If $0\neq t, x\in\mathbb{R}:\sum_{n=0}^\infty \sin^2(tx^n)<\infty$ (i.e. x is a Pisot number) is then $\sum_{n=0}^\infty-\log(1-\sin^2(tx^n))<\infty$?

Question

Suppose that $x\in\mathbb{R}$ is a Pisot number, i.e. for some $0\neq t\in\mathbb{R}:\sum_{n=0}^\infty \sin^2(tx^n)<\infty$. Do we then know that also $\sum_{n=0}^\infty -\log(1 - \sin^2(tx^n)) < \infty$ that is that if $\sum_{n=0}^\infty \sin^2(tx^n)$ converges, does also $\sum_{n=0}^\infty \log(\cos^2(tx^n))$ converge? A book I am reading on basically assumes this connection without proving it. Is there some suitable asymptotic bound that we could use for the logarithm?

Answer 1

Yes, in fact this is true for any sequence $a_n$ such that

$$\sum_{n=0}^\infty a_n<\infty$$

Note that this implies $a_n\to 0$. We also have

$$\lim_{n\to \infty} \frac{\log(1-a_n)}{a_n}=-\lim_{n\to \infty}\frac{\log(1+a_n)}{a_n}=-\lim_{h\to 0}\frac{\log(1+h)}{h}=-1$$

This implies

$$\sum_{n=0}^\infty -\log(1-a_n)\text{ converges iff }\sum_{n=0}^\infty a_n\text{ converges}$$

Also, you have $\cos^2(x)=1-\sin^2(x)$ which implies the second series is the negative of the first series.