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

*

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


*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.
 A: 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.
