Papapetros Vaggelis wrote:Find which of the following real numbers is the greatest :
\(\displaystyle{1\,,\sqrt{2}\,,\sqrt[3]{3}\,,\sqrt[4]{4},...,\sqrt[n]{n}}\),
where \(\displaystyle{n\in\mathbb{N}}\) .
All time classic!
By distinguishing cases we have:
\( \color{gray} \bullet \) Let \( n =1 \) . Then this case is boring since we only have one number. Thus , in this case the only great number is \( 1 \).
\( \color{gray} \bullet \) Let \( n =2 \). This is case is also boring. It is immediately seen that \( \sqrt{2} > 1 \) and the result follows.
So, what is interesting to examine is the case \( n \geq 3 \). But , now by defining \( x^{1/x}, \; x \in [1, +\infty) \) we note that it differentiable and the derivative is given by \( \displaystyle x^{1/x}\cdot \frac{1-\ln x}{x^2}, \; x \geq 1 \). Solving the equation \( f'(x)=0 \) we get that equality holds when \( x =e \). Hence \( f \) is strictly decreasing in \( [e, +\infty) \) and strictly increasing in \( [1, e]\). Therefore maximum is attained at \( x =e \) and \( f(e)= e^{1/e} \).
However,
$$1< 2 < e< 3< \cdots < n $$
Hence:
$$\sqrt[3]{3}> \sqrt[4]{4}>\cdots > \sqrt[n]{n}$$
(from monotony)
Also an easy check reveals that the greatest number of the first three is \( \sqrt[3]{3} \).
Conclusion: The greatest number in this case is \( \sqrt[3]{3} \).