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 $$x_n = \left(\frac{-\alpha}{\beta}\right)^n x_0 + \frac{1}{\beta}\sum_{k=0}^{n-1} (-1)^{k} y_{n-k} \left(\frac{\alpha}{\beta}\right)^k $$

The result follows.

  • $\begingroup$ How can you prove that the sum is convergent too? $\endgroup$ – Ansper Nov 22 '20 at 20:37
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    $\begingroup$ $$\left|(-1)^{k} y_{n-k} \left(\frac{\alpha}{\beta}\right)^k\right| \leq M \left(\frac{\alpha}{\beta}\right)^k$$ for a certain constant $M$, since $(y_n)$ is bounded (because it is convergent). So the series is absolutely convergent because $\beta > \alpha> 0$. $\endgroup$ – TheSilverDoe Nov 22 '20 at 20:38
  • $\begingroup$ 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. $\endgroup$ – Ansper Nov 22 '20 at 23:25
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    $\begingroup$ @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 ! $\endgroup$ – TheSilverDoe Nov 23 '20 at 7:43

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