Is there any subsequence of the sequence $(\frac{\cos(\alpha - n \beta) - \lambda \cos(\alpha + n \beta)}{ \cos^n(\beta) })$ that converges?

Question

Let's take $\alpha$ and $\beta$ two reals in $(0, \frac{\pi}{2})$. Let's take $\lambda \in (0,1)$. Let's define the sequence $(u_n)$ as follows: \begin{eqnarray} u_n & = & \frac{\cos(\alpha - n \beta) - \lambda \cos(\alpha + n \beta)}{ \cos^n(\beta) } \end{eqnarray}

1. Can we extract a subsequence $(u_{\phi(n)})$ that converges to a real number ? ( where $\phi$ is a strictly increasing application from $\mathbb{N} \to \mathbb{N}$ )

2. If not, can we prove that the limit of $|u_n|$ is $+\infty$?

My intuition: I think It is possible to construct such a subsequence since we can always construct $\phi$ such that $\cos(\alpha - \phi(n) \beta) - \lambda \cos(\alpha + \phi(n) \beta)$ converges to $0$. The problem is that I am struggling with is how then to construct $(u_{\phi(n)})$ such that it converges.

Also, I had the idea is to use the Tschebyscheff polynomials (Chebyshev Polynomials)(https://en.wikipedia.org/wiki/Chebyshev_polynomials) as we can always express $u_n$ in function of chebyshev polynomials. I don't know yet how to move it forward so far. So if anyone has any idea, please share it here. Thank you in advance.

Answer 1

Here is an idea of solution that I would like your comment on. Let's first write the following: \begin{eqnarray} \hspace{0.0cm} A & = & \cos (\alpha) (1-\lambda) \\ B & = & \sin (\alpha) (1 + \lambda) \\ \cos (\alpha - \beta n ) - \lambda \, \cos (\alpha + \beta n ) & = & A \cos (\beta n) + B \sin ( \beta n ) \hspace{1.cm} \\ & = & \sqrt{A^2 + B^2} \, \sin ( a_0 + \beta n ) \end{eqnarray} Where $a_0 \in (0, \frac{\pi}{2})$ ( since $\alpha \in (0, \frac{\pi}{2})$ ) . \begin{eqnarray} \hspace{0.0cm} \sin ( a_0 ) & = & \frac{ A }{ \sqrt{A^2 + B^2} } \\ \cos ( a_0 ) & = & \frac{ B }{ \sqrt{A^2 + B^2} } \end{eqnarray}

And therefore \begin{eqnarray} \hspace{0.0cm} u_n & = & \frac{ \sqrt{A^2 + B^2} \, \sin ( a_0 + \beta n ) } { \cos^n (\beta ) } \hspace{1.cm} \end{eqnarray}

We have two cases:
Case 1: $\beta = \frac{\pi p - a_0 }{q}$ and $a_0 = \frac{r \pi}{s}$ where $(r,s) = 1$, $p, q, r, s \in \mathbb{N}$.

In this case we have for each $n \in \mathbb{N}$: \begin{eqnarray} \sin ( a_0 + q (n s + 1) \beta ) & = & (-1)^{(n s + 1)p + 1} \sin ( \pi n r ) = 0 \hspace{1.0cm} \end{eqnarray} And therefore for each $n \in \mathbb{N}$: \begin{eqnarray} u_{ q (n s + 1) } & = & 0 \hspace{1.cm} \end{eqnarray} And hence a subsequence of $(u_n)$ that converges to zero.

Case 2: $\beta = \lim_{m \to +\infty } \frac{s_m p_m - r_m }{s_m q_m} \pi $ and $a_0 = \lim_{m \to +\infty } \frac{r_m }{s_m} \pi $ where $(r_m,s_m) = 1$, $p_m, q_m, r_m, s_m \in \mathbb{N}$.

Thanks to the density of $\mathbb{Q}$ in $\mathbb{R}$, we can find the sequences $(\frac{\pi p_m - a_0 }{q_m})$ and $(\frac{\pi r_m }{s_m})$ that converges respectively to $\beta$ and $a_0$.

Without loss of generality, we can assume that the sequence $(q_m)$ is increasing.

Intuitively, if it works in Case 1, it should work in Case 2 thanks to the density of $\mathbb{Q}$ in $\mathbb{R}$.

For each $n$, we can write that: \begin{eqnarray} \sin ( a_0 + n \beta ) & = & \lim_{m \to +\infty } \sin ( \frac{r_m }{s_m} \pi + n \frac{s_m p_m - r_m }{s_m q_m} \pi ) \hspace{1.0cm} \\ u_n & = & \lim_{m \to +\infty } \frac{ \sqrt{A^2 + B^2} \, \sin ( \frac{r_m }{s_m} \pi + n \frac{s_m p_m - r_m }{s_m q_m} \pi ) } { \cos^n (\beta ) } \hspace{1.cm} \end{eqnarray} And \begin{eqnarray} \hspace{-1.cm} \lim_{n \to +\infty } u_n & = & \lim_{n \to +\infty } \lim_{m \to +\infty } \frac{ \sqrt{A^2 + B^2} \, \sin ( \frac{r_m }{s_m} \pi + n \frac{s_m p_m - r_m }{s_m q_m} \pi ) } { \cos^n (\beta ) } \hspace{1.cm} \end{eqnarray} If we take $n$ of the form $n_m = q_m ( 1 + m \prod^{m}_{i=1} s_i )$, we will have (subject to justification!): \begin{eqnarray} \hspace{-1.cm} \lim_{m \to +\infty } \sin ( a_0 + n_m \beta ) & = & \lim_{m \to +\infty } \underbrace{ \sin ( \frac{r_m }{s_m} \pi + q_m ( 1 + m \prod^{m}_{i=1} s_i ) \frac{s_m p_m - r_m }{s_m q_m} \pi ) }_{ = 0} \hspace{1.0cm} \\ & = & \lim_{m \to +\infty } 0 = 0 \end{eqnarray} And \begin{eqnarray} \hspace{-1.12cm} \lim_{m \to +\infty } u_{ n_m } & = & \lim_{m \to +\infty } \underbrace{ \frac{ \sqrt{A^2 + B^2} \, \sin ( \frac{r_m }{s_m} \pi + q_m ( 1 + m \prod^{m}_{i=1} s_i ) \frac{s_m p_m - r_m }{s_m q_m} \pi ) } { \cos^{n_m} (\beta ) } }_{ = 0} \hspace{1.cm} \\ & = & \lim_{m \to +\infty } 0 = 0 \end{eqnarray} Note that the sequence $(n_m)$ is strictly increasing since the sequence $(q_m)$ is increasing and the sequence $ ( 1 + m \prod^{m}_{i=1} s_i )$ is strictly increasing. And hence a subsequence of $(u_n)$ that converges to zero.

Please let me know if there is any problem with such solution. Thank you.