approaches $(1/2) \ln 2 \pi e$, as $n \to \infty$, I'm looking for the error terms of this expansion.