Is the sequence $a_{n}= {{n}\choose{[n/2]}}^{1/n}$ increasing?

Question

Let the sequence $a_{n}= {{n}\choose{[n/2]}}$$^{1/n}$ for every $n\in\mathbb{N}$, where $[x]=\max\{k \in \mathbb{Z} | k\leq x\}$ the floor function. How can I show that this sequence is increasing? I have tried using logarithms and gamma functions (for derivative) but I didn't get far. Any ideas?

Answer 1

You can use the identity $k\cdot \binom{m}{k} = m\cdot \binom{m-1}{k-1}$ and apply induction over n by showing $a_{n+2} \ge a_n$.

For $n=2 $ & $ n=3:$$$ a_1 \le a_2 \le a_3 \Leftrightarrow \binom{1}{0} \le \sqrt{\binom{2}{1}} \le \binom{3}{1}^{1/3} \Leftrightarrow 1 \le \sqrt{2} \le 3^{1/3}$$

Given that the assumption is proven for n, we now need to show $a_{n+2} \ge a_n$. First reshape $a_{n+2}$: $$ a_{n+2} = \binom{n+2}{[\frac{n}{2}+2]}^{\frac{1}{n+2}} = \bigg(\frac{[n/2]+1}{[n/2]+1}\binom{n+2}{[\frac{n}{2}]+1}\bigg)^{\frac{1}{n+2}} = \bigg((n+2) \binom{n+1}{[\frac{n}{2}]}\bigg)^{\frac{1}{n+2}} = \bigg((n+2)\frac{(n+1)}{(n+1) - [n/2])} \binom{n}{[\frac{n}{2}]}\bigg)^{\frac{1}{n+2}} $$

Now check whether the inequality holds: $$ a_{n+2} \ge a_n\\ \Leftrightarrow \bigg((n+2)\frac{(n+1)}{(n+1) - [n/2])} \binom{n}{[\frac{n}{2}]}\bigg)^{\frac{1}{n+2}} \ge \binom{n}{[n/2]}^\frac{1}{n} \\\Leftrightarrow \bigg(\bigg((n+2)\frac{(n+1)}{(n+1) - [n/2])} \binom{n}{[\frac{n}{2}]}\bigg)^{\frac{1}{n+2}} \bigg)^{n+2} \ge \bigg(\binom{n}{[n/2]}^\frac{1}{n}\bigg)^{n+2} \\ \Leftrightarrow (n+2)\frac{(n+1)}{(n+1) - [n/2])} \binom{n}{[\frac{n}{2}]} \ge \binom{n}{[n/2]} \cdot \binom{n}{[n/2]}^\frac{2}{n} \\ \Leftrightarrow (n+2)\frac{(n+1)}{(n+1) - [n/2])} \ge \binom{n}{[n/2]}^\frac{2}{n} $$

In this last inequality, the right hand side is smaller than 4, because $$ a_n \le \bigg(\sum_{i=0}^n \binom{n}{i}\bigg)^{1/n} = (2^n-1)^{1/n} \le (2^n)^{1/n} = 2 $$

And because in the left hand side in the inequality $(n+2) \ge 4$ for $n\ge 2$, and because the other factor is $\ge 1$, the statement is proofen.

Answer 2

Use a stronger version of $$\left( \frac n k \right)^k < \binom n k $$ for $k < n$.

Alternatively, by Stirling's approximation $$ \binom {2n} n \sim \frac{4^n}{\sqrt{\pi n}} $$