# Prove: ${a_n}$ converges to $a$

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

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

Or,

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

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

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 $a_n \rightarrow a$.