# 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).

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 $$\ell/(1-\alpha)$$.