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Mendicant_Bias
 one year ago
(Introductory Real Analysis) I'm trying to prove that the product of a bounded sequence and a sequence that converges to zero, itself converges to zero. I have the basic idea to start but don't understand some of the work my Professor did. Could someone help?
Mendicant_Bias
 one year ago
(Introductory Real Analysis) I'm trying to prove that the product of a bounded sequence and a sequence that converges to zero, itself converges to zero. I have the basic idea to start but don't understand some of the work my Professor did. Could someone help?

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Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0\[\text{Given}\ x_n \rightarrow 0, \ y_n \leq M,\]\[\text{Prove that:} \ x_ny_n \longrightarrow 0\]

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0\[\text{Given} \ x_n \rightarrow 0, \]\[(\forall \lambda > 0)(\exists \ J \in N)(\forall n>J), \ \ \ x_nc<\lambda\]\[(\exists \ M > 0)(\forall n \in N), \ \ \ y_n\leq M\]

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Applying the above to our problem: \[x_n  0<\lambda\]\[x_n<\lambda\]

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0My confusion is here: \[(\exists \ J \ \epsilon \ N)(\forall n > J), \ \ \ x_n<\frac{\varepsilon}{M}\]

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0How does that last part come about? @Zarkon @phi

phi
 one year ago
Best ResponseYou've already chosen the best response.1you have \[ (\forall \lambda > 0)(\exists \ J \in N)(\forall n>J), \ \ \ x_nc<\lambda \] and that is true for all lambda's it looks like they decided to let lambda be epsilon/M .

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0I had to move on, but I'd love to come back to this question sometime for my own sake, I frankly still don't get it.
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