Find the least positive integer

[anemone]

Find the least positive integer $k$ such that $\displaystyle {2n\choose n}^{\small\dfrac{1}{n}}<k$ for all positive integers $n$.

 

[Opalg]

[sp]Since \(\displaystyle {2n+2 \choose n+1} = \frac{(2n+2)(2n+1)}{(n+1)^2}{2n \choose n} = 4\frac{n+\frac12}{n+1}{2n\choose n} < 4{2n\choose n}\), and \(\displaystyle {2\choose 1} = 2 < 4\), it follows by induction that \(\displaystyle {2n\choose n} < 4^n\) and therefore \(\displaystyle {2n\choose n}^{\!\!1/n}<4.\)

On the other hand, \(\displaystyle {34\choose 17}^{\!\!1/17} >3.006 >3.\) So the least value of $k$ is $4$.[/sp]

 

[anemone]

Thanks, Opalg for participating and your solution! Your method is a nice one, I enjoy reading it!

Solution by other:

Spoiler

Note that $\displaystyle {2n\choose n}<{2n\choose 0}+{2n\choose 1}+\cdots+{2n\choose 2n}=(1+1)^{2n}=4^n$

and for $n=5$, $\displaystyle {10\choose 5}=252>3^5$, we can conclude that $k=4$.