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

[runningeagle]

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.

 

[BigGlenntheHeavy]

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

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

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

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

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

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

 

[runningeagle]

Thanks!

So, -1/n^3 is less than (that thing) is less than 1/n^3 ?

 

[BigGlenntheHeavy]

\(\displaystyle See \ my \ revision \ above.\)