Find the supermum and infimum of a set I Got the set
$\sqrt[k]{k+1}$
I have found the supermum by the Inequality of arithmetic and geometric means.
And the result is 2.
I dont have a way for solving the infimum .
I tried to solve it by move to to some eqaution and try by Binomial theorem, but i didnt success.
I know that the infimum is 1, but how can i proove it ?
Thanks.

Note that for any positive integer $k$, we have $\sqrt[k]{k+1}\gt 1$.  We want to show that given any $\epsilon\gt 0$ we can find an $N$ such that $\sqrt[N]{N+1}\lt 1+\epsilon$.
By the Binomial Theorem, if $n\ge 2$ then $(1+\epsilon)^n \gt \frac{n^2}{2}\epsilon^2$. If $n$ is large enough, we therefore have
$$(1+\epsilon)^n \gt n+1.$$
(What "large enough" means can be easily made explicit.)
Taking $n$-th roots, we conclude that for such an $n$ we have
$$\sqrt[n]{n+1}\lt 1+\epsilon.$$
