Infimum and supremum of set $A=\{\sqrt[n]{\frac{1}{n}}, n \in \mathbb{N}, n \ge 3 \}$

Question

I have to find the infimum and supremum of the set $A=\{\sqrt[n]{\frac{1}{n}}, n \in \mathbb{N}, n \ge 3 \}$. $ \sqrt[n]{\frac{1}{n}} = \frac{1}{\sqrt[n]{n}}$. $\lim_{n \to \infty}\sqrt[n]{n} = 1$ so $\lim_{n \to \infty}\sqrt[n]{\frac{1}{n}} = 1$. So supremum of $A$ is $1$. I guess infimum of $A$ is $\frac{1}{\sqrt[3]{3}}$. Am I right?

Answer 1

Let $a_n=n^{-1/n}$. Then $\ln a_n=-f(n)$ where $$f(x)=\frac{\ln x}x.$$ Investigate this function for $x>0$. What are its maxima/minima? Where does it increase? decrease?

Answer 2

Let $$L=\lim_{n\to \infty } \, \left(\frac{1}{n}\right)^{1/n}$$ We compute $$\log L=\log \left(\lim_{n\to \infty } \, \left(\frac{1}{n}\right)^{1/n}\right)=\lim_{n\to \infty } \, \log \left(\left(\frac{1}{n}\right)^{1/n}\right)=\lim_{n\to \infty } \, \frac{\log \left(\frac{1}{n}\right)}{n}=\lim_{n\to \infty } \, \left(-\frac{\log n}{n}\right)=$$

$=0$ and if $\log L=0$ then $L=1$ $$\lim_{n\to \infty } \, \sqrt[n]{\frac{1}{n}}=1$$ Which is $$\sup \left\{\sqrt[n]{\frac{1}{n}}\right\}_{n\in\mathbb{N}}=1$$ Infimum is also minimum (because it is an element of the sequence, while $1$ is not) and it is $$\inf \left\{\sqrt[n]{\frac{1}{n}}\right\}_{n\in\mathbb{N}}= \sqrt[3]{\frac{1}{3}}$$

The sequence is increasing. Indeed $$\left(\frac{1}{n+1}\right)^{\frac{1}{n+1}}>\left(\frac{1}{n}\right)^{1/n}$$ because taking $\log$ of both side, for $n\ge 3$, we have $$\frac{\log \left(\frac{1}{n+1}\right)}{n+1}>\frac{\log \left(\frac{1}{n}\right)}{n}$$ $$-\frac{\log (n+1)}{n+1}>-\frac{\log (n)}{n}$$ $$\frac{\log (n+1)}{n+1}<\frac{\log (n)}{n}$$ $$n \log (n+1)<(n+1) \log (n)$$ $$\log \left((n+1)^n\right)<\log \left(n^{n+1}\right)$$ $$(n+1)^n<n^{n+1}$$ $$\left(\frac{1}{n}+1\right)^n n^n<n\cdot n^n$$ $$\left(\frac{1}{n}+1\right)^n<n$$ as $n\to\infty$ the RHS keeps less than $e$ while the RHS goes to $\infty$

We have finally proved that the sequence is increasing.