# Test the convergence of the series $\sum \sin [\pi(\sqrt5+2)^n]$ [duplicate]

I have just approached the following series

$$\sum_{n=1}^\infty \sin [\pi(\sqrt5+2)^n]$$ And I already have a question. The $\lim_{n \to \infty}\pi(\sqrt5+2)^n=+\infty$. And the $\lim_{n\to \infty} \sin[\pi(\sqrt5+2)^n]$ does not exist.

Can $\sum a_n$ converge if $\lim_{n\to \infty}a_n$ does not exist? I guess the answer is yes because I know that $\sum a_n$ cannot converge if $\lim_{n\to \infty}a_n\neq 0$. Am I right?

The second part of my question concerns the strategy to use to test convergence of such a series. I can test convergence of series with standard tests. But it seems I need to develop a more sophisticated approach in order to successfully deal with series like this. For example, here, should I use some sort of expansion? If yes, what kind of expansion?

## marked as duplicate by Aqua, erfink, user99914, qwr, JMPOct 14 '17 at 6:05

• If $\sum a_n$ converges, then $a_n \to 0$. – Travis Sep 5 '14 at 13:53
• If the limit of the sequence is not equal to zero (and you showed that already) then certainly the corresponding series cannot be convergent – imranfat Sep 5 '14 at 13:54
• Even though $(\sqrt5+2)^n\to\infty$, it could possibly be that this number is very close to an integer when $n$ is large, in which case the sum could still converge (but I don't think it will). – Harald Hanche-Olsen Sep 5 '14 at 13:54
• $\sin(\pi(2+\sqrt{5})^n) = -\sin(\pi(2-\sqrt{5})^n)$ and the series converges. The proof will be very similar to what you can find in this answer – achille hui Sep 5 '14 at 14:04

As $n$ increases, $(2+\sqrt{5})^n$ gets closer and closer to an integer, since: $$(2+\sqrt{5})^n + (2-\sqrt{5})^n \in \mathbb{Z}.$$ If $[x]$ is the distance of $x$ from the closest integer, we have: $$[(2+\sqrt{5})^n] = (\sqrt{5}-2)^n =\frac{1}{(2+\sqrt{5})^n}<\frac{1}{4^n},$$ hence: $$|\sin(\pi(2+\sqrt{5})^n)| \leq \frac{\pi}{4^n},$$ giving that the original series is convergent.
• Why $(2+\sqrt5)^n+(2-\sqrt5)^n \in \Bbb Z$? – Charlie Sep 5 '14 at 14:52
• Because $A_n = (2+\sqrt{5})^n + (2-\sqrt{5})^n$ satisfies the recurrence relation $$A_0=2,\quad A_1=4,\quad A_{n+2}= 4 A_{n+1}-A_n.$$ Or just use the binomial theorem and see that all the terms $k\sqrt{5}$ cancel out. – Jack D'Aurizio Sep 5 '14 at 15:00
• I am also studying, at the same time, the other answer by @achillehui (read comments under my question). I cannot understand how you derive the fact that our $A_n$ satisfies a linear recurrence relation. Could you explicit the logic please? – Charlie Sep 5 '14 at 15:17
• Every sequence defined by $$A_n = k_1 \sigma^n + k_2 \bar{\sigma}^n,$$ where $\sigma,\bar{\sigma}$ are the roots of a second-degree polynomial $t^2-st+p$, satisfies the recurrence relation $$A_{n+2} = s A_{n+1} - p A_{n},$$ where $s=\sigma+\bar{\sigma}$ and $p=\sigma\bar{\sigma}$ due to Vieta's formulas. – Jack D'Aurizio Sep 5 '14 at 15:20
• In a neighbourhood of a zero (i.e. in a neighbourhood of a point belonging to $\pi\mathbb{Z}$) the sine function behaves almost like a linear function. To be clear: $$z\in\pi\mathbb{Z},|t|<\pi/2 \Longrightarrow |\sin(z+t)|\leq |t|.$$ – Jack D'Aurizio Sep 5 '14 at 15:51