0

$\left\{n^{\frac{1}{n}}\;\middle\vert\;n\in\mathbb{N}\right\}$

What is the supremum and infimum of the above set?

The set is $\left\{1, 2^{\frac{1}{2}}, 3^{\frac{1}{3}},....\right\}$

Now, $n^{\frac{1}{n}}\geq0$ Which implies that 0 is a lower bound of the set. Now if I can show that for some $\varepsilon>0$, there exists $k\in\mathbb{N}$ such that $0<k^{\frac{1}{k}}<0+\varepsilon$ then we may conclude that 0 is the infimum of the set, but I do not know how can I show the above.

Please anyone help me solve this problem. Thanks in advance.

José Carlos Santos
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user587389
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1 Answers1

5

Actually, you only have $n^{1/n}\leqslant1$ when $n=1$; for all other numbers, $n^{1/n}>1$. On the other hand, if $f(x)=x^\frac{1}{x}$, then $f'(x)=x^{x-2}\left(1-\log(x)\right)$ and therefore $f$ attains its maximum at $e$. It's not hard to deduce from this that$$\sup\left\{n^{\frac1n}\,\middle|\,n\in\mathbb N\right\}=\sqrt[3]3.$$

BarzanHayati
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José Carlos Santos
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  • What about the infimum? – user587389 May 17 '19 at 16:36
  • It is $1$, since $\sqrt[n]n=1$ if $n=1$ and it is greater than $1$ otherwise. – José Carlos Santos May 17 '19 at 16:37
  • Ok, thank you Sir. – user587389 May 17 '19 at 16:43
  • If my answer was useful, perhaps that you could mark it as the accepted one. – José Carlos Santos May 17 '19 at 16:47
  • In many of resource like https://math.stackexchange.com/questions/1755360/derivative-of-xx-and-the-chain-rule derivative of $$ f(x)=x^x$$ is $$f^{'}(x)=x^x(1+logx)$$ but @JoséCarlosSantos used another equation and you reached $e$ as answer of the $f^{'}(x)=0$ – BarzanHayati May 17 '19 at 17:05
  • @JoséCarlosSantos How can I deduce $3^{\frac{1}{3}}$ as the Supremum of the given set from the fact that $x^{\frac{1}{x}}$ is maximised at $x=e$? I still cannot get it. – user587389 May 23 '19 at 12:10
  • Since $x^{\frac1x}$ increases to the left of $e$ and decreases to its right, the maximum value that $n^{\frac1n}$ can take is at one of the natural numbers which are closer to $e$, which are $2$ and $3$. And it turns out that $2^{\frac12}<3^{\frac13}$. – José Carlos Santos May 23 '19 at 13:28