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I think it doesn't exists but I can't find out how to prove it; note that $n$ is integer.

I know that there are infinite many integers $a_n$ s.t. $\sin(a_n)>0$ and then limsup is $+ \infty$.

I want to prove that liminf $\ne +\infty$ but I don't know how.

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  • $\begingroup$ How efficiently can you approximate $\pi$ with rationals? $\endgroup$
    – anomaly
    Commented May 25, 2017 at 18:31
  • $\begingroup$ I know that there are sequences of integers s.t. $|p_m - \pi q_m| < 1/q_m$ $\endgroup$
    – Bremen000
    Commented May 25, 2017 at 18:44
  • $\begingroup$ You want to show the $\lim\sup$ is different than $\lim\inf$. Think about subsequences of the naturals where $\sin$ is very close to $1$, and where $\sin$ is very close to $-1$. $\endgroup$
    – Steve D
    Commented May 25, 2017 at 19:18
  • $\begingroup$ @SteveD he knows that, the actual proof is the difficult part. $\endgroup$
    – Mickey
    Commented May 25, 2017 at 19:21
  • $\begingroup$ @Mickey: The fact "he knows that" was not evident from the question at the time of my comment. $\endgroup$
    – Steve D
    Commented May 25, 2017 at 19:43

2 Answers 2

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The limit doesn't exist. We are going to prove it the hard way.

  1. $\limsup\limits_{n\to\infty}n(\sin n +1 ) = +\infty$

    If one make a plot of $\max( \sin(x),\sin(x+3) )$ over $[0,2\pi]$, one will notice it is bounded from below by $-2\cos(1.5) \approx -0.0707372$. For any $n$, at least one of $\sin n + 1$ or $\sin(n+3) + 1$ will be greater than $0.9$. As a result $$\limsup_{n\to\infty} n(\sin n+1) \ge \limsup_{n\to\infty} 0.9n = \infty$$

  2. $\liminf_\limits{n\to\infty} n(\sin n+1) = 0$.

    We will use a theorem by Minkowski,

    Given any $\theta \notin \mathbb{Q}$ and $\alpha \notin \mathbb{Z}$ such that $x - \theta y - \alpha = 0$ has no integer solutions.
    For any $\epsilon > 0$, there exists infinitely many pairs of integers $p,q$ such that $$|q(p - \theta q - \alpha)| < \frac14\quad\text{ and }\quad |p - \theta q - \alpha| < \epsilon$$

    Apply this to $\theta = \frac{1}{2\pi}$ and $\alpha = -\frac34$, there are infinitely many pairs of non-zero $p,q$ such that $$\left|q - 2\pi\left(p+\frac34\right)\right| < \frac{\pi}{2|q|}$$ If $q$ is positive, let $n = q, m = p$. If $q$ is negative, let $n = -3q$ and $m = -3(p+1)$.
    Above inequality reduces to $$\left|n - 2\pi\left(m + \frac34\right)\right| < \frac{\pi}{2n} \times \begin{cases}1, & q > 0\\ 9, & q < 0\end{cases}$$ Together with another easy to verify inequality, $$0 \le \sin\left(\frac{3\pi}{2}+\theta\right) + 1 \le \frac{\theta^2}{2}$$ we obtain infinitely many positive $n$ such that $$0 \le n(\sin(n) + 1) \le \frac{81\pi^2}{8n}$$ This implies $$\liminf_{n\to\infty} n(\sin n+1) \le \liminf_{n\to\infty} \frac{81\pi^2}{8n} = 0$$ Since that limit inferior has to be non-negative, it has to be zero.

Combine $1)$ and $2)$, we can conclude $\;\lim_{n\to\infty}n(\sin n + 1)\;$ doesn't exist.

Notes

About this particular theorem (Minkowski has tons of theorems), Minkowski first proved it around $1901$ using some geometrical methods he invented. Nowadays, those methods are known as Geometry of numbers. Look at any book under this subject, you should find a proof of that.

I learned this stuff from a book The Geometry of Numbers by Olds, Lax and Davidoff's. It is easy to read and has a proof at Chapter 10. Another proof can be found at Chapter 2 of Ivan Niven's classic Diophantine Approximations.

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  • $\begingroup$ Thank you! Do you know where I can find a proof of Minkowsky theorem? $\endgroup$
    – Bremen000
    Commented May 25, 2017 at 20:36
  • $\begingroup$ @Bremen000 I updated my answer with some refs. $\endgroup$ Commented May 25, 2017 at 21:15
  • $\begingroup$ @achillehui Can you take a look at my answer and see whether it is true or not ? $\endgroup$
    – S.H.W
    Commented May 25, 2017 at 21:17
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    $\begingroup$ The sequence $1,2,3,\dots$ lands in each of $(0,\pi), (2\pi,3\pi), (4\pi,5\pi),\dots,$ which gives $\sin n > 0$ for infinitely many $n.$ $\endgroup$
    – zhw.
    Commented May 26, 2017 at 15:21
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    $\begingroup$ @Bremen000 No, we can't. We don't know what is the value of $\mu(\pi)$ ( the irrationality measure for $\pi$ ). We just know $2 \le \mu(\pi) < 7.6063$. If we know the value of $\mu(\pi)$, depends on whether $p$ is less than or greater than $2(\mu(\pi)-1)$, it might be possible to show $\liminf_{n\to\infty} n^p(1+\sin(n)) = 0$ or $+\infty$ using a similar technique.... $\endgroup$ Commented May 30, 2017 at 19:47
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consider $$u_n=v_n.w_n $$ with $\lim v_n=+\infty $ and $\lim w_n $ doesn't exist, so to compute $\lim u_n $, we get an ideterminate form.

$$\{\sin (n),n\in\mathbb N\} $$ is dense in $[-1,1] $ .

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  • $\begingroup$ You haven't disproved that the limit is $\infty.$ $\endgroup$
    – zhw.
    Commented May 25, 2017 at 18:55
  • $\begingroup$ @zhw. Ok. i will edit it now. $\endgroup$ Commented May 25, 2017 at 18:56
  • $\begingroup$ I already knew that $\lim_{n \to \infty} \sin(n)$ doesn't exists but i do not understand how "we conclude..." $\endgroup$
    – Bremen000
    Commented May 25, 2017 at 19:11

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