# Infimum and Supremum of a bounded sequence

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 $$Sup \langle f_{n}\rangle$$ ?

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 $$\sup_{n\in\mathbb N}f_n=3^{1/3}.$$
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.