# limit $\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)$

$\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)$

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 $n\in \mathbb{N}$

• 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. – Clement C. Aug 23 '16 at 21:32
• Does it even exist? – iamvegan Aug 23 '16 at 21:33
• I am not sure, but I think $\{n\sqrt{3}\}$ (the fractional part) fluctuates on $(0,1)$. This might lead a divergent sequence here. – iamvegan Aug 23 '16 at 21:39
• Well it turns out it does exist :) – iamvegan Aug 23 '16 at 21:53
• @Dr.MV Doesn't it also require that $\left|(k-\sqrt{l})\right| < 1$? $k^2 > l$ is not enough to guarantee that. – Michael Anderson Aug 24 '16 at 2:47

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}
• @SimpleArt restrict $x$ to be an integer. – YoTengoUnLCD Aug 24 '16 at 1:24
• @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
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.
since $$K(n)=(2+\sqrt3)^n+(2-\sqrt3)^n \in \mathbf{N}$$ $$\lim_{n\to ∞}(2-\sqrt3)^n=0$$ You can do the following: $$\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)=\lim_{n\to ∞}\sin(\pi(K(n)-(2-\sqrt3)^n))=-\lim_{n\to ∞}\sin(\pi(2-\sqrt3)^n)=0$$