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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 $n\in \mathbb{N}$

jimjim
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Debarun Mukherjee
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3 Answers3

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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}

xpaul
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    (+1) This is interesting! The function looks like fluctuating but apparently it is not. – iamvegan Aug 23 '16 at 21:52
  • Then how do you explain the graph?!?!?!? https://www.desmos.com/calculator/cmsnuraqfr – Simply Beautiful Art Aug 24 '16 at 00:41
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    @SimpleArt restrict $x$ to be an integer. – YoTengoUnLCD Aug 24 '16 at 01:24
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    @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. – CiaPan Aug 24 '16 at 10:25
  • @SimpleArt Anytime when $n$ is used as a variable, it is quietly assumed that we are dealing with integer inputs... – imranfat Aug 24 '16 at 15:39
  • This is confusing because the function does not converge over the reals, but it does converge over the integers. – user3294068 Aug 24 '16 at 20:05
  • @user3294068 Well, different properties for the integers, different simplifications and all. – Simply Beautiful Art Aug 24 '16 at 20:12
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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.

Jack D'Aurizio
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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: $$\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)=\lim_{n\to ∞}\sin(\pi(K(n)-(2-\sqrt3)^n))=-\lim_{n\to ∞}\sin(\pi(2-\sqrt3)^n)=0 $$

Frank
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