Consider convergence of series: $\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$ Consider convergence of series:
$$\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$$
My tried:
We have
$$\sum_{n=1}^{\infty }\sin(\pi (2+\sqrt{3})^{n})=\sum_{n=1}^{\infty}\sin\left(\pi[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}]-\pi(2-\sqrt{3})^{n}\right)\, (*)$$
Because $$(2+\sqrt{3})^{n}=\sum_{k=0}^{n}C_{n}^{k}2^{n-k}3^{\frac{k}{2}}$$
                $$(2-\sqrt{3})^{n}=\sum_{k=0}^{n}(-1)^{k}C_{n}^{k}2^{n-k}3^{\frac{k}{2}}$$
Hence $$(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}=\left\{\begin{matrix} 0&,k=2l+1 \\ m\in N&,k=2l \end{matrix}\right.$$
       $$\Rightarrow (1)=\sum_{n=1}^{\infty}\sin\left(m\pi-\pi(2-\sqrt{3})^{n}\right)=\sum_{n=1}^{\infty}(-1)^{m+1}\sin\frac{\pi}{(2+\sqrt{3})^{n}}<\sum_{n=1}^{\infty}\sin\frac{\pi}{(2+\sqrt{3})^n}$$
$\sum_{n=1}^{\infty}\sin\frac{\pi}{(2+\sqrt{3})^n}$ converge
Hence series is converge.
True or False?
 A: A simple way to show convergence is to note that $$(2+\sqrt{3})^2=7+4\sqrt{3}=2*7-(2- \sqrt{3})^2$$
$$(2+\sqrt{3})^3=26+15\sqrt{3}=2*26-(2-\sqrt{3})^3$$
$$(2+\sqrt{3})^4=97+56\sqrt{3}=2*97-(2-\sqrt{3})^4$$
and in general
$$(2+\sqrt{3})^n=2k-(2-\sqrt{3})^n$$
with k integer.
The summation then reduces to $$-\sum_{n=1}^{\infty}\sin\left[\pi\left(2-\sqrt{3}\right)^n\right]$$
where the terms clearly tend to zero as n tends to infinity. Convergence can therefore easily be shown using the comparison test, observing that $$\sum_{n=1}^{\infty}\sin\left[\pi\left(2-\sqrt{3}\right)^n\right]<\sum_{n=1}^{\infty}\sin[\pi(1/2)^n]<\sum_{n=1}^{\infty}\pi(1/2)^n=\pi$$.
A: As you noticed,
$$\sin(\pi(2+\sqrt{3})^n+\pi(2-\sqrt{3})^n)=0$$
so$$\sin(\pi(2+\sqrt{3})^n)=-\dfrac{\sin(\pi(2-\sqrt{3})^n)\cos(\pi(2+\sqrt{3})^n)}{\cos(\pi(2-\sqrt{3})^n)}$$
or for $n$ large enough
$$\begin{cases}  \sin(\pi(2-\sqrt{3})^n)=\pi(2-\sqrt{3})^n+o(\pi(2-\sqrt{3})^n)\\ \cos(\pi(2-\sqrt{3})^n)\ge \frac12 \end{cases}$$
and
$$|\cos(\pi(2+\sqrt{3})^n)|\le1$$
Therefore
$$|n^2\sin(\pi(2+\sqrt{3})^n)|\le 2\pi n^2(2-\sqrt{3})^n+o(n^2(2-\sqrt{3})^n)$$
$$|n^2\sin(\pi(2+\sqrt{3})^n)|\le 2\pi \exp(2\ln(n)-n\ln(2-\sqrt{3}))+o(n^2(2-\sqrt{3})^n)\to 0$$
so your serie converge.
A: Let $T_n$ be the sequence $(2 + \sqrt{3})^n + (2-\sqrt{3})^n$ for $n \in \mathbb{N}$.
Since 
$$(\lambda - (2 + \sqrt{3}))(\lambda - (2-\sqrt{3})) = \lambda^2 - 4\lambda + 1,$$
$T_n$ satisfies a linear recurrence relation:
$$T_{n+2} - 4T_{n+1} + T_{n} = 0$$
Notice $T_0 = 2$ and $T_1 = 4$ are both even, this implies $T_n$ are even for all $n \in \mathbb{N}$ and hence
$$\sin\big(\pi(2 + \sqrt{3})^n\big) 
= \sin\big(\pi( T_n - (2-\sqrt{3})^n)\big)
= - \sin\big(\pi(2 - \sqrt{3})^n\big)
$$
for all $n \ge 1$.
Since $0 < 2-\sqrt{3} < 1$, the sequence all has negative sign. 
Since $0 < \sin x < x$ for $x \in (0,\pi)$, we have
$\displaystyle\;\;-\pi(2-\sqrt{3})^n < \sin\big(\pi(2+\sqrt{3})^n\big) < 0$.
This means the sequence is always negative and bounded below by the negative of a geometric series. As a result, the series 
$\displaystyle\;\sum_{n=1}^\infty \sin\big(\pi(2+\sqrt{3})^n\big)\;$ converges to 
some negative number greater than $-\frac{\pi(2-\sqrt{3})}{1-(2-\sqrt{3})} = -\frac{\pi}{\sqrt{3}+1} \sim -1.149902719556431$.
Random Notes
I have done a little bit of online search. It turns out in certain sense, the convergence of $\sum\limits_{n=1}^\infty \sin\big(\pi(2+\sqrt{3})^n\big)$ is exceptional!
It is known that for almost all $\alpha > 1$ (i.e except a set of Lebesgue measure $0$), 
$\{ \alpha^n \}$, the fractional part of $\alpha^n$ is an equidistributed sequence. A consequence of this is for
almost all $\alpha > 1$, the sequence $\sin(\pi \alpha^n)$ does not converge to $0$ and
hence the series $\sum\limits_{n=1}^\infty \sin(\pi \alpha^n)$ diverges.
There are known exceptions to this. In particular, it is known that 
$\{ \alpha^n \}$ is not equidistributed mod 1 if $\alpha$ is a 
PV number.
i.e. an algebraic integer $\alpha > 1$ and all other roots of its minimal polynomials lie strictly inside the unit circle. 
Quoting wiki, the powers of PV numbers have a very "biased" distribution (mod 1). 

If $\alpha$ is a PV number and $\lambda$ is any algebraic integer in the field $\mathbb{Q}(\alpha)$ then the sequence $\|\lambda\alpha^n\|$, where $\|x\|$ denotes the distance from the real number $x$ to the nearest integer, approaches $0$ at an exponential rate.

This implies $\sum\limits_{n=1}^\infty \sin(\pi\alpha^n)$ converges whenever $\alpha$ is a PV number. Since $2 + \sqrt{3}$ is a PV number, the corresponding series do converge.
