Convergence of a sequence $y_k$ defined by recurrence relation

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Question:

Let $|a|<1$ and let $\left(x_k\right)_{k\ge 1}$ be a sequence that converges to zero. Define a sequence $(y_k)_{k\ge 0}$ given by the following relation

$y_k=x_k+ay_{k-1}$.

Determine whether $y_k\to 0$.

My attempt: It can be shown that $y_k=x_k+ax_{k-1}+a^2x_{k-2}+\dotsb+a^{k-1}x_1+a^ky_0$ for each $k\ge 1$. The first and last term of RHS goes to zero. But till now I am unable to estimate the terms $ax_{k-1}+a^2x_{k-2}+\dotsb+a^{k-1}x_1$ whether it converges to zero. Help is appreciated.

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That's what I found so far. Do you think it may work?

As you wrote $$y_n = \sum_{k=0}^na^{n-k}x_k,$$ where $x_0=y_0$.

The sequence $(x_n)$ is bounded, so that $|x_n| < M$, for all $n$ and some positive $M$.

Fix some $\varepsilon >0$. Since $(x_n) \to 0$, for $k>N$ we have $$|x_k| < \varepsilon(1-|a|).$$

Then choose $n$ large enough (say $n>N_1$) so that $$|a^n|< \frac{\varepsilon |a^N|}{M(N+1)}$$

We can write, for $n>\max(N,N_1)$,

\begin{eqnarray} |y_n| &=& \left|\sum_{k=0}^{N}a^{n-k}x_k + \sum_{k=N+1}^na^{n-k}x_k\right|<\\ &<&M\sum_{k=0}^{N}\left|a^{n-k}\right| + \varepsilon (1-|a|) \sum_{k=N+1}^n\left| a^{n-k}\right|<\\ &<&M(N+1)|a^{n-N}|+\varepsilon<\\ &<&2\varepsilon. \end{eqnarray}