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Using only the delta definition of a limit, how can we prove that the sequence $\{a_n\}$, where $a_n = \sin n$, as $n$ tends to infinity does not have a limit?

Thanks!

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    Follows from stronger result at http://math.stackexchange.com/questions/9319/sinn-subsequence-limits-set; Related: http://math.stackexchange.com/questions/22047/sinn-is-not-u-d-mod-1; 2 more versions of first link: http://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1, http://math.stackexchange.com/questions/7252/is-sinn-for-all-n-in-the-intergers-dense2011-03-15
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    This question is asking for something weaker than the density asked for in previous problems, and correspondingly is easier to answer, and therefore I am unsure whether it should be considered a duplicate.2011-03-15
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    See http://www.mathkb.com/Uwe/Forum.aspx/math/16396/how-to-prove-that-lim-n-oo-sin-n-does-not-exist-but-in-special-way (you'll probably find at least one satisfying answer there).2011-03-15
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    @Arturo Magidin There was something obviously wrong with my notation the first time I edited this question. For future reference, what was it, so I can be more observant in my own problem-solving?2011-03-16
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    @Billare: $\{a_n\}$ is interpreted as a sequence (or a family of terms); $\{a_n\}=\sin n$ just looks wrong: $\sin n$ is itself not a sequence or a set of values.2011-03-16

5 Answers 5

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No need for $\epsilon$ actually. If $\sin(n) \rightarrow l$, then $\sin(n+1)$ also, and $\sin(n+1)=\sin(n)\cos(1)+\sin(1)\cos(n)$. Since both $\sin(n)$ and $\sin(n+1)$ have limit $l$ and $\sin(1) \neq 0$, $\cos(n) \rightarrow \frac{l(1-\cos(1))}{\sin(1)}$, and so $e^{in}=\cos(n)+i \sin(n)$ has a limit. But $e^{i(n+1)}$ must then have the same limit (call it $x$), which implies $x=e^{i} x$, and since $e^{i} \neq 1$, $x$ has to be zero, which is a contradiction with the fact that $|e^{in}|=1$.

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Assume $\lim \sin(n) = l$. Then so is $\lim \sin(2n) = l$. So $\lim \cos(2n) = 1 - 2l^2$, but so does the limit of $\cos(2(n + 1))$. Now apply the sum-formula to $\sin(2(n + 1) - 2n)$.

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    hey jonas, can you please explain me the transformation fron sin to cos and what to do afterwards a little bit more?2011-03-15
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    There exists infinitely many integers $n$ such that $\sin(n)$, $\sin(n+1)$ and $\sin(n+2)$ are all in the interval $[\frac12,1]$, so I fail to see the conclusion of the argument in your last paragraph.2011-03-15
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    @Nir: That is just a double angle formula.2011-03-15
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    @Didier: Err... I will fix that. Well, anyway, I see that Henry added what I mean.2011-03-15
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The following are true, based on standard trigonometric identities and $\sin(1) \approx 0.84147$ and $\sin(3) \approx 0.14112$:

$$\begin{align} \textrm{if } \sin(n) \le -0.4, & \textrm{ then } 0 < \sin(n+3) ; \\ \textrm{if } -0.4 \le \sin(n) \le 0.4, & \textrm{ then } \sin(n+1) < -0.4 \textrm{ or } 0.4 < \sin(n+1) ; \\ \textrm{if } 0.4 \le \sin(n),& \textrm{ then } \sin(n+3) < 0; \end{align}$$

so there is no value $L$ where for any positive $\varepsilon < 0.2$ you have all of $\sin(n), \sin(n+1), \sin(n+3)$ and $\sin(n+4)$ within $\varepsilon$ of $L$.

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Let if possible $\sin n\rightarrow x$. Then $\sin k=\sin(n+k-n)=\sin(n+k)\cos k-\cos(n+k)\sin n\rightarrow x(\cos k-\sqrt{1-x^2})$ for each positive integer k. Now as $k\rightarrow \infty$ implies that $x=x.0=0$ which shows that $\sin k=0$ for all k...a contradiction.

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In any interval of the form $ [k\pi +\frac{\pi}{3},k\pi +\frac{2\pi}{3}]$, where $k $ is any natural number, there is at least a natural number $n_{k}$. The reason is that any such interval has length $\pi/3$ that is greater than 1. Since those intervals are mutually disjoint then the sequence $\{n_{k}\}$ is a sub-sequence of the sequence $\{n\}$ and, obviously, $|\sin n_k|\geq \frac{\sqrt{3}}{2}$. In a similar way, considering the intervals of the form $[k\pi -\frac{\pi}{6},k\pi +\frac{\pi}{6}]$, we can construct another sub-sequence $\{m_k\},$ of the sequence $\{n\},$ such that $|\sin m_k|\leq \frac{1}{2}$. Assume that $\lim_{n\to \infty}\sin{n}$ exists and is the number $l$. Using both sub-sequences defined above, we obtain that $|l|\geq \frac{\sqrt{3}}{2}$ and $|l|\leq \frac{1}{2}$, and this is a contradiction.