How to bound oscillating sequences for the squeeze theorem?
#1
I keep getting stuck when I try to use the squeeze theorem on sequences that oscillate, like with a sine or cosine term. I understand the basic idea of bounding a function, but finding the two simpler sequences that actually converge to the same limit for an oscillating one feels tricky. My textbook's examples seem to pick perfect bounds out of thin air.
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#2
I get stuck on that too. The trick is to build an envelope that tends to the limit you want. For something like a_n = sin(n)/n the two sides can be -1/n and 1/n, since -1 <= sin(n) <= 1, so -1/n <= sin(n)/n <= 1/n. Both outer sequences go to 0, so you get limit 0. It feels almost magical until you realize you just dropped the oscillation into a decaying factor.
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#3
I did cos(n)/n last semester and used the same bound. The bound -1/n and 1/n felt obvious once I remembered sin and cos are bounded by 1.
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#4
I wonder if your issue is that you're trying to squeeze sin(n) itself. It doesn't have a limit, so you can't force one with those bounds unless you attach a vanishing multiplier.
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#5
Sometimes the problem is the target; you might be dealing with a_n = f(n) * sin(n) where f(n) -> 0. Then you bound by |f(n)|, so -|f(n)| <= a_n <= |f(n)|. That gives a limit 0.
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#6
I remember chasing bounds that felt like they came out of nowhere. One time I tried to bound sin(n) between -1 and 1 and then draw a line to a constant, but that only works if something is shrinking.
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#7
I once plotted the sequence and saw it bounce but get closer to 0 after dividing by n; the visual helped me trust the envelope idea.
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#8
If you want, share a specific sequence and I can try to bound it with a couple of convergent envelopes.
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