limit (-1)^n(1/n^3)cos(n^3)

runningeagle

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Oct 3, 2009
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MSP221197ebh6091210g420000519hc638b6fe878f


Hi,

I am trying to prove this limit of this sequence.

I have tried to do a proof from the seires (showing that the series converges, therefore the sequence converges), I have tried to do l'hopital's rule (I get 0 times a limit that does not exist) and I have tried breaking the sequence up as a product of two or three limits, but I still run into a problem with the (-1)^n limit not existing.

Thank you.
 
The Marqui cant help you on this one, ergo well resort to the squeeze theorem.\displaystyle The \ Marqui \ can't \ help \ you \ on \ this \ one, \ ergo \ we'll \ resort \ to \ the \ squeeze \ theorem.

1  (1)ncos(n3)  1, n > 0.\displaystyle -1 \ \le \ (-1)^{n}cos(n^{3}) \ \le \ 1, \ n \ > \ 0.

1n3  (1)ncos(n3)n3  1n3\displaystyle \frac{-1}{n^{3}} \ \le \ \frac{(-1)^{n}cos(n^{3})}{n^{3}} \ \le \ \frac{1}{n^{3}}

limn1n3  limn(1)ncos(n3)n3  limn1n3\displaystyle \lim_{n\to\infty} \frac{-1}{n^{3}} \ \le \ \lim_{n\to\infty} \frac{(-1)^{n}cos(n^{3})}{n^{3}} \ \le \ \lim_{n\to\infty} \frac{1}{n^{3}}

0  limn(1)ncos(n3)n3  0\displaystyle 0 \ \le \ \lim_{n\to\infty}\frac{(-1)^{n}cos(n^{3})}{n^{3}} \ \le \ 0

Hence, only one way to go: limn(1)ncos(n3)n3 = 0\displaystyle Hence, \ only \ one \ way \ to \ go: \ \lim_{n\to\infty}\frac{(-1)^{n}cos(n^{3})}{n^{3}} \ = \ 0
 
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