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Prove that if (a_n) is a null sequence and (b_n) is a bounded sequence then the sequence (a_nb_n) is null:
from definitions if b_n is bounded then ## \exists H \in \mathbb{R} ## s.t. ## |b_n| \leq H ## if a_n is a null sequence it converges to 0 (from my book), i.e. given ## \epsilon ' > 0 ## ## \exists N \in \mathbb{R} ## s.t. n>N ## \Rightarrow |a_n| < \epsilon ' ## set ## e' = \dfrac{\epsilon}{|H|} ## then for ## n > N \Rightarrow |a_nb_n | < \epsilon ' |H| = \epsilon ## i.e. a_nb_n also converges to 0.
I'm slightly worried about this, if H is negative, then can we just multiply ## |a_n| ## and ## |b_n| ## as afaik from the axioms if ## a>0 ## ## x>y## then ## xa>ya ## and I'm not sure if H is negative or if |a_n| is negative if we can simply just multiply the two inequalities. Any help, explaining this, thanks.
from definitions if b_n is bounded then ## \exists H \in \mathbb{R} ## s.t. ## |b_n| \leq H ## if a_n is a null sequence it converges to 0 (from my book), i.e. given ## \epsilon ' > 0 ## ## \exists N \in \mathbb{R} ## s.t. n>N ## \Rightarrow |a_n| < \epsilon ' ## set ## e' = \dfrac{\epsilon}{|H|} ## then for ## n > N \Rightarrow |a_nb_n | < \epsilon ' |H| = \epsilon ## i.e. a_nb_n also converges to 0.
I'm slightly worried about this, if H is negative, then can we just multiply ## |a_n| ## and ## |b_n| ## as afaik from the axioms if ## a>0 ## ## x>y## then ## xa>ya ## and I'm not sure if H is negative or if |a_n| is negative if we can simply just multiply the two inequalities. Any help, explaining this, thanks.