$\left\{n^{\frac{1}{n}}\;\middle\vert\;n\in\mathbb{N}\right\}$
What is the supremum and infimum of the above set?
The set is $\left\{1, 2^{\frac{1}{2}}, 3^{\frac{1}{3}},....\right\}$
Now, $n^{\frac{1}{n}}\geq0$ Which implies that 0 is a lower bound of the set. Now if I can show that for some $\varepsilon>0$, there exists $k\in\mathbb{N}$ such that $0<k^{\frac{1}{k}}<0+\varepsilon$ then we may conclude that 0 is the infimum of the set, but I do not know how can I show the above.
Please anyone help me solve this problem. Thanks in advance.
Actually, you only have $n^{1/n}\leqslant1$ when $n=1$; for all other numbers, $n^{1/n}>1$. On the other hand, if $f(x)=x^\frac{1}{x}$, then $f'(x)=x^{x-2}\left(1-\log(x)\right)$ and therefore $f$ attains its maximum at $e$. It's not hard to deduce from this that$$\sup\left\{n^{\frac1n}\,\middle|\,n\in\mathbb N\right\}=\sqrt[3]3.$$