Is the sequence $a_{n}= {{n}\choose{[n/2]}}^{1/n}$ increasing? Let the sequence $a_{n}= {{n}\choose{[n/2]}}$$^{1/n}$ for every $n\in\mathbb{N}$, where $[x]=\max\{k \in \mathbb{Z} | k\leq x\}$ the floor function. How can I show that this sequence is increasing? I have tried using logarithms and gamma functions (for derivative) but I didn't get far. Any ideas?
 A: You can use the identity $k\cdot \binom{m}{k} = m\cdot \binom{m-1}{k-1}$ and apply induction over n by showing $a_{n+2} \ge a_n$.
For $n=2 $ & $ n=3:$$$ 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
$$
And because in the left hand side in the inequality $(n+2) \ge 4$ for $n\ge 2$, and because the other factor is $\ge 1$, the statement is proofen.
A: Use a stronger version of $$\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}}  $$
