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Find the following limit: $$\lim_{k\rightarrow \infty} \frac{\left(1+\sin\alpha_{k+1}\right)^{\alpha_{k+1}}}{\left(1+\sin\alpha_{k}\right)^{\alpha_{k}}}$$ on $\left\{\alpha_k\in\mathbb{N}\ \ |\ \ k\in \mathbb{N}: \sin\alpha_{k}>\sin\alpha_{k-1} \right\}=\{1, 2, 8, 14, 33, 322, 366, 699,...\}$, i.e. Numbers n where sin(n) increases monotonically to 1.

I've tried to use Squeeze theorem: $$1\leq\frac{\left(1+\sin\alpha_{k+1}\right)^{\alpha_{k}}}{\left(1+\sin\alpha_{k}\right)^{\alpha_{k}}}\leq\frac{\left(1+\sin\alpha_{k+1}\right)^{\alpha_{k+1}}}{\left(1+\sin\alpha_{k}\right)^{\alpha_{k}}}\leq \frac{\left(1+\sin\alpha_{k+1}\right)^{\alpha_{k+1}}}{\left(1+\sin\alpha_{k}\right)^{\alpha_{k+1}}}$$ and $$\lim_{k\rightarrow \infty}\frac{\left(1+\sin\alpha_{k+1}\right)^{\alpha_{k+1}}}{\left(1+\sin\alpha_{k}\right)^{\alpha_{k+1}}}=1$$ because $\sin\alpha_{k+1}, \sin\alpha_{k}\rightarrow 1$.

Am I correct? Or did I misinterpret the solution?

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More information about the sequence $a_n$ are required. –  Yiorgos S. Smyrlis Jan 12 at 18:36
    
Please do not use MathJax-only question titles and refrain from using display style in the titles. –  AlexR Jan 12 at 18:37
    
In your squeeze inequality you have contradicting arguments from second to third and from third to fourth. Is $a_k$ increasing? Decreasing? What's it's starting value? You need to go through those questions. –  user88595 Jan 12 at 20:06
    
$\alpha_k$ is increasing... –  Mark Jan 12 at 23:38
    
If $\alpha_k $ is increasing by assumption, please edit your post. –  Fabio Lucchini Jan 13 at 23:02

1 Answer 1

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I haven't got a complete answer, but only some remarks which may be useful in order to solve your problem.

Put $\beta_k=\sin(\alpha_k)\in [-1,1]$ and $a_k=\arcsin(\beta_k)\in [-\frac\pi 2,\frac\pi 2]$. Then $a_k$ and $\beta_k$ are strictly increasing, hence there exists $\beta\in (-1,1]$ such that $\beta_k\uparrow\beta$ and $a\in(-\frac\pi 2,\frac\pi 2]$ such that $a_k\uparrow a$. Moreover, $\sin(\alpha_k)=\sin(a_k)$, hence there exists a sequence of integers $n_k\in\mathbb Z$ such that $\alpha_k=(-1)^{n_k}a_k+n_k\pi$.

I claim that $\alpha_k\to +\infty$ and $n_k\to +\infty$. Suppose, on contrary, that a subsequence $\alpha_{k_h}$ is bounded. Then also $|n_{k_h}|$ is bounded because so is $a_{k_h}$. Then only finitely many values for $(\alpha_{k_h},n_{k_h})$ are possible. Consequently only finitely many values for $a_{k_h}$ are possible. This contradicts the fact that $a_k$ is stricly increasing.

From this it follow $\alpha_k\sim\pi n_k $. Now taking $\log $ of your limit we get: $\alpha_{k+1}\log (1+\beta_{k+1})-\alpha_k\log (1+\beta_k ) $.

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