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?
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If A has an min or max element the inf or sup must equal it. Prove a_n are decreasing for n >= 3 so a_3 is max element. As a_n are decreasing and prove 0 < a_n. So decreasing and bounded means lim a_n = inf. Prove n-root (n)- > 1 so lim a_n=0. – fleablood Oct 27 '16 at 21:25
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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 $n>e$?
MrYouMath
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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.
sTertooy
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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 $n=3$.
Arthur
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