TITLE: Does the series $\sum_{n=0}^{+\infty} \sin((1+\sqrt{2})^n\pi)$ converge?
QUESTION [6 upvotes]: I am trying the following exercise,

Convergence of the series $\sum_{n=0}^{+\infty} \sin((1+\sqrt{2})^n\pi)$

I tried like the method for $\sum_{n=0}^{+\infty}  \sin((2+\sqrt{3})^n\pi)$,
with $u_n=\sin((1-\sqrt{2})^n\pi)$ unfortunately $(1-\sqrt{2})<0$ so I cannot use theorem for positivness.
There is an another trick for this one ?
Thank you in advance

REPLY [12 votes]: Let $b_n=(1+\sqrt{2})^n +(1-\sqrt{2})^n$.  Then $b_n$ is an integer. (As pointed out by Aaron, that can be seen by expanding, using the Binomial Theorem, and noting the cancellations. It can also be seen by noting that the expression for $b_n$ is invariant under the mapping that takes $\sqrt{2}$ to $-\sqrt{2}$.)  
So $\pi(1+\sqrt{2})^n$ differs from $\pi b_n$ ($\pi$ times an integer) by $\pi(1-\sqrt{2})^n$. Note that if $n$ is large, then the absolute value of $\pi(1-\sqrt{2})^n$ is very close to $0$.
The $n$-th term of our series has absolute value $|\sin(\pi(1-\sqrt{2})^n)|$, which is less than $\pi(\sqrt{2}-1)^n$. 
But $\sum \pi(\sqrt{2}-1)^n$ converges, since it is a geometric series with common ratio $\lt 1$. Thus by Comparison the series of the problem converges absolutely, and hence converges.  
Remark: Whenever $1+\sqrt{2}$ has a problem, its buddy $1-\sqrt{2}$ is ready to help.