# Find limit supremum from 3 sequence theorem

Find limit $$\limsup_{n\rightarrow \infty} \sqrt[n]{\left| \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right) \right)^p\right |}$$

dependency on value of $$p$$

I think that $$p$$ doesn't matter there (by doesn't matter I mean that I can compute limit without knowledge about p and on the end power result to p) because it is just const value so I can write that as

$$\limsup_{n\rightarrow \infty} \left(\sqrt[n]{\left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right|} \right)^p$$ and just calculate $$\limsup_{n\rightarrow \infty} \sqrt[n]{\left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right|}$$ I suspect that this sequence has normal $$lim$$ so probably $$\limsup_{n\rightarrow \infty} \sqrt[n]{\left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right|} = \lim_{n\rightarrow \infty} \sqrt[n]{\left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right|}$$

Ok now I want to use three sequence theorem: $$... \le \sqrt[n]{\left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right|} \le \sqrt[n]{2}$$ But I don't know how to estimate that from below...

Compose Taylor series for large values of $$n$$ to get $$2-2 \cos\left(\frac{1}{n}\right)-\frac 1n{\sin \left(\sin \left(\frac{1}{n}\right)\right)}=\frac{1}{4 n^4}-\frac{7}{72 n^6}+O\left(\frac{1}{n^8}\right)$$
• By $O\left(\frac{1}{n^8}\right)$ you mean asymptotic value? – trolley Mar 26 at 12:25
• @trolley. The big $O$ notation. To see your problem simpler, let $x=\frac 1n$ and work with $x\to 0$. – Claude Leibovici Mar 26 at 12:33
• Ok, do you suggest that $\frac{1}{4 n^4}-\frac{7}{72 n^6}+O\left(\frac{1}{n^8}\right)$ is less than $\left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right|$ why? – trolley Mar 26 at 19:34
• @trolley. Not at all ! It is just an asymptotics expression. But, checking $$4n^4 \, \left| 2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n} \right)\right| < 1$$ – Claude Leibovici Mar 27 at 5:43