Suppose that $a_n$ and $b_n$ are two convergent sequences with limits $a$ and $b$ respectively, and $(\frac{a_n}{b_n})^2$ converges to some limit $L$ as well. Can we conclude that $\frac{a_n}{b_n}$ converges (to one of $\pm\sqrt{L}$)?

If $b$ is nonzero then this is obviously true, so we just need to consider the case where $b=0$ (and hence $a=0$). In this case, can we necessarily conclude that $\frac{a_n}{b_n}$ converges? I can't find a counterexample nor a proof for this. If the sequence may not converge, are there any conditions we can put on $L$ (e.g. $L=0$ or $L \ne 0$) that will make the conclusion true?

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    $\begingroup$ $a_n = \frac{1}{n}$, $b_n=\frac{1}{n^2}$? You can even choose $b_n = \frac{(-1)^n}{n^2}$ if you want to rule out $+\infty$ as a limit. $\endgroup$ – Clement C. Sep 7 '16 at 16:52
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    $\begingroup$ $$a_n=\frac1n\qquad b_n=\frac{(-1)^n}{n}$$ $\endgroup$ – Did Sep 7 '16 at 16:52
  • $\begingroup$ Both of you are right, of course - can't believe I missed it. Thanks! $\endgroup$ – Richard Sep 7 '16 at 17:01

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