# Prove that $(x_n)\rightarrow 0$ if $(y_n)\rightarrow 0$

I want to prove the following theorem:

Let $$q$$ be a real number such that $$0 ≤ q < 1$$. Let $$(x_n)$$ be a bounded sequence and let $$(y_n)$$ be the sequence defined by $$∀n ∈ \mathbb{N}, y_n = x_n − qx_{n+1}$$. If $$y_n$$ converges to $$0$$, then $$x_n$$ converges to $$0$$.

I can show that when $$\lim x_n$$ exists, $$(x_n)\rightarrow 0$$, but my trouble is showing that $$\lim x_n$$ in fact exists. My strategy is to show that $$\limsup x_n = \liminf x_n$$ so that I may conclude that $$\lim x_n$$ exists. Of course, when $$q=0$$, the result is immediate.

Here's what I have so far, when $$0,

\begin{align*} \limsup y_n & = \limsup(x_n-qx_{n+1})\\ & \leq \limsup(x_n) + \limsup(-(qx_{n+1})\\ & = \limsup(x_n) -\liminf(qx_{n+1})\\ & = \limsup(x_n) - q\liminf(x_{n+1})\\ & = \limsup(x_n) - q\liminf(x_n), \end{align*}

and by similar reasoning I've been able to show

\begin{align*} \liminf y_n \geq \liminf x_n -q\limsup x_n. \end{align*}

You can assume that each line follows from a result I proved in a previous problem. Given what I have so far, is there a way to reach $$\limsup x_n \leq \liminf x_n$$ so that I may conclude $$\limsup x_n = \liminf x_n$$? What am I missing?

I want to use the fact that $$y_n$$ converges to $$0$$, but whenever I try using this fact with my inequalities, I just arrive at what we already know: $$\liminf x_n \leq \limsup x_n$$.

Jaca's answer is fine but here's an argument following your line of thought in case you need it.

So, the fact that $$(x_n)$$ is a bounded sequences allows us to play with both $$\limsup x_n$$ and $$\liminf x_n$$ as numbers, what I want to say is that both limits are finite.

Remember that, since $$\lim y_n=0$$, we have both limits superior and inferior equal to zero.

We have $$x_n=y_n+qx_{n+1}$$, taking $$\limsup$$ on both sides we get \begin{align} \limsup x_n &\leq \limsup y_n + q\limsup x_{n+1} \\ &= \limsup y_n + q\limsup x_{n} . \end{align} So that $$(1-q)\limsup x_n \leq 0$$ Since $$1-q>0$$ (we're in the interesting case $$0) we must have $$\limsup x_n\leq 0$$. Now take $$\liminf$$ in the equation $$x_n=y_n+qx_{n+1}$$ so that $$\liminf x_n \geq q\liminf x_n \quad\Rightarrow\quad (1-q)\liminf x_n\geq 0 ,$$ and similiarly, $$1-q>0$$ implies $$\liminf x_n\geq 0$$ and we obtain $$\liminf x_n=\limsup x_n =0$$.

$$0\le q<1$$ then $$-1<-q$$ iff $$-|x_{n+1}|\leq-q|x_{n+1}|$$ iff $$|x_n|-|x_{n+1}|\leq|x_n|-q|x_{n+1}|\leq|x_n-qx_{n+1}|=|y_n|\to0$$

Therefore $$|x_n|$$ is a Cauchy's sequence. Suppose $$|x_n|\to c$$. Applying limit in $$|x_n|-q|x_{n+1}|\leq|y_n|$$ we obtain $$(1-q)c=0$$. As $$1-q \neq0$$, we get $$c=0$$, i.e., $$|x_n|\to0$$. Therefore $$x_n\to0$$.

• Thanks for this. I follow what you're doing, but I haven't encountered Cauchy sequences yet. Could you provide some hint for me based on the work I've shown already? Feb 27, 2020 at 3:56