Question #159890

Prove that a sequence that diverges to +∞ (resp. −∞) is divergent


1
Expert's answer
2021-02-03T04:28:07-0500

A sequence {an}n=1+\{a_n\}_{n=1}^{+\infty} that diverges to +∞ or -∞ is unbounded:

an+a_n\to+\infty means M>0NNn>Nan>M\forall M>0\,\exist N\in\mathbb{N}\, \forall n>N a_n>M that implies supnan=+\sup\limits_n| a_n|=+\infty

ana_n\to-\infty means M>0NNn>Nan<M\forall M>0\,\exist N\in\mathbb{N}\, \forall n>N a_n<-M that implies supnan=+\sup\limits_n| a_n|=+\infty

A sequence {an}n=1+\{a_n\}_{n=1}^{+\infty} that converges to a finite number A is bounded:

ε>0N(ε)Nn>N(ε)anA<ε\forall \varepsilon>0\,\exist N(\varepsilon)\in\mathbb{N}\, \forall n>N(\varepsilon) |a_n-A|<\varepsilon

If we take ε=1\varepsilon=1 then n>N(1)anA<1\forall n>N(1) |a_n-A|<1, an<A+1|a_n|<|A|+1 and supnanmax{A+1,a1,a2,,aN(1)}<+\sup\limits_n |a_n|\leq \max\{|A|+1,|a_1|, |a_2|,\dots,|a_{N(1)}|\}<+\infty

Therefore, any sequence that diverges to +∞ or -∞ is divergent.


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