Alternative ways to evaluate a limit of this sequence Find the limit $$\lim\limits_{n\to \infty} \sin \left( (2 + \sqrt 3 )^n\pi\right)$$ for $n \in \mathbb N$.
I know that this question has been asked earlier here.
However, I cannot convince myself about the answer given. I understood that $(2 + \sqrt 3 )^n$ approaches an even integer, but are there any other ways to show that this is true without using its conjugate? (like adding or subtracting integral multiples of $\pi$)
I tried doing that but couldn't show that the resulting sequence converges. Can anyone help me out with this? Thanks in advance.
 A: Expanding, $(2+\sqrt3)^n= \sum_{k=0}^n {n\choose k}2^{n-k}\sqrt3^k$.
Using the fact that, $\sin n\pi=0$ for $n\in \mathbb Z$ we can write
\begin{align}\sin ((2+\sqrt3)^n\pi)
&=\sin\left(\left(\sum_{k=0}^n {n\choose k}2^{n-k}\sqrt3^k\right)\pi\right)\\
&=\color{blue}{-\sin\left(-\left(\sum_{k=0}^n {n\choose k}2^{n-k}\sqrt3^k\right)\pi\right)}\\
&\color{blue}{=-\sin\left(\left(\sum_{k=0}^n {n\choose k}2^{n-k}(-\sqrt3)^k\right)\pi\right)}\\
&=-\sin ((2-\sqrt3)^n\pi)
\end{align}
Hence we have $\lim\limits_{n\to \infty} \sin \left( (2 + \sqrt 3 )^n\pi\right)=\lim\limits_{n\to \infty} -\sin \left( (2-\sqrt 3 )^n\pi\right)$
But $\sin x$ is continuous and $\lim\limits_{n\to \infty}(2-\sqrt 3 )^n\pi=0$ $\implies \lim\limits_{n\to \infty} -\sin \left( (2-\sqrt 3 )^n\pi\right)=0$

proof of $\color{blue}{\text{3rd}}$ equality:
$$\sin\left(-\left(\sum_{k=0}^n {n\choose k}2^{n-k}\sqrt3^k\right)\pi\right)=\sin\left(-\sum_{k=even} {n\choose k}2^{n-k}(\sqrt3)^k\pi - \sum_{k=odd} {n\choose k}2^{n-k}(\sqrt3)^k\pi\right)$$
(but since $\sqrt3^k$ is an integer when k is even, and $\sin n\pi=0 $ for $n\in \mathbb Z$)
$$=(-1)^{n+1}\sin\left(- \sum_{k=odd} {n\choose k}2^{n-k}(\sqrt3)^k\pi\right)=(-1)^{n+1}\sin\left(\sum_{k=odd} {n\choose k}2^{n-k}(-\sqrt3)^k\pi\right)$$
(again using the same facts)
$$=(-1)^{n+1}(-1)^{n+1}\sin\left(\sum_{k=odd} {n\choose k}2^{n-k}(-\sqrt3)^k\pi+\sum_{k=even} {n\choose k}2^{n-k}(-\sqrt3)^k\pi\right)=\sin\left(\left(\sum_{k=0}^n {n\choose k}2^{n-k}(-\sqrt3)^k\right)\pi\right)$$
