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

Solution 1:

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.

Solution 2:

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

Solution 3:

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.