Evaluating $\lim_{n \to \infty}\left({^n\mathrm{C}_0}{^n\mathrm{C}_1}\dots{^n\mathrm{C}_n}\right)^{\frac{1}{n(n+1)}}$

Question

$\lim_{n \to \infty}\left({^n\mathrm{C}_0}{^n\mathrm{C}_1}\dots{^n\mathrm{C}_n}\right)^{\frac{1}{n(n+1)}}$ is equal to:

a) $e$

b) $2e$

c) $\sqrt e$

d) $e^2$

Though it looks really innocent at first sight, it certainly isn't.

Attempt: It's $\infty^{\infty}$ form.

I had tried taking the product raised to the power $\frac{1}{n(n+1)}$ as function $f(n)$. Then I took logarithm of both sides to see if things simplifying. Even after factoring out the extra factorials it wasn't easy.

Note that ${^n\mathrm{C}_x} = \binom{n}{x}$.

Answer 1

Use AM-GM inequality as a bit of trick, $${(C_0^nC_1^nC_2^n \ldots C_{n-1}^nC_n^n)}^{\frac{1}{n+1}} \leq \frac{C_0^n+C_1^n+C_2^n +\ldots +C_{n-1}^n+C_n^n}{n+1}=\frac{2^n}{n+1}$$ $$\lim_{n\rightarrow \infty}{(C_0^nC_1^nC_2^n \ldots C_{n-1}^nC_n^n)}^{\frac{1}{n(n+1)}} \leq \lim_{n\rightarrow \infty}\frac{2}{\sqrt[n]{n+1}}= 2$$ So the only possible answer is c).

Or $$\lim_{n\rightarrow \infty}{(C_0^nC_1^nC_2^n \ldots C_{n-1}^nC_n^n)}^{\frac{1}{n(n+1)}} \leq \lim_{n\rightarrow \infty}((2^n)^{n+1})^{\frac{1}{n(n+1)}}= 2$$

Answer 2

Using Stolz theorem 2 times: $$\lim_{n\to\infty} \ln[(\binom{n}{0}\cdot...\binom{n}{n})^{1/(n(n+1))}] = \lim_{n\to\infty} \frac{\ln(\binom{n}{0}\cdot...\cdot\binom{n}{n})}{n^2+n} =\\ \lim_{n\to\infty} \frac{\ln(\binom{n+1}{0}\cdot...\cdot\binom{n+1}{n+1})-\ln(\binom{n}{0}\cdot...\cdot\binom{n}{n})}{(n+1)^2+(n+1)-(n^2+n)} = \lim_{n\to\infty} \frac{\ln(\frac{(n+1)^{n-1}}{n!})}{2n} =\\ \lim_{n\to\infty} \frac{\ln(\frac{(n+2)^n}{(n+1)!})-\ln(\frac{(n+1)^{n-1}}{n!})}{2} = \lim_{n\to\infty} \frac{\ln(\frac{n+2}{n+1})^n}{2} = \frac{1}{2} $$ So that the limit is $e^{\frac{1}{2}}$

Answer 3

$$\prod_{k=0}^{n}\binom{n}{k}=\frac{n!^{n}}{\prod_{k=0}^{n}k!^2}=\frac{n!^n}{\left[\prod_{k=1}^{n}k^{n+1-k}\right]^2}=\frac{n!^n}{n!^{2n+2}}\prod_{k=1}^{n}k^{2k} \tag{1}$$ hence the outcome depends on the asymptotic behaviour of the hyperfactorial.
Since by Riemann sums $$ \lim_{n\to +\infty} \frac{1}{n}\sum_{k=1}^{n}\log\frac{k}{n}=\int_{0}^{1}\log(x)\,dx = -1, $$ $$ \lim_{n\to +\infty} \frac{1}{n}\sum_{k=1}^{n}\frac{k}{n}\log\frac{k}{n}=\int_{0}^{1}x\log(x)\,dx = -\frac{1}{4}\tag{2} $$ we have

$$ \frac{1}{n^2}\log\prod_{k=0}^{n}\binom{n}{k} = \frac{1}{n^2}\left[2\sum_{k=1}^{n}k\log k-(n+2)\log n!\right]\to \frac{1}{2}\tag{3} $$ and the correct option is c).