Limit of sequence with $a_{n+1} = b_n - q a_n$

Question

Let the sequences $\{a_n\}$ and $\{b_n\}$ satisfy the relation $a_{n+1} = b_n - q a_n$ where $0 < q < 1$, and assume that $\lim\limits_{n\to\infty}b_n = b$. Prove that $\lim\limits_{n\to\infty} a_n$ also exists.

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All I can do is to show that $a_{n}$ is bounded:

$\left| b_n \right|\leq M$

$a_{n+1}=\sum_{k=0}^{n-1}{\left( -1 \right)^{k}q^{k}b_{n-k}+\left( -1 \right)^{n}q^{n}a_{1}}$

$\left| a_{n+1} \right|\leq M\left( 1+q+q^{2}+...+q^{n-1} \right)+q\left| a_{1} \right|\leq\frac{M}{1-q}+q\left| a_{1} \right|$

Is there a way to prove that this sequence is monotonic?

Answer 1

We have $$ a_{n+1}-\frac b{1+q}=b_n-b-qa_n+b-\frac b{1+q}=(b_n-b)-q\Big(a_n-\frac b{1+q}\Big). $$

Hence we just need to consider the case where $b=0$. Otherwise by letting $a'_n=a_n-\frac b{1+q}$ and $b'_n=b_n-b$ we have $a_{n+1}'=b'_n-qa_n'$ and $b'_n\to0$.

We will prove $a_n\to0$ when $b=0$.

By recursion, $$ \begin{aligned} |a_n|&\leqslant |b_{n-1}|+q|a_{n-1}|\leqslant|b_{n-1}|+q|b_{n-2}|+q^2|a_{n-2}|\leqslant\cdots\\ &\leqslant |b_{n-1}|+q|b_{n-2}|+\dots+q^{n-1}|b_0|+q^n|a_0|. \end{aligned}$$

For any $k$, there exists $N$ s.t. when $n-k>N$, $|b_{n-k}|<\varepsilon$. Therefore $$ |b_{n-1}|+q|b_{n-2}|+\dots+q^{k-1}|b_{n-k}|< \varepsilon\cdot\frac1{1-q}. $$ Also, since $\{b_n\}$ is bounded, $$ q^k|b_{n-k-1}|+\cdots+q^{n-1}|b_0|+q^n|a_0|< Mq^k\cdot\frac1{1-q}. $$

Combining both results we have $$ |a_n|<\varepsilon\cdot\frac1{1-q}+Mq^k\cdot\frac1{1-q}. $$ We can select sufficiently small $\varepsilon$ and then sufficiently large $k$ such that RHS is less than $\varepsilon'$. Now given any $\varepsilon'>0$, we have $|a_n|<\varepsilon'$ for sufficiently large $n$, indicating $a_n\to0$.

The proof is done.

Answer 2

Use your relation $a_{n+1}=\sum_{k=0}^{n-1}{\left( -1 \right)^{k}q^{k}b_{n-k}+\left( -1 \right)^{n}q^{n}a_{1}}$ to show, that $a_n$ is a Cauchy-sequence.

To do so, let $l\in \mathbb N$ and use $|b_n| \leq M$: $$\begin{eqnarray*}\left|a_{n+l}-a_n\right| & = & \left|\sum_{\color{blue}{k=n-1}}^{n+l-2}{\left( -1 \right)^{k}q^{k}b_{n-k}+\left( -q \right)^{n+l-1}a_{1}} - \left( -q \right)^{n-1}a_{1}\right| \\ & \leq & q^{\color{blue}{n-1}}M\sum_{\color{blue}{k=0}}^{l-1}q^k + q^{n-1}|a_1||q^l-1|\\ & \leq & \left(\frac M{1-q} + |a_1|\right)q^{n-1}\quad (\star)\end{eqnarray*}$$ ($\star$ ) is independent of $l$ and has limit $0$. Consequently, $a_n$ is Cauchy.