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Given a sequence $ f_{n} = \big\{n^{1/n}, n \in N \big\} $ . Prove that $f_{n}$ is bounded, hence find supremum and infimum

  1. Now i can work out that the sequence is convergent and hence, it is bounded
  2. I can show that $f_{n} \geq1, \forall n \in N$ and $1 \in \langle f_{n} \rangle \Longrightarrow Inf \langle f_{n}\rangle = 1$

But how do i proceed to find $Sup \langle f_{n}\rangle $ ?

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2 Answers 2

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Since $\log$ is increasing and $f_n>0$ for all $n\in\mathbb N$, we have that $$\log \sup_{n\in\mathbb N} f_n=\sup_{n\in\mathbb N} \log f_n=\sup_{n\in\mathbb N}\frac{\log n}{n}. $$ Consider the function $g(x)=\frac{\log x}{x}$ on the interval $[1,\infty)$. Observe that $$ g'(x)=\frac{1-\log x}{x^2} $$ which is positive on $[1,e)$ and negative on $(e,\infty)$. This means $g(x)$ is increasing on $[1,e)$ and decreasing on $(e,\infty)$, in particular its maximum occurs at $x=e$. But $e\not\in \mathbb N$, so the supremum will be attained at either $n=2$ or $n=3$. In fact $$ \frac{\log 2}{2}<\frac{\log 3}{3}\iff 3\log 2<2\log 3\iff \log 8<\log 9, $$ and therefore $$ \sup_{n\in\mathbb N}\frac{\log n}{n}=\frac{\log 3}{3}. $$ Therefore $\sup_{n\in\mathbb N}f_n=3^{1/3}.$

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The function $f(x)=\frac {\log x} x$ is decreasing in $[3,\infty)$ because its derivative is negative: $1-\log x \leq 1-\log 3<0$ for $x \geq 3$. Hence $f (3) , f (4),...$ is decreasing sequence . The maximum of the sequence is therefore one of the numbers $f(1), f(2),f(3)$. I will let find out which one is the largest.

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