Calculus Problem by LOGA 1

Note that $(2 + \sqrt{3})^n + (2 - \sqrt{3})^n \; = \; 2\sum_{0 \le m \le \frac12n}\binom{n}{2m} 2^{n-2m} 3^m \; =\; 2K_n$ is an even integer, and hence $\sin\left((2 + \sqrt{3})^n\pi + \tfrac16\pi\right) \; = \; \sin\left(2K_n\pi + \tfrac16\pi - (2 -\sqrt{3})^n\pi\right) \; = \; \sin\left(\tfrac16\pi - (2 - \sqrt{3})^n\pi\right)$ Since $0 < 2 - \sqrt{3} < 1$ we deduce that $(2 - \sqrt{3})^n \to 0$ as $n \to \infty$ , and hence $\lim_{n \to \infty}\sin\left((2 + \sqrt{3})^n\pi + \tfrac16\pi\right) \; = \; \sin\tfrac16\pi = \tfrac12$

This is just an approximate and intuitive way to solve this problem:

First, note that choosing a bigger integer value of $n$ will result in a closer approximation of $(2+ \sqrt{3})^{n}$ to an even integer. For example:

As the value of $(2+ \sqrt{3})^{n}$ get nearer to an even number when $n$ increases, $\pi (2+ \sqrt{3})^{n}$ can be approximated and treated like $2k\pi$ (with $k \in \Z$ ) so that:

$\displaystyle \sin(\pi (2+ \sqrt{3})^{n} + \frac{\pi}{6}) \approx \sin(2k\pi + \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}$

$\implies \displaystyle \lim_{n \to \infty} \sin ( \pi (2+ \sqrt{3})^{n} + \frac{\pi}{6}) = \green{\boxed{\frac{1}{2}}}$