Given the sequence ${b_n}$, let $lim_{n \to \infty}\ b_n = b$.

Suppose that the sequence ${a_n}$ and the number $a$ have the property for which there exists $M\in \mathbb{R}$ and there exists $N \in \mathbb{N}$ such that

$$|a_n - a| \leq M\cdot |b_n - b|, \ \forall n\in \mathbb{N}: \ n \geq N$$

Prove that the $\lim_{n \to \infty} \ a_n = a$.

I need to show that:

$$\forall \epsilon > 0 \ \ \exists N \in \mathbb{N}: \ \forall n \geq N: |a_n - a| < \epsilon$$

I know how to set $\epsilon$ such that $\epsilon > 0$. I’m lost from here. Because ${b_n}$ converges I know

$|b_n - b| < \epsilon$

And I think it’s safe to assume:

$|b_n - b| \leq M\cdot |b_n - b|$.

So I could prove this either by showing

$$|a_n - a| \leq |b_n - b|$$


$$M \cdot |b_n - b| < \epsilon$$

But I’m not sure how to start either way. Any suggestions?


2 Answers 2


Let $\epsilon>0$. Since $\lim_{n\to\infty }b_n=b$ then there's $n_0\in\Bbb N$ such that

$$|b_n-b|\le\frac {\epsilon}M\quad\text{whenever}\; n\ge n_0$$ so for $n\ge n_0$ we have

$$|a_n-a|\le M|b_n-b|\le\epsilon$$ and the result follows.


Take $\epsilon$ fixed since $b_n \rightarrow b$,

\begin{align*} \exists N_0 \in \mathbb{N} \quad \text{ s.t. } \forall n \geq N_0, \quad \vert b_n - b \vert \leq \frac{\epsilon}{M}. \end{align*} It follows that $\forall n \geq N_0$, $\vert a_n - a \vert \leq \epsilon$.

As $\epsilon$ was chosen arbitrarily, it follows that $a_n \rightarrow a$.


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