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