> 
> Let $(u\_n)\_{n\in \mathbb{N}}$ defined by : $u\_0=1, u\_1=1$ and for all integer $u\_{n+1}=3u\_n-u\_{n-1}$
> 
> 
> Study the convergence of $$\displaystyle\sum\_{n=0}^\infty \frac{1}{n+1}\sin\left(\frac{p\pi u\_n}{q}\right)$$ with $p,q \in \mathbb{N}^\*$
> 
> 
> 

I have $ S\_N = \sum\_{n=0}^N \frac{1}{n+1} \sin \left( \frac{2 \pi p u\_n}{q} \right)
 = \frac{a\_N}{N+1} - \sum\_{n=0}^{N-1} \frac{a\_n}{(n+1)(n+2)}$

where $$a\_n = \sum\_{k=0}^n \sin \left( \frac{2 \pi p u\_k}{q} \right)$$

Now If I shows that $a\_n$ is bounded, then
$$\frac{a\_N}{N+1} \xrightarrow[N \to \infty]{} 0 \mbox{ and } \frac{a\_n}{(n+1)(n+2)} \underset{n \to \infty}{=} \mathcal{O} \left( \frac{1}{n^2} \right)$$

Unfortunately I was not managed to prove it.

I try an another method :

$$\forall n \in \mathbb{N}, \left(
 \begin{array}{c}
 u\_n \\ u\_{n+1}
 \end{array}
\right) =
 A \left(
 \begin{array}{c}
 u\_{n-1} \\ u\_n
 \end{array}
 \right)$$

with
$$
A = \left[
 \begin{array}{ c c }
 0 & 1 \\
 -1 & 3
 \end{array} \right]
$$

Then,
$$
\forall n \in \mathbb{N}, \left(
 \begin{array}{c}
 u\_n \\ u\_{n+1}
 \end{array}
 \right)
 =
 A^n \left(
 \begin{array}{c}
 0 \\ 1
 \end{array}
 \right)
$$

As before I don't see how can I continue

Thank you in advance