a is a real sequence $ a: \mathbb{N} \rightarrow \mathbb{R}$ defined as $a(0) = 0, a(n):= n^\frac{1}{n} $ $ \forall n \in \mathbb{N^*}$. What's the Supremum of $a(\mathbb{N})$ ?
I've calculated the first derivative of the function $ f(x) = x^\frac{1}{x}$ and determined the critical point. My result is max(a($\mathbb{N}$)) = e. Can I now treat the maximum equivalent to the supremum or is there an other way ?
While $e$ maximises the value of $f(x)$, $a(n)$ is only defined for $n$ a natural number. But what you have shown is that $a(n)$ increases for $n<e$ and decreases for $n>e$, so the two values of $n$ to investigate are 2 and 3, since $2 < e < 3$.
$a(2) = \sqrt{2}$
$a(3) = \sqrt[3]{3}$
As $(\sqrt[3]{3})^6 = 9$ and $(\sqrt{2})^6 = 8$, $a(3)$ is maximal and the supremum is therefore $\sqrt[3]{3}$.