I have the set $A=\{\sqrt[n]{n}: n \in \mathbb{N}\}$. I can see that $\inf A$ is $1$ as $\sqrt[1]{1}=1$ but I am having trouble with $\sup A$. I know that it is $\sqrt[3]{3}$ because it only decreases from then onwards. But I have trouble proving that is $\sqrt[3]{3}$. I have tried induction but I can't get it to work. Any strategies appreciated.
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3The derivative is $x^{1/x-2}(1-\log x)$, so $x^{1/x}$ has a global maximum at $x=e$. Then it decreases. – Aig Mar 10 '24 at 04:10
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The derivative of $f(x)=x^{1/x}$ is equal to $x^{1/x-2}(1-\log x)$. It is negative when $0<x<e$ and positive when $x>0$. It is $0$ if $x=e$. So $x=e$ is the global maximum. It means that we only need to check $n=2$ and $3$. Since $\sqrt[3]3>\sqrt 2$, we have that $$\sup A = \sqrt[3]3.$$
Aig
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Note that the ratio $$ \begin{align} \frac{\frac{(n+1)^n}{n^{n+1}}}{\frac{n^{n-1}}{(n-1)^n}} &=\frac{\left(n^2-1\right)^n}{n^{2n}}\tag{1a}\\ &=\left(1-\frac1{n^2}\right)^n\tag{1b}\\[12pt] &\lt1\tag{1c} \end{align} $$ Inequality $(1)$ means that $\frac{(n+1)^n}{n^{n+1}}$ is decreasing.
When $n=3$, we have $$ \begin{align} \frac{(n+1)^n}{n^{n+1}} &=\frac{64}{81}\tag{2a}\\[6pt] &\lt1\tag{2b} \end{align} $$ Thus, for $n\ge3$, we have $$ (n+1)^{\frac1{n+1}}\lt n^{\frac1n}\tag3 $$
When $n=2$, we have $$ \begin{align} \frac{(n+1)^n}{n^{n+1}} &=\frac98\tag{4a}\\[6pt] &\gt1\tag{4b} \end{align} $$ Thus, for $n\le2$, we have $$ (n+1)^{\frac1{n+1}}\gt n^{\frac1n}\tag5 $$
Inequalities $(3)$ and $(5)$ show that $n^{\frac1n}$ is greatest when $n=3$.
robjohn
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Part of this answer was previously given in $(5)$-$(7)$ of this answer. – robjohn Mar 10 '24 at 19:41