# Show that $\lim_{n \rightarrow \infty}\frac{a_n}{b_n}=1 \implies \lim_{n \rightarrow \infty}{(a_n-b_n)}=0$

I have a question concerning a special limit

$$\lim_{n \rightarrow \infty}\frac{a_n}{b_n}=1$$

Can I always conclude from this that

$$\lim_{n \rightarrow \infty}{(a_n-b_n)}=0$$

Even if $a_n$ and $b_n$ are divergent? I tried to find some counter examples but I couldn't.

• Add condition $b_n$ bounded. No need for it to converge. – GEdgar Sep 13 '15 at 20:37

False. Take $a_n = 1+n$ and $b_n = n$. Then $\frac{a_n}{b_n} \to 1$ and $a_n -b_n \to 1 \neq 0$.
• It seems to me that we must have $b_n \to b \in \Bbb R$. So we can do: $$a_n - b_n = b_n\left(\frac{a_n}{b_n}-1\right) \to b \cdot 0 = 0.$$If $b_n$ diverges then we'll have some $\infty \cdot 0$, and we can't conclude anything. – Ivo Terek Sep 13 '15 at 20:34
• If $b_n$ is convergent also $a_n$ must be convergent to have a limit of 1 for the ratio. I think something like order of divergence of $b_n$ less than the order of Convergence of $\frac{a_n}{b_n}-1$ could used. – MrYouMath Sep 13 '15 at 20:41