# Assuming $(\alpha x_{n-1}+\beta x_n)_{n\geq 1}$ converges and $\beta > \alpha$, show that $(x_n)_{n\geq 0}$ converges.

Let $$\alpha,\beta$$ be non-zero positive real numbers such that $$\alpha < \beta$$, and $$(x_n)_{n\geq 0}$$ a sequence of real numbers and $$(y_n)_{n\geq 1}$$ be the sequence defined for all $$n\in \mathbb{N}^{\ast}$$ : $$y_n=\alpha x_{n-1} +\beta x_n$$ Show that if $$(y_n)_n$$ converges, then $$(x_n)_n$$ converges as well.
I thought of using the definition of the limit and trying to show that $$x_n\to \frac{\ell}{\alpha+\beta}$$ where $$\ell$$ is the limit of $$(y_n)_n$$. However, in the process, I find that there is a symmetry between $$\alpha$$ and $$\beta$$, respectively between $$x_{n-1}$$ and $$x_n$$ which is not supposed to be the case due to the hypothesis. So I find myself stuck, I would really like some hints.

Hint : Show inductively that for all $$n \geq 0$$, you have $$x_n = \left(\frac{-\alpha}{\beta}\right)^n x_0 + \frac{1}{\beta}\sum_{k=0}^{n-1} (-1)^{k} y_{n-k} \left(\frac{\alpha}{\beta}\right)^k$$
• $$\left|(-1)^{k} y_{n-k} \left(\frac{\alpha}{\beta}\right)^k\right| \leq M \left(\frac{\alpha}{\beta}\right)^k$$ for a certain constant $M$, since $(y_n)$ is bounded (because it is convergent). So the series is absolutely convergent because $\beta > \alpha> 0$. – TheSilverDoe Nov 22 '20 at 20:38
• Doesn't that just give that the general term converges to $0$? It doesn't necessarily imply that the series must converge too. If this uses a theorem about absolute convergent series, I would like to know what is it as I still haven't learned about it. – Ansper Nov 22 '20 at 23:25
• @Ansper No, because you bound it by $(\alpha/\beta)^k$ which (not only converges to $0$), but is also the general term of a convergent series. The theorem on absolute series I use is that if the series $\sum |a_n|$ converges, then the series $\sum a_n$ converges. You will see when you learn about it, this is very useful ! – TheSilverDoe Nov 23 '20 at 7:43