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(1) It is easy to prove that $\lim\limits_{n\to\infty}{\sin(n)}$ does not exist.
(2) I want to ask how to prove that $\lim\limits_{n\to\infty}{\sin(n^2)}$ does not exist.
(3) Furthermore, $\lim\limits_{n\to\infty}{\sin(n^k)}$ does not exist. ($k$ is a positive integer.)
(4) In addition, $\lim\limits_{n\to\infty}{\sin(n^{\alpha})}$ does not exist. ($\alpha$ is a positive real number.)

Gary
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gaoxinge
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    Please put your own thoughts on this problem to avoid closure. I personally want interesting problems like this to be discussed. – Sungjin Kim Oct 15 '13 at 07:18
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    Try to prove that if $\lim_{n\to\infty}f(n)$ exists, then so does, $\lim_{n\to\infty}f(n^\alpha)$, and $\lim_{n\to\infty}f(n)=\lim_{n\to\infty}f(n^\alpha)$. Conclude from this that one limit exists if and only if the other does. – Baby Dragon Oct 15 '13 at 07:47
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    @BabyDragon Not in this case. $n$ represents natural number. Thus, the limit is taken over natural number $n$ goes to infinity. – Sungjin Kim Oct 15 '13 at 07:49
  • possible duplicate of How prove this $\lim_{n\to \infty}\sin{n^m}$ divergent. – user37238 Oct 15 '13 at 08:09
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    @user37238 Not exactly. The question on the link does not contain case (4). – Sungjin Kim Oct 15 '13 at 08:27
  • (1)If $\lim\limits_{n\to\infty}{sin(n)}=\lim\limits_{n\to\infty}{sin(n+1)}$ exists, then $sin(n+1)-sin(n-1)=2sin(n)sin(1)$ can deduce $0=\lim\limits_{n\to\infty}{sin(n+1)-sin(n-1)}=\lim\limits_{n\to\infty}{2sin(n)sin(1)}$. It means $\lim\limits_{n\to\infty}{sin(n)}=0$ – gaoxinge Oct 15 '13 at 14:14
  • (2)If $\lim\limits_{n\to\infty}{sinn^2}=\lim\limits_{n\to\infty}{sin(n+1)^2}=\lim\limits_{n\to\infty}{sin(n+2)^2}=a$ exists, then $\lim\limits_{n\to\infty}{sin(n^2+1)con(2n)}=\lim\limits_{n\to\infty}{sin(n^2+4)cos(4n)=a} and \lim\limits_{n\to\infty}{cos(n^2+1)sin(2n)}=\lim\limits_{n\to\infty}{cos(n^2+4)sin(4n)=0}$. Then we have $0=\lim\limits_{n\to\infty}{cos(n^2+4)sin(4n)}=2(\lim\limits_{n\to\infty}{cos(n^2+1)sin(2n)cos(2n)cos(3)}-\lim\limits_{n\to\infty}{sin(n^2+1)cos(2n)sin(2n)sin(3)})$.
    It means $\lim\limits_{n\to\infty}{sinn^2}=0$     – gaoxinge Oct 15 '13 at 14:26
  • (3)There is a link to this question. Although there is no answer to the (4), I think this is awesome. – gaoxinge Oct 15 '13 at 14:34
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    @gaoxinge I do not see what you are doing in above comments. Especially the second one. Did you notice that I posted an answer to (1),(2),(3), and partially (4)? – Sungjin Kim Oct 15 '13 at 19:15
  • The tag ([tag:limit-theorems]) is intended for questions about limit theorems in probability theory and not for questions about determining limits of sequences or functions, see the tag-wiki and the tag-excerpt. (The tag-excerpt is also shown when you are adding a tag to a question.) – Martin Sleziak Nov 29 '13 at 17:30

2 Answers2

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Concepts I use here is "Weyl's Equidistribution Criterion"

Reference: http://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/

(1) The limit does not exist simply because the sequence $\{e^{in}\}$ is equidistributed on the unit circle.

(2), (3) follows from Van der Corput's lemma: We can bring down exponent $k$-case to exponent $1$-case. Better way to put it is:

For any integer $k\geq 1$, the sequence $\{e^{in^{k}}\}$ is equidistributed on the unit circle.

(4) For $0<\alpha<1$, the sequence $\{e^{in^{\alpha}}\}$ is dense in the unit circle, so the limit of $\sin(n^{\alpha})$ does not exist.

For $\alpha>1$, and $\alpha=\frac{q}{p}\in\mathbb{Q}^{+}$, consider $\{\sin(m^p)^{\frac{q}{p}} \}$. This is a subsequence of $\{\sin (n^{\frac{q}{p}})\}$. The subsequence $\{\sin(m^q)\}$ diverges by (3). Thus the original sequence diverges as well.

Remaining case is now $\alpha>1$ and irrational.

This case can be settled by the lemma:

(Lemma) For a sequence of real numbers $\{x_n\}$, suppose that the set of all subsequential limits of $\{\exp(ix_n)\}$ is finite. Denote $D$ the difference sequence operator on the space of real sequences. Then for $\{y_n:=Dx_n=x_{n+1}-x_n\}$, we have also that the set of all subsequential limits of $\{\exp(iy_n)\}$ is finite.

If $x_n=n^{\alpha}$, and $\lfloor\alpha\rfloor=m$, then $z_n:=D^m x_n \sim cn^{\alpha-m}$ for some positive constant $c$. Thus, by the same reason for the case $0<\alpha<1$, we have $\{\exp(iz_n)\}$ is dense in the unit circle.

If $\{\sin(n^{\alpha})\}$ has a limit, then $\{\exp(ix_n)\}$ has a finite set of sequential limits. Then by lemma applied $m$-times, $\{\exp(iz_n)\}$ also has a finite set of sequential limits. This contradicts above.

Hence, $\{\sin(n^{\alpha})\}$ does not have a limit.

Sungjin Kim
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  • I'm sorry. Your opinion is fascinating.
    Weyl's Equidistribution Criterion
    A real sequence ${x_{n}}$ in $[0,1)$ is equidistributed if and only if for all integers $k\not=0$ one has $\frac{1}{N}\sum_{n=1}^{N}e^{2{\pi}ikx_{n}}\rightarrow{0}$, as $N\rightarrow\infty$.
    I have tried to estimate the value of $\frac{1}{N}\sum_{n=1}^{N}e^{2{\pi}ikx_{n}}$, but I failed. Could you tell me how you estimate the value when $k$ is a positive integer and $0<\alpha<1$? Thank you.     – gaoxinge Oct 16 '13 at 08:08
  • (3) and (4) are my conjectures. – gaoxinge Oct 16 '13 at 08:46
  • Follow the link(T.Tao's lecture note): (3) is explained in Lemma 4, Corollary 5, and Corollary 6. – Sungjin Kim Oct 16 '13 at 18:13
  • (4) $0<\alpha<1$ case is because $n^{\alpha}$ increases to infinity, but the difference $(n+1)^{\alpha}-n^{\alpha}$ decreases to $0$. – Sungjin Kim Oct 16 '13 at 18:15
  • Thank you for insisting on solving this problem. – gaoxinge Oct 28 '13 at 06:25
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We can prove that $\exists \{c_i\}_{i=1}^{\infty} \subseteq \mathbb N$ and $c_1<c_2<c_3<...$ such that $\lim_{i\rightarrow \infty}\sin (c_i^{\alpha})=c$, $\forall c\in [-1,1]$,where $\alpha $ is a positive rational number.

chenyuandong
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