Supremum and infimum nth root of n

Question

I'm looking at $A = \{ \sqrt[n]{n} - {1 \over \sqrt[n]{n}}: n \in \mathbb{N} \}$. I find $inf \text{A} = 0$ and $sup \text{A} = \sqrt[3]{3} - {1 \over \sqrt[3]{3}}$. While the infimum is relatively easy to "see," the supremum is not. Is there a way to find it analytically rather than by inspection?

Answer 1

Hint: Use the derivative of $n^{1/n}=e^{\ln n / n}$. The second term is the same but with a minus sign in the power. The derivative has a zero at $n=e$, what happens for $n<e$ and $n>e$?

Answer 2

Consider the function $$f: \mathbb{R}^+ \to \mathbb{R}^+: x \mapsto x^{\frac{1}{x}} - x^{-\frac{1}{x}}$$ This has derivative $$f'(x) = x^{-\frac{1}{x}-2}(1+x^{\frac{2}{x}})(1-\ln(x))$$ Hence the derivative is positive when $x<e$, is zero when $x = e$ and is negative when $x > e$. Since $2 < e < 3$, this tells you the supremum would be either at $n = 2$ or $n = 3$, which you can then find by calculating and comparing both values.

Answer 3

Note that $\sqrt[n]n$ has a maximum for $n=3$, which means that $\frac1{\sqrt[n]n}$ has a minimum at the same point. Together this means that $\sqrt[n]n-\frac1{\sqrt[n]n}$ has a maximum for $n=3$.