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Supose two sequences os positive real numbers $\{a_n\}$, $\{b_n\}$ such that the limit $\lim\limits_{n \mapsto \infty}b_n$ exists and is equal to $b > 0$. It is true that $\limsup a_nb_n = \limsup a_n\cdot\lim b_n$? Even in case of $\limsup a_n = \infty$?

This question was treated here: lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $

Is necessary that $\{a_n\}$ to be bounded for the proffs? Thanks.

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Yes, it is true. If $a=\limsup_na_n$, then there is a subsequence $(a_{n_k})_{k\in\mathbb N}$ of $(a_n)_{n\in\mathbb N}$ such that $\lim_ka_{n_k}=a$. Then $\lim_ka_{n_k}b_{n_k}=ab$ and therefore $\limsup_na_nb_n\geqslant ab$. On the other hand, suppose that $\limsup_na_nb_n=L>ab$. Then there is a sequence $(a_{n_k}b_{n_k})_{k\in\mathbb N}$ whose limit is $L$, but this implies that the limit of $(a_{n_k})_{k\in\mathbb N}$ is $\frac Lb>a$, which is absurd.

This proof assumes that $b\in(0,+\infty)$ and that $(a_n)_{n\in\mathbb N}$ is bounded above. The proof in the case in which $b=+\infty$ or in which $(a_n)_{n\in\mathbb N}$ is unbounded (above) is similar (simpler, actually).

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