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2. If {an} and {bn} are convergent and there exists a constant k>0 such that |bn| > k for all n=1, 2, ...., then {an/bn} is also convergen

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- Thread starter darthprince
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2. If {an} and {bn} are convergent and there exists a constant k>0 such that |bn| > k for all n=1, 2, ...., then {an/bn} is also convergen

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CompuChip

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E.g. you can try to show that

[tex]\forall_{\epsilon > 0} \exists_{N_+} : n > N_+ \implies |(a_n + b_n) - L| < \epsilon[/tex]

where L = a + b is the postulated limit. Of course you already know that given such an [itex]\epsilon[/itex] you can find N

[tex]n > N_a \implies |a_n - a| < \epsilon[/tex]

and

[tex]n > N_b \implies |b_n - b| < \epsilon[/tex]

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