MHB Find the least positive integer

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The least positive integer \( k \) such that \( {2n \choose n}^{\frac{1}{n}} < k \) for all positive integers \( n \) is determined to be 4. The discussion shows that \( {2n \choose n} < 4^n \) through induction, establishing that \( {2n \choose n}^{\frac{1}{n}} < 4 \). Additionally, it is noted that \( {34 \choose 17}^{\frac{1}{17}} > 3.006 \), confirming that \( k \) must be greater than 3. Therefore, the conclusion is that the least value of \( k \) is indeed 4. The discussion highlights the significance of combinatorial identities in deriving this result.
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Find the least positive integer $k$ such that $\displaystyle {2n\choose n}^{\small\dfrac{1}{n}}<k$ for all positive integers $n$.
 
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[sp]Since $${2n+2 \choose n+1} = \frac{(2n+2)(2n+1)}{(n+1)^2}{2n \choose n} = 4\frac{n+\frac12}{n+1}{2n\choose n} < 4{2n\choose n}$$, and $${2\choose 1} = 2 < 4$$, it follows by induction that $${2n\choose n} < 4^n$$ and therefore $${2n\choose n}^{\!\!1/n}<4.$$

On the other hand, $${34\choose 17}^{\!\!1/17} >3.006 >3.$$ So the least value of $k$ is $4$.[/sp]
 
Thanks, Opalg for participating and your solution! Your method is a nice one, I enjoy reading it!

Solution by other:
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 $k=4$.
 
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