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I'm studying convergent sequences at the moment.

And I came across this question in the section of Stolz Theorem.

I realised that $\{x_n\}$ is monotonously decreasing and has a lower bound of $0$, so $\{x_n\}$ must be convergent, and the limit is $0$ (let $L=\sin(L)$, then $L=0$).

So to prove the original statement, I just need to prove lim n(Xn)^2 → 3, and in order to prove that, I just need to prove $\lim \frac{1}{x_n^2} - \frac{1}{{x_{n-1}}^2} \to \frac{1}{3}$ by Stolz Theorem

but I have no clue what to do from there.

PS: $x_{n+1}$ is $x$ sub $n+1$, and $x_n$ is outside the square root.

Thanks guys

JJS
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1 Answers1

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We have $x_{n+1}=\sin x_n=x_n\left(1-\frac{1}{6}x^2_n+\epsilon_1(x_n)\right),$ says Taylor, so $x^2_{n+1}=x^2_n\left(1-\frac{1}{3}x^2_n+\epsilon_2(x_n)\right),$ where $|\epsilon_i(x)|\le C_ix^4$ for $x$ small enough. That means $$\frac{3}{x^2_{n+1}}=\frac{3}{x^2_n\left(1-\frac{1}{3}x^2_n+\epsilon_2(x_n)\right)}=\frac{3}{x^2_n}+\frac{1-3\epsilon_2(x_n)/x^2_n}{1-\frac{1}{3}x^2_n+\epsilon_2(x_n)}.$$ Can you move on from here?

  • I'm still 100 pages from Taylor Series Chapter, but Thank you professor. I'm going to save your answer and revisit when I move to that Chapter. – JJS Jun 12 '17 at 12:50
  • Well, you didn't mention that in your question. It's very technical to do without Taylor. But if you just take my word for it, can you complete the proof? –  Jun 12 '17 at 12:56
  • I'm not too sure. What's the meaning of ϵ1and ϵ2? Is it epsilon as in ε-N language? – JJS Jun 13 '17 at 01:57
  • They are just error terms which can be estimated, as written in my answer. BTW, one could simply prove $1/\sin^2x−1/x^2\rightarrow1/3$ as $x\rightarrow0$, but maybe you've never heard of l'Hospital and derivatives, either. –  Jun 13 '17 at 04:15
  • I think I'm starting to get it, thanks a lot. I still have 80 pages to go till l'Hospital and derivatives. – JJS Jun 13 '17 at 08:14