For $\alpha \in ]-1, 1[$, how to show that sequence $(u_{n + 1} - \alpha u_n)$ converges implies $(u_n)$ converges? Let be $\alpha \in ]-1, 1[$ and $(u_n)_{n \in \mathbb{N}}$ a real-valued sequence.
Let us suppose that $(u_{n + 1} - \alpha u_n)_{n \in \mathbb{N}}$ converges. How to show that $(u_n)_{n \in \mathbb{N}}$ converges?
What I tried:


*

*Work on $u_{n + 1} - u_n = (\alpha - 1) u_n + L + o(1)$ for some $L \in \mathbb{R}$ which would be the limit of $(u_{n + 1} - \alpha u_n)_n$ and use some kind of Cesaro sommation technique.

*Consider the set of limits of all subsequences of $(u_n)_n$ and show that its lower bound is its upper bound (unsuccessful).

 A: Write $a_n = u_n - \alpha u_{n-1}$ and notice that for any $0 < k < n$,
$$ u_n = a_n + \alpha a_{n-1} + \cdots + \alpha^{k-1}a_{n-k+1} + \alpha^{k} u_{n-k}. $$
Since $(a_n)$ converges, it is bounded by some $M_1 > 0$. Then the above formula with $k = n-1$ shows that
$$ \lvert u_n \rvert \leq \frac{M_1}{1-\alpha} + \lvert u_0 \rvert =: M_2, $$
which proves that $(u_n)$ is also bounded. Now let $\ell$ the limit of $(a_n)$ and fix a positive integer $N$. Then
\begin{align*}
\left| u_n - \frac{\ell}{1-\alpha} \right|
&= \left| \alpha^N u_{n-N} + \sum_{k=0}^{N-1} \alpha^k (a_{n-k} - \ell) - \frac{\alpha^N \ell}{1-\alpha} \right| \\
&\leq \alpha^N \left(M_2 + \frac{\lvert\ell\rvert}{1-\alpha}\right) + \sum_{k=0}^{N-1} \alpha^k \lvert a_{n-k} - \ell \rvert.
\end{align*}
Taking limsup as $n\to\infty$, we have
$$ \limsup_{n\to\infty} \left| u_n - \frac{\ell}{1-\alpha} \right| \leq \alpha^N \left(M_2 + \frac{\lvert\ell\rvert}{1-\alpha}\right). $$
Since the left-hand side is independent of $N$, we may let $N\to\infty$ to show that the limsup is indeed $0$. Therefore $u_n$ converges to $\ell/(1-\alpha)$.
