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Let us consider a sequence: $$ \left\{\alpha_{k}\in N\ \ |\ \ k\in N: \sin\alpha_{k}>\sin\alpha_{k-1} \right\} $$ Calculate the following limits $$ 1)\lim_{k\rightarrow \infty} \frac{2^{\alpha_k}}{2^{\alpha_{k+1}}} $$ $$ 2)\lim_{k\rightarrow \infty} \frac{\alpha_k}{\alpha_{k+1}} $$ $$ 3)\lim_{k\rightarrow \infty} \left(\frac{2^{\alpha_k}}{2^{\alpha_{k+1}}}\right) \left(\frac{\alpha_k}{\alpha_{k+1}}\right) $$

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  • $\begingroup$ $\lim_{k \to \infty }\sin(\alpha _{k})=\alpha $ which exists. So $0< \sin(\alpha _{k+1}) - \sin(\alpha _{k})=\cos(\varphi _{k})\cdot (\alpha _{k+1} -\alpha _{k})< \varepsilon $, where $\varphi _{k} \in (\alpha _{k}, \alpha _{k+1})$ ... no further progress though ... $\endgroup$
    – rtybase
    Dec 19, 2013 at 19:30
  • $\begingroup$ and $\lim_{k\rightarrow \infty }\cos(\varphi _{k})=0$ $\endgroup$
    – rtybase
    Dec 19, 2013 at 19:50

1 Answer 1

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There is nowhere nearly enough information. For example, in the first question, $\alpha_k= \pi/2-1/k$ is a possibility (whereupon the limit is $1.$) But $\alpha_k = \pi/2 -1/k + 2\pi k^2 $is another poossibility, whereupon the limit is $0,$ and $\alpha_k = \pi/2 -1/k -2\pi k^2$ yet another, when the limit is $\infty.$

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  • $\begingroup$ I just modified the text and the title of my question! $\endgroup$
    – Lely
    Dec 12, 2013 at 20:29
  • $\begingroup$ @Lely It’s not clearer now $\endgroup$ Dec 15, 2013 at 12:23
  • $\begingroup$ I think the condition that $\alpha_k\in\mathbb{N}$ may play a big role here. At least it does lead to $\lim_{k\to\infty} \alpha_k \to \infty$. However, I'm not sure how to proceed beyond that now. $\endgroup$
    – MoonKnight
    Dec 19, 2013 at 19:03

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