Mendicant_Bias
  • Mendicant_Bias
(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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Mendicant_Bias
  • Mendicant_Bias
\[\text{Given}\ x_n \rightarrow 0, \ |y_n| \leq M,\]\[\text{Prove that:} \ x_ny_n \longrightarrow 0\]
Mendicant_Bias
  • Mendicant_Bias
\[\text{Given} \ x_n \rightarrow 0, \]\[(\forall \lambda > 0)(\exists \ J \in N)(\forall n>J), \ \ \ |x_n-c|<\lambda\]\[(\exists \ M > 0)(\forall n \in N), \ \ \ |y_n|\leq M\]
Mendicant_Bias
  • Mendicant_Bias
Applying the above to our problem: \[|x_n - 0|<\lambda\]\[|x_n|<\lambda\]

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Mendicant_Bias
  • Mendicant_Bias
My confusion is here: \[(\exists \ J \ \epsilon \ N)(\forall n > J), \ \ \ |x_n|<\frac{\varepsilon}{M}\]
Mendicant_Bias
  • Mendicant_Bias
How does that last part come about? @Zarkon @phi
phi
  • phi
you have \[ (\forall \lambda > 0)(\exists \ J \in N)(\forall n>J), \ \ \ |x_n-c|<\lambda \] and that is true for all lambda's it looks like they decided to let lambda be epsilon/M .
Mendicant_Bias
  • Mendicant_Bias
I 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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