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Given that the sequence $(a_n b_n)$ converges and $a_n \to 0$, are there conditions which can be placed on $(b_n)$ and/or $(a_n b_n)$ so that $(b_n)$ converges as well?

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If $(b_n)$ converges, then the limit of $(a_n b_n)$ is $0$. So you have to assume that at least. But if $(a_n b_n)$ and $(a_n)$ converge to $0$, many different things could happen for $(b_n)$. You can take $a_n = 2^{-n}$ and $b_n = (-1)^n$ then $b_n$ has no limit. Or you can also take $a_n = 3^{-n}$ and $b_n = 2^{n}$ then $b_n$ diverges to infinity... –  Joel Cohen Jul 6 '11 at 20:01
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up vote 2 down vote accepted

Well if you are willing to drop $\langle a_n \rangle$ converging to $0$, then something can be said.

In general, if $\langle a_n \rangle$ converges to a positive number $a$ and the sequence $\langle b_n \rangle$ is ultimately non-negative, then: $\limsup_{n \to \infty}~a_n b_n=a.\limsup_{n \to \infty}~b_n$. And same for $\liminf$. So, if $\langle a_nb_n \rangle$ does converge, then so will $\langle b_n \rangle$.

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Let $c = \lim_{n\to \infty} a_nb_n$. If $c\neq 0$ then $|b_n|$ is unbounded, and hence does not converge. So if $b_n$ converges then $c=0$.

Now the sequence $b_n$ doesn't have to satisfy any particularly strong constraint, it must merely grow slowly enough. To me this suggests we're unlikely to find any useful sufficient conditions for convergence of $b_n$.

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