# Does $\lim \frac {a_n} {b_n}$ exist and $\lim a_n \neq 0$ imply $\lim b_n$ exist?

Suppose $\lim_{n \rightarrow \infty} \frac {a_n} {b_n}$ exist and $(a_n)$ converges to some number $k \neq 0$. Is it then possible to conclude that $(b_n)$ converges ?

Also, suppose $\lim_{n \rightarrow \infty} \frac {a_n} {b_n}$ exist and $(b_n)$ converges to some number $k \neq 0$. Is it then possible to conclude that $(a_n)$ converges ?

I am well aware of the statement that if $(a_n)$, $(b_n)$ converges and $b_n\neq 0$ for $n \ge N$ then $\lim_{n \rightarrow \infty} \frac {a_n} {b_n} = \frac {\lim_{n \rightarrow \infty} a_n} {\lim_{n \rightarrow \infty} b_n}$. I've tried to use the contrapositive of this statement to prove my hypotheses. It should be said, that this is not an exercise, but something I've been wondering about.

• $$\frac{a_n}{a_n/b_n} = \,?$$ – Daniel Fischer Oct 15 '14 at 18:37
• Thanks ! I cannot believe I haven't thought of that. – Shuzheng Oct 15 '14 at 18:41
• For the second statement, I must also assume $a_n \neq 0$ for $n \ge N$ ?? – Shuzheng Oct 15 '14 at 18:43
• Isn't there an extra case when $\lim_{n\to\infty}{\frac{a_n}{b_n}}=0$? If $a_n\to 0$, we could still have divergence for $b_n$, unless you count $\infty$ as a proper limit. Very pretty way to go about the problem, though. – Some Math Student Oct 15 '14 at 18:44
• No, $a_n$ can be $0$ without problem, there we look at $b_n\cdot \frac{a_n}{b_n}$. – Daniel Fischer Oct 15 '14 at 18:44