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

Question

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?

What you had in your head when writing this was impeccable. What you wrote in the last few lines is total gibberish. — fedja, Jun 16 '14 at 02:33
$m$ is always even and the terms in your series are negative for $n \ge 1$. — achille hui, Jun 16 '14 at 02:56
Then is that right lim sin [π(2+√3)^n] --> 0 ? In my opinion , the limit , probably , does not converge . But I can't prove it. — user128766, Jun 16 '14 at 03:02
@user128766 I would say it's very hard to prove it. Take a look at this. I'll be waiting for someone to post an answer, I'm not even going to try. — Git Gud, Jun 16 '14 at 03:03
@user128766, the limit is $0$ and the sum converges to $-\sum_{n=1}^\infty \sin\left(\frac{\pi}{(2+\sqrt{3})^n}\right)$. — achille hui, Jun 16 '14 at 03:12
@GitGud This one isn't that one; the connection between powers of quadratics and integer recurrence relations lets one bound the distance between individual terms here and multiples of $\pi$ by an exponentially decreasing value. — Steven Stadnicki, Jun 16 '14 at 04:24
Possible duplicate of Does $\sum_{n=1}^{\infty} \sin(\pi(2+\sqrt{3})^n)$ converge? Converge absolutely? — Martin Nicholson, Oct 13 '17 at 15:30

achille hui · Answer 1 · 2018-03-07T23:49:35.947

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.

I have understood everything in your answer. The only point I have not already got is why $T_n$ satisfies that linear recurrence relation. Would you mind if I ask you to explain to me the logic of that statement? @achillehui — Charlie, Sep 05 '14 at 15:59
@Charlie, Given any two distinct numbers $\alpha, \beta$, the sequence $a_n$ defined by $a_n = \alpha^n$ satisfies $a_{n+1} - \alpha a_{n} = 0$, so $$a_{n+2} - (\alpha+\beta)a_{n+1} + \alpha\beta a_{n} = (a_{n+2} - \alpha a_{n+1}) - \beta (a_{n+1} - \alpha a_n) = 0 - \beta 0 = 0$$ Similarly, the sequences $b_n$ defined by $b_n = \beta^n$ satisfies $$b_{n+2} - (\alpha+\beta)b_{n+1} + \alpha\beta b_{n} = 0$$ Combine these two, we find any linear combination of $a_n$ and $b_n$, i.e. $c_n = A\alpha^n + B\beta^n$ satisfies: $$c_{n+2} - (\alpha+\beta) c_{n+1} + \alpha\beta c_{n} = 0$$ — achille hui, Sep 05 '14 at 16:15

Anatoly · Answer 2 · 2014-06-16T14:46:16.327

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

Fabien · Answer 3 · 2014-06-16T09:56:19.330

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.

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

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