## On the geometric mean of the binomial coefficients

Today we prove a remarkable asymptotic result regarding the binomial coefficients. Namely, let $G_n=\left(\prod_{k=1}^n\binom{n}{k}\right)^{\tfrac{1}{n+1}}$. We will show

$\displaystyle \lim_{n\to\infty}G_n^{\frac{1}{n}}=\sqrt{e}.$

Before getting to the proof, I should say that this is most definitely not my creation; I believe the first publication of this proof may be credited to Polya (but I am not sure). In any case, I was shown this proof at MSU in my Putnam training seminar. It was presented by Dr. Ignacio Uriarte-Tuero. Any unintended errors introduced are mine alone.

To start, we use the definition of $\binom{n}{k}$ to write

\begin{aligned} G_n^{n+1}&=\prod_{k=1}^n\binom{n}{k}\\&=\frac{(n!)^{n+1}}{\prod_{k=1}^n(k!)^2}\end{aligned}

Now for $1\leq j\leq n$, we want to examine the power of $j$ which appears in the fraction. So, for instance, $n$ appears $n+1$ times in the numerator and twice in the denominator from $(n!)^2$, $n-1$ appears $n+1$ times in the numerator and four times in the denominator from $(n!)^2$ and $((n-1)!)^2$, etc. We get the formula

$G_n^{n+1}=\prod_{k=1}^n(n+1-k)^{n+1-2k}$

We now multiply by $1$ in a clever way. Noting that $\sum_{k=1}^n(n+1-2k)=0$ we obtain

$G_n^{n+1}=\prod_{k=1}^n(n+1-k)^{n+1-2k}=\prod_{k=1}^n\left(\frac{n+1-k}{n+1}\right)^{n+1-2k}$

as the denominator has total exponent $0$. We can now write

$G_n=\prod_{k=1}^n\left(\frac{n+1-k}{n+1}\right)^{1-\frac{2k}{n+1}}=\prod_{k=1}^n\left(1-\frac{k}{n+1}\right)^{1-\frac{2k}{n+1}}$

We want to examine $G_n^{\frac{1}{n}}$. Taking logarithms, we get

$\frac{1}{n}\ln(G_n)=\frac{1}{n} \sum_{k=1}^n(1-\frac{2k}{n+1})\ln(1-\frac{k}{n+1})$

We now recognize this expression as a Riemann sum of the function $g(x)=(1-2x)\ln(1-x)$ on the interval ${}[0,1]$. (Ok, it is not exactly a Riemann sum; we should be dividing by $n+1$, not $n$. But in the limit this will not matter.) Taking limits, it follows that

$\lim_{n\to\infty}\frac{\ln(G_n)}{n}=\lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^n(1-\frac{2k}{n+1})\ln(1-\frac{k}{n+1})=\int_0^1(1-2x)\ln(1-x)dx$

Evaluating the integral is now easy via integration by parts; it turns out that

$\int_0^1(1-2x)\ln(1-x)dx=\frac{1}{2}$

and thus $\lim_{n\to\infty}\ln(G_n^{\frac{1}{n}})=\frac{1}{2}$ and $\lim_{n\to\infty} G_n^{\frac{1}{n}}=\sqrt{e}$.

And we’re done! As always, thanks for reading!