Finding $\lim_{n\rightarrow \infty} \sqrt[n]{\binom{\alpha n}{n}}$

Question

I've seen in my book a solution to the case $\alpha=2$ by showing $a_n=\binom{2 n}{n}=\prod_{i=1}^n\frac{(2i-1)(2i)}{i^2}(=b_i)$, so $\lim \sqrt[n]{a_n}=\lim b_n=4$

for $\alpha=3$, by denoting $a_n=\binom{3n}{n}$ they show $\frac{a_{n+1}}{a_n}=\frac{(3n+1)(3n+2)(3n+2)}{(n+1)(2n+1)(2n+2)}\rightarrow \frac{27}{4}$ which is also the limit of $\sqrt[n]{a_n}$ (there was a question before that proved $\lim_{n \rightarrow \infty} \sqrt[n]{a_n} = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$ prior).

Is there some generalization of this for all $\alpha\in\mathbb{N}$?

Answer 1

Note that by Stirling's formula, $n! \sim (\frac n e)^n\sqrt{2\pi n}$
Therefore, $$\begin{align} {an \choose n}&\sim \frac {(\frac{an}e)^{an}\sqrt{2\pi an}} {(\frac{n}e)^{n}\sqrt{2\pi n}(\frac{(a-1)n}e)^{(a-1)n}\sqrt{2\pi (a-1)n}}\\ &\sim (\frac a {a-1})^{an}(a-1)^n\sqrt{\frac a {2\pi n(a-1)}} \end{align} $$ Since $\lim_{n \to \infty}({\frac a {2\pi n(a-1)}})^{\frac 1 {2n}} =1$, $$\lim_{n \to \infty}{an \choose n}^{{\frac 1 n}}=(a-1)(\frac a {a-1})^a$$

Answer 2

Yes, there is a general result for $\alpha\in\mathbb{N}$.

We have that $$\frac{a_{n+1}}{a_n}= \frac{(\alpha n)^{\alpha}+o(n^{\alpha})}{(n+1)\cdot\left(((\alpha-1)n)^{\alpha-1}+o(n^{\alpha-1})\right)} =\frac{\alpha^{\alpha}+o(1)}{(\alpha-1)^{\alpha-1}+o(1)}\to \frac{\alpha^{\alpha}}{(\alpha-1)^{\alpha-1}}$$ where $a_n=\binom{\alpha n}{n}$. Hence $$\lim_{n\rightarrow \infty} \sqrt[n]{\binom{\alpha n}{n}} =\lim_{n\rightarrow \infty} \sqrt[n]{a_n} =\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_n} =\frac{\alpha^{\alpha}}{(\alpha-1)^{\alpha-1}}. $$

Answer 3

In the same spirit as Aforest's answer $$A_n=\binom{a n}{n}^{\frac{1}{n}}\implies \log(A_n)=\frac 1 n \left(\log ((a n)!)-\log (((a-1) n)!)-\log(n!) \right)$$ Now, apply Stirling approximation as Aforest did that is to say $$\log(m!)\approx m (\log (m)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left(m\right)\right)$$

Then use Taylor series for infinitely large values of $n$ to get, after a few simplifications, $$\log(A_n)\approx a \log \left(\frac{a}{a-1}\right)+\log(a-1)+\frac{\log \left(\frac{a}{2 \pi (a-1) n}\right)}{2 n}$$ which gives the limit and also how it is approached.

This seems to work quite well even for small values of $n$ as shown in the following table for a few values of $a$ $$\left( \begin{array}{cccc} a & n & A_n\text{(approx)} & A_n\text{(exact)} \\ 2 & 10 & 3.36669 & 3.36248 \\ 2 & 15 & 3.51791 & 3.51596 \\ 2 & 20 & 3.60666 & 3.60554 \\ \cdots \\ 3 & 10 & 5.60015 & 5.59471 \\ 3 & 15 & 5.87983 & 5.87729 \\ 3 & 20 & 6.04263 & 6.04116 \\ \cdots \\ 4 & 10 & 7.82013 & 7.81308 \\ 4 & 15 & 8.22682 & 8.22352 \\ 4 & 20 & 8.4629 & 8.46099 \\ \cdots \\ 5 & 10 & 10.0357 & 10.0269 \\ 5 & 15 & 10.5689 & 10.5648 \\ 5 & 20 & 10.8781 & 10.8757 \\ \cdots \\ 6 & 10 & 12.2492 & 12.2387 \\ 6 & 15 & 12.9089 & 12.9039 \\ 6 & 20 & 13.2910 & 13.2881 \end{array} \right)$$

Very interesting problem, indeed !

Answer 4

We have $$ \binom{\alpha n}{n}=\frac{\Gamma\left(\alpha n+1\right)}{\Gamma\left((\alpha-1)n+1\right)n!} $$ so applying Gautschi's Inequality, we get $$ \begin{align} \frac{\binom{\alpha n+\alpha}{n+1}}{\binom{\alpha n}{n}} &=\frac{\Gamma\left(\alpha n+\alpha+1\right)}{\Gamma\left((\alpha-1)n+\alpha\right)(n+1)!}\frac{\Gamma\left((\alpha-1)n+1\right)n!}{\Gamma\left(\alpha n+1\right)}\\ &=\frac1{n+1}\frac{\Gamma\left(\alpha n+\alpha+1\right)}{\Gamma\left(\alpha n+1\right)}\frac{\Gamma\left((\alpha-1)n+1\right)}{\Gamma\left((\alpha-1)n+\alpha\right)}\\ &\sim\frac1n(\alpha n)^\alpha((\alpha-1)n)^{1-\alpha}\\ &=\frac{\alpha^\alpha}{(\alpha-1)^{\alpha-1}} \end{align} $$