So the sequence is bounded.
Problem states:
Let be a sequence in with , and n.
Prove that = 0.
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Note: The following had to be proven previously, so I assume it will help in completing the above proof: "Let and be sequences of real numbers. If is bounded and = 0,
then = 0."
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My ideas so far:
Since { } → 0, I know is bounded, and by the theorem if I can show sin (1/ ) → 0 then I'm done.
Or, since → 0, if I can show sin (1/ ) is bounded, then I'm done.
My question now is, which of the two approaches is correct, and how would I go about doing that?
I thought I could do this:
I know sin (n) is bounded by [-1, 1] so,
-1 ≤ sin (n) ≤ 1 so dividing all by n gives,
-1/n ≤ sin (n)/n ≤ 1/n
I know both -1/n and 1/n converge to zero, so by squeeze theorem,
sin (n) / n converges to 0 too. But my classmate pointed out that
sin (n) / n is not the same as sin (1/ ). So I can't do that. :| Therefore I am stuck.
Any help, suggestions, corrections, ideas are greatly appreciated.
Thank you for your time!
(Don't forget DST ends this weekend!)