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Evaluate

$$\lim\limits_{n\to \infty} \frac{\sqrt n}{n^{\sin n}}$$

I wanted to say that the limit does not exist because $\sin n$ sometimes negative and sometimes positive and sometimes equal to zero, can you help me with that please? also I think that L'Hopital can't help in this case.

Error 404
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  • Are you familiar with the definitions of $\liminf$ and $\limsup$ ? – christina_g Mar 17 '16 at 21:51
  • @christina_g Yes, I am – Error 404 Mar 17 '16 at 21:53
  • Cool. So, we can have a subsequence $a_{n_k}$ of $a_n = \sqrt{n} / n^{\sin n}$ such that $ \forall n_k $ $ \sin n_k < -1/2$ (by taking $n_k$ sufficiently close to $2k \pi - \pi /6$, for some $k$ ). We can also find another subsequence $a_{n_l}$ such that $ \sin n_k >1/2$, by using a similar argument.

    So $\liminf a_n \leq -1/2$ and $\limsup a_n \geq 1/2 \Rightarrow \liminf a_n \neq \limsup a_n$.

     – christina_g Mar 17 '16 at 22:03
  • My bad the last part is: $$\liminf a_n \leq \lim \frac{ \sqrt{n}}{n^{1/2}}=1 $$ and $\limsup a_n \geq \lim \frac{ \sqrt{n}}{n^{-1/2}}= \infty \Rightarrow \liminf a_n \neq \limsup a_n$. – christina_g Mar 17 '16 at 22:10
  • @christina_g thank you – Error 404 Mar 17 '16 at 22:27

2 Answers2

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The sequence $\{\sin(n)\}$ is dense in $[-1,1]$ (see this answer: Sine function dense in $[-1,1]$)

This implies that for any $N\in\mathbb{N}$, I can find $n>N$ such that $\sin(n)$ is very close to 1, hence $\frac{\sqrt{n}}{n^{\sin n}}$ is approximately $n^{-\frac{1}{2}}$.

On the other hand, I can find $n>N$ such that $\sin(n)$ is very close to $-1$, so $\frac{\sqrt{n}}{n^{\sin n}}$ is approximately $n^{\frac{3}{2}}$.

Together, these facts imply that the sequence $\frac{\sqrt{n}}{n^{\sin n}}$ diverges.

Community
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carmichael561
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Since $\pi\gt3$, each interval of the form $[{(2k+1)\pi\over2}-{\pi\over6},{(2k+1)\pi\over2}+{\pi\over6}]$ contains an integer $n_k$ (because each such interval is of length $\pi/3\gt1$). When $k$ is even, $\sin n_k\gt\sin{\pi\over3}={\sqrt3\over2}$ so that

$${\sqrt{n_{2k}}\over n_{2k}^{\sin n_{2k}}}\lt{1\over n_{2k}^{(\sqrt3-1)2}}\to0$$

When $k$ is odd, $\sin n_k\lt-\sin{\pi\over3}=-{\sqrt3\over2}$ so that

$${\sqrt{n_{2k+1}}\over n_{2k+1}^{\sin n_{2k+1}}}\gt n_{2k+1}^{(\sqrt3+1)2}\to\infty$$

This is enough to conclude that limit $\lim_{n\to\infty}{\sqrt n\over n^{\sin n}}$ does not exist.

Barry Cipra
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