# Calculation of limits with $\alpha_k,2^{\alpha_k}$ on $\left\{\alpha_{k}\in N\ \ |\ \ k\in N: \sin\alpha_{k}>\sin\alpha_{k-1} \right\}$

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 $$1)\lim_{k\rightarrow \infty} \frac{2^{\alpha_k}}{2^{\alpha_{k+1}}}$$ $$2)\lim_{k\rightarrow \infty} \frac{\alpha_k}{\alpha_{k+1}}$$ $$3)\lim_{k\rightarrow \infty} \left(\frac{2^{\alpha_k}}{2^{\alpha_{k+1}}}\right) \left(\frac{\alpha_k}{\alpha_{k+1}}\right)$$

• $\lim_{k \to \infty }\sin(\alpha _{k})=\alpha$ 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 ... Dec 19, 2013 at 19:30
• and $\lim_{k\rightarrow \infty }\cos(\varphi _{k})=0$ Dec 19, 2013 at 19:50

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 $\infty.$
• 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. Dec 19, 2013 at 19:03