Asymptotically $\binom{2n}{n}\approx\frac{2^{2n}}{\sqrt{\pi n}}$ or $\binom{2n}{n}\approx2^{2n}$? The Wikipedia entry about the binomial coefficient has the following (asymptotic) estimation for $n$ and $k$ that grow in the same rate (under "Bounds and asymptotic formulas")
$$\log\binom{n}{k}\approx nH_{2}\left(\frac{k}{n}\right)$$
where $H_{2}\left(\alpha\right)=-\alpha\log\left(\alpha\right)-\left(1-\alpha\right)\log\left(1-\alpha\right)$ is the binary entropy function
Which would imply
$$\binom{2n}{n}\approx2^{2nH_{2}\left(\frac{1}{2}\right)}=2^{2n}$$
However, the same paragraph has the following asymptotic estimation of the central binomial coefficient
$$\binom{2n}{n}\approx\frac{2^{2n}}{\sqrt{\pi n}}$$
Aren't these two contradictory? If not, why not and if so, which one is correct?
 A: You're using the approximation symbol $\approx$ where Wikipedia uses the asymptoticity symbol $\sim$. The former is an informal way of saying that two expressions are approximately equal without specifying in which way or to what extent. By contrast, the latter makes a well-defined statement about the two sides: $f(x)\sim g(x)$ is defined to mean
$$
\lim\frac{f(x)}{g(x)}=1\;,
$$
where a particular limit process, typically $x\to\infty$, is specified or assumed from the context.
The two asymptotic statements that you cite, spelled out properly including the limit process, are
$$
\binom{2n}n\sim_{n\to\infty}\frac{2^{2n}}{\sqrt{\pi n}}
$$
and
$$
\log\binom n{\lambda n}\sim_{n\to\infty}-n\left(\lambda\log\lambda+(1-\lambda)\log(1-\lambda)\right)
$$
(where the logarithm can be taken for any base).
There is no contradiction because
$$
\log_2\left(\frac{2^{2n}}{\sqrt{\pi n}}\right)=2n-\frac12\log_2(\pi n)\sim_{n\to\infty}2n
$$
since
$$
\lim_{n\to\infty}\frac{\log n}n=0\;.
$$
By contrast, omitting the factor $\frac1{\sqrt{\pi n}}$ in the first result without taking logarithms would make it false.
