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$\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)$

I tried to write it as $\sin (n\pi - \theta)$ to get the form $∞-∞$ form within $\sin$ function. But could not proceed after that. How should I do it?

Edit:I am sorry, I forgot to mention $n\in \mathbb{N}$

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  • $\begingroup$ An idea (haven't pursued the line of reasoning further): Maybe rewriting $(2+\sqrt{3})^n = \sum_{k=0}^n\binom{n}{k} 2^{n-k} \sqrt{3}^k$ and splitting the sum depending on the parity of $k$ may help. $\endgroup$
    – Clement C.
    Aug 23, 2016 at 21:32
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    $\begingroup$ Does it even exist? $\endgroup$
    – iamvegan
    Aug 23, 2016 at 21:33
  • $\begingroup$ I am not sure, but I think $\{n\sqrt{3}\}$ (the fractional part) fluctuates on $(0,1)$. This might lead a divergent sequence here. $\endgroup$
    – iamvegan
    Aug 23, 2016 at 21:39
  • $\begingroup$ Well it turns out it does exist :) $\endgroup$
    – iamvegan
    Aug 23, 2016 at 21:53
  • $\begingroup$ @Dr.MV Doesn't it also require that $\left|(k-\sqrt{l})\right| < 1$? $k^2 > l$ is not enough to guarantee that. $\endgroup$ Aug 24, 2016 at 2:47

3 Answers 3

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Note $|2-\sqrt{3}|<1$ and hence $\lim_{n\to\infty}(2-\sqrt{3})^n=0$. Since $(2+\sqrt{3})^n+(2-\sqrt{3})^n$ is an integer, one has \begin{eqnarray} &&\lim_{n\to\infty}\sin[\pi(2+\sqrt{3})^n]\\ &=&\lim_{n\to\infty}\sin\bigg[\pi\big[(2+\sqrt{3})^n+(2-\sqrt{3})^n\big]-\pi(2-\sqrt{3})^n\bigg]\\ &=&\lim_{n\to\infty}\bigg\{\sin\bigg[\pi\big[(2+\sqrt{3})^n+(2-\sqrt{3})^n\big]\bigg]\cos\big[\pi(2-\sqrt{3})^n\big]\\ &&-\cos\bigg[\pi\big[(2+\sqrt{3})^n+(2-\sqrt{3})^n\big]\bigg]\sin\big[\pi(2-\sqrt{3})^n\big]\bigg\}\\ &=&0. \end{eqnarray}

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    $\begingroup$ (+1) This is interesting! The function looks like fluctuating but apparently it is not. $\endgroup$
    – iamvegan
    Aug 23, 2016 at 21:52
  • $\begingroup$ Then how do you explain the graph?!?!?!? desmos.com/calculator/cmsnuraqfr $\endgroup$ Aug 24, 2016 at 0:41
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    $\begingroup$ @SimpleArt restrict $x$ to be an integer. $\endgroup$ Aug 24, 2016 at 1:24
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    $\begingroup$ @SimpleArt You can replace $x$ with $\operatorname{floor}(x)$, as YoTengoUnLCD says. However that makes a diverging graph too, for $x$ greater than twenty-something, apparently due to a limited precision of power calculation. Which makes your answer a good example why experimental methods (graphs among them) do not work as proofs in mathematics. $\endgroup$
    – CiaPan
    Aug 24, 2016 at 10:25
  • $\begingroup$ @SimpleArt Anytime when $n$ is used as a variable, it is quietly assumed that we are dealing with integer inputs... $\endgroup$
    – imranfat
    Aug 24, 2016 at 15:39
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Hint(s): $(2+\sqrt{3})^n$ is closer and closer to an integer as $n$ increases, since $$ (2+\sqrt{3})^n+(2-\sqrt{3})^n $$ is an integer and $|2-\sqrt{3}|<\frac{1}{3}$. The sine function is a Lipschitz-continuous function and $\sin(\pi m)=0$ for any integer $m$, hence your limit equals zero.

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since $$ K(n)=(2+\sqrt3)^n+(2-\sqrt3)^n \in \mathbf{N} $$ $$\lim_{n\to ∞}(2-\sqrt3)^n=0$$ You can do the following: $$\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)=\lim_{n\to ∞}\sin(\pi(K(n)-(2-\sqrt3)^n))=-\lim_{n\to ∞}\sin(\pi(2-\sqrt3)^n)=0 $$

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