Infimum of $n$-th root of $n$

Question

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.

Answer 1

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 $n>1$.

Answer 2

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 ...

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$$n=e^{\ln{n}} \Rightarrow \sqrt[n]{n}=e^{\frac{\ln{n}}{n}}$$ 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.

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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 $n\geq 5$