TITLE: Consider convergence of series: $\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$
QUESTION [10 upvotes]: 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?

REPLY [5 votes]: 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 $$\sum_{n=1}^{\infty}\sin\left[\pi\left(2-\sqrt{3}\right)^n\right]<\sum_{n=1}^{\infty}\sin[\pi(1/2)^n]<\sum_{n=1}^{\infty}\pi(1/2)^n=\pi$$.