1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ How can you prove that the sum is convergent too? $\endgroup$ – Ansper Nov 22 '20 at 20:37
  • 2
    $\begingroup$ $$\left|(-1)^{k} y_{n-k} \left(\frac{\alpha}{\beta}\right)^k\right| \leq M \left(\frac{\alpha}{\beta}\right)^k$$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.