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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.

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    $\begingroup$ $$\frac{a_n}{a_n/b_n} = \,?$$ $\endgroup$ – Daniel Fischer Oct 15 '14 at 18:37
  • $\begingroup$ Thanks ! I cannot believe I haven't thought of that. $\endgroup$ – Shuzheng Oct 15 '14 at 18:41
  • $\begingroup$ For the second statement, I must also assume $a_n \neq 0$ for $n \ge N$ ?? $\endgroup$ – Shuzheng Oct 15 '14 at 18:43
  • $\begingroup$ 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. $\endgroup$ – Some Math Student Oct 15 '14 at 18:44
  • $\begingroup$ No, $a_n$ can be $0$ without problem, there we look at $b_n\cdot \frac{a_n}{b_n}$. $\endgroup$ – Daniel Fischer Oct 15 '14 at 18:44

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