Supremum/Maximum and Infimum/minimum of a given set Determine $\sup E$, $\inf E$, and (where possible) $\max E$, $\min E$ for the set $E = \{ \sqrt[n]{n}: n \in \mathbb{N}\}$.
Attempt: I've written that $\inf E = 1 = \min E$.
When it comes to finding $\sup E$, I've noticed punching in increasing values of n on my calculator, the elements of $E$ seem to never go past about $1.4\ldots$, but I still don't know what $\sup E$ is. How do I figure this out?
 A: The function $f(x)=x^{\frac{1}{x}}$ has derivative
$$ f^{\prime}(x)=x^{\frac{1}{x}}\frac{1-\log x}{x^2}$$
Therefore $f$ has its global maximum on $[1,\infty)$ at $x=e$, and is increasing on $[1,e)$ and decreasing on $(e,\infty)$. Therefore the only values of $n$ you need to check are $n=2$ and $n=3$. And $3^{\frac{1}{3}}>\sqrt{2}$.
A: Note that $$\sqrt[n]{n}=n^{1/n}=\exp(\log(n)/n).$$ Thus, $n^{1/n}$ will be maximum when $\log(n)/n$ is maximum (this is a standard property of exponents since the base of the exponent is greater than $1$, i.e., $e>1$).
Consider the more-general setting $f(x)=\log(x)/x$. We can calculate the derivative to be $$f'(x)=\frac{1}{x^2}-\frac{\log x}{x^2}=\frac{1}{x^2}\cdot(1-\log x).$$ This has a $0$ at $x=e$ and is the only $0$ of the derivative, thus this is either a maximum or minimum. Plugging in $x=1$ and $x=e^2$ (these choices aren't particularly relevant, you just need something less than $e$ and greater than $e$... you could just as well take the second derivative and check for concavity), we see that the derivative goes from positive to negative so that the point corresponds to a maximum.
Since the point $x=e$ is a maximum, we only need to consider the nearest integers to $e$, which are $n=2$ and $n=3$. Without a calculator, we still need to figure out if $2^{1/2}<3^{1/3}$ or if $3^{1/3}<2^{1/2}$. Since both values are positive, we can raise both sides to the sixth power and preserve the inequality, giving $(2^{1/2})^6=2^3=8$ and $(3^{1/3})^6=3^2=9$. Since $8<9$, we have $2^{1/2}<3^{1/3}$; thus the maximum occurs when $n=3$.
