Question #326318

Show that the sequence (an) is bounded iff |an| is bounded .


1
Expert's answer
2022-04-12T12:14:25-0400

ANSWER

A sequence {an}\left \{a _{n} \right \} is called bounded if there are real numbers ll and bb , such that

lanbl\leq a_{n}\leq b for all nNn\in N . (1)

1) Let {an}\left \{a _{n} \right \} be a bounded sequence, denote M=max{l,b}M=max\left \{ |l |, \: |b|\right \}

Since bM,lM|b|\leq M, |l|\leq M , then MlM,MbM-M\leq l\leq M, -M\leq b\leq M . Hence

MlanbM-M\leq l\leq a_{n}\leq b\leq M for all nNn\in N .

It means, that

0anM0\leq |a_{n}|\leq M (2)


Those sequence {an}\left \{|a _{n}| \right \} is bounded.

2) Conversely , let the sequence {an}\left \{|a _{n}| \right \} is bounded. Then exists M>0M>0 such that inequality (2)

holds for all nNn\in N ..Since (2) is equivalent to the inequality

ManM-M\leq a_{n}\leq M ,

then (1) is satisfied for l=M,b=Ml=-M, b=M . Hence, the sequence {an}\left \{a _{n} \right \} is bounded.


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