Infimum of $n$-th root of $n$ Let  $A = \left \{ \sqrt[n]{n} \mid n \in \mathbb{N}\right \}$. I need to find and prove the infimum of $A$.
Because $n \in \mathbb{N},$ we can know for sure that $\sqrt[n]{n} \geq 1$. (Does that need a deeper proof?)
Then to prove that $1$ is the infimum, for $\forall \varepsilon>0$, I need to find a certain $n \in \mathbb{N}$ that makes $$\sqrt[n]{n} < 1 + \varepsilon$$
How do I "engineer" that $n \in \mathbb{N}$ ?
What's the trick? Raising $\sqrt[n]{n} < 1 + \varepsilon $ to the power of $n$ doesn't really help.
 A: The infimum of $A$ is $1$ because $1\in A$ (just take $n=1$) and $n>1\implies\sqrt[n]n>1$.
On the other hand, you can write $\sqrt[n]n$ as $1+\varepsilon_n$ and then\begin{align}n&=\sqrt[n]n^n\\&=(1+\varepsilon_n)^n\\&=1+n\varepsilon_n+\frac{n(n-1)}2{\varepsilon_n}^2+\cdots\\&>\frac{n(n-1)}2{\varepsilon_n}^2\end{align}and therefore$$\varepsilon_n<\sqrt{\frac2{n-1}},$$if $n>1$.
A: If you are not allowed to use the fact that $\lim\limits_{n\rightarrow\infty} \sqrt[n]{n}=1$ (which immediately leads to $\sqrt[n]{n} < 1+\varepsilon$ from the definition of limit), then try the following trick ...

$$n=e^{\ln{n}} \Rightarrow \sqrt[n]{n}=e^{\frac{\ln{n}}{n}}$$
It is not too difficult to show that $0<\frac{\ln{n}}{n} < 1$ and thus, we can apply Bernoulli's inequality
$$e^{\frac{\ln{n}}{n}}=(1+e-1)^{\frac{\ln{n}}{n}}\leq 1+\frac{(e-1)\ln{n}}{n}$$
or 
$$\sqrt[n]{n}\leq 1+\frac{(e-1)\ln{n}}{n}$$
and because $\frac{n}{\ln{n}}>\sqrt{n}$ for $n>1$ we have
$$\sqrt[n]{n}\leq 1+\frac{(e-1)\ln{n}}{n}<1+\frac{e-1}{\sqrt{n}}$$
you can squeeze $\frac{e-1}{\sqrt{n}} < \varepsilon$ now.

Now, why

$\frac{n}{\ln{n}}>\sqrt{n}$ for $n>1$

Because, using induction
$$\frac{2}{\ln{2}} > \sqrt{2},\space \frac{3}{\ln{3}} > \sqrt{3},\space \frac{4}{\ln{4}} > \sqrt{4}$$
and $$\frac{n}{\ln{n}}>\sqrt{n} \iff 
\color{red}{\sqrt{n} > \ln{n}} \iff 
\sqrt{n} + \ln{\left(1+\frac{1}{n}\right)} > \ln{n} + \ln{\left(1+\frac{1}{n}\right)}=\ln{(n+1)}$$
using $\ln{(1+x)}<x$ from here
$$\color{red}{\sqrt{n+1}>}\sqrt{n} +\frac{1}{n}>\sqrt{n} + \ln{\left(1+\frac{1}{n}\right)} >\color{red}{\ln{(n+1)}}$$
for $n\geq 5$
