Asymptotics of the Central Binomial Coefficient

The coefficient

$$\binom{2n}{n}$$

is the largest binomial coefficient in the expansion of $(1+x)^{2n}$. It appears throughout combinatorics, probability, number theory, and analysis. Although it is an integer defined exactly by factorials, its size for large $n$ is more usefully described by an asymptotic formula.

The main result is

$$\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}},$$

where $a_n \sim b_n$ means that $a_n/b_n \to 1$ as $n \to \infty$.

Derivation from Stirling’s formula

Start with the factorial expression:

$$\binom{2n}{n} = \frac{(2n)!}{(n!)^2}.$$

Stirling’s formula gives

$$k! \sim \sqrt{2\pi k}\left(\frac{k}{e}\right)^k.$$

Apply it to $(2n)!$ and $n!$:

$$(2n)! \sim \sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}$$

and

$$(n!)^2 \sim \left[\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right]^2 = 2\pi n\left(\frac{n}{e}\right)^{2n}.$$

Taking the ratio gives

$$\begin{aligned} \binom{2n}{n} &\sim \frac{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}} {2\pi n\left(\frac{n}{e}\right)^{2n}} \\ &= \frac{\sqrt{4\pi n}}{2\pi n}\,2^{2n} \\ &= \frac{4^n}{\sqrt{\pi n}}. \end{aligned}$$

The factors involving $n/e$ cancel except for the factor $2^{2n}=4^n$. The remaining square-root factor produces the denominator $\sqrt{\pi n}$.

A sharper asymptotic expansion

The leading equivalent is often enough, but the Stirling expansion provides corrections:

$$\binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} \left(1-\frac{1}{8n}+\frac{1}{128n^2}+O\left(\frac{1}{n^3}\right)\right).$$

In particular, retaining only the first correction gives

$$\binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} \left(1-\frac{1}{8n}+O\left(\frac{1}{n^2}\right)\right).$$

This says that the basic approximation $4^n/\sqrt{\pi n}$ is slightly too large for positive $n$, because its first relative correction is negative.

To see where this comes from, use the logarithmic form of Stirling’s expansion:

$$\log(k!) = \left(k+\frac12\right)\log k-k+\frac12\log(2\pi)+\frac{1}{12k}+O\left(\frac{1}{k^3}\right).$$

Substituting $k=2n$ and $k=n$, then forming $\log(2n)!-2\log(n!)$, yields

$$\log\binom{2n}{n} = n\log 4-\frac12\log(\pi n)-\frac{1}{8n}+O\left(\frac{1}{n^3}\right).$$

Exponentiating produces the correction expansion above.

Worked example: $n=10$

The exact value is

$$\binom{20}{10}=184756.$$

The leading asymptotic approximation is

$$\frac{4^{10}}{\sqrt{10\pi}} = \frac{1,048,576}{\sqrt{10\pi}} \approx 187079.$$

This is close, but it is about $1.26\%$ larger than the exact answer. Including the first two correction terms gives

$$\frac{4^{10}}{\sqrt{10\pi}} \left(1-\frac{1}{80}+\frac{1}{12800}\right) \approx 184755,$$

which differs from the exact value by roughly one unit after rounding. The correction terms become increasingly effective as $n$ grows.

What the formula means about growth

The dominant factor is $4^n$, so the central binomial coefficient grows exponentially. The factor $1/\sqrt{\pi n}$ is a slowly varying reduction. Thus

$$\binom{2n}{n}=\Theta\left(\frac{4^n}{\sqrt n}\right).$$

This is stronger information than simply saying the coefficient is $O(4^n)$. It identifies the precise polynomial factor separating the central coefficient from $4^n$.

There is also a useful probabilistic interpretation. For a fair coin, the probability of exactly $n$ heads in $2n$ tosses is

$$\frac{\binom{2n}{n}}{2^{2n}} = \frac{\binom{2n}{n}}{4^n}.$$

Therefore the asymptotic result says

$$\frac{\binom{2n}{n}}{4^n} \sim \frac{1}{\sqrt{\pi n}}.$$

So the probability of exactly half heads and half tails decreases like $n^{-1/2}$, rather than exponentially.

Common mistakes

Confusing $\binom{2n}{n}$ with $\binom{n}{2}$

These are different quantities. The central binomial coefficient is

$$\binom{2n}{n}=\frac{(2n)!}{n!\,n!},$$

whereas $\binom{n}{2}=n(n-1)/2$ grows only quadratically.

Dropping the square-root factor

It is not correct to write $\binom{2n}{n}\sim 4^n$. The ratio of the two sides behaves like $1/\sqrt{\pi n}$ and tends to zero. The factor $\sqrt{n}$ is essential.

Treating $\sim$ as equality

The notation $\sim$ describes a limiting ratio, not an exact identity. For finite $n$, use the factorial formula or a correction expansion when numerical accuracy matters.

Using Stirling’s formula inconsistently

Applying Stirling’s approximation only to $(2n)!$, while leaving $n!$ exact, does not produce the standard asymptotic simplification. Both factorials must be handled consistently, or the resulting error must be tracked separately.