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Suppose that $\{a_n\}, \{b_n\}$ are sequences of nonnegative real numbers with lim $b_n = \beta \ne 0$ and limsup $a_n = \alpha$. Prove that limsup $a_nb_n = \alpha\beta$.

The text gave part of the proof: "There exists a subsequence $\{a_{n_k}\}$ of $\{a_n\}$ such that $a_{n_k} \to \alpha$. With the convergence of $\{b_n\}$, we have $a_{n_k}b_{n_k} \to \alpha\beta$. So, $\alpha\beta$ is a subsequential limit of $\{a_nb_n\}$, entailing that limsup$\{a_nb_n\} \ge \alpha\beta$." Afterward, the author claims that the condition $\beta \ne 0$ is crucial. I do not see how it is necessary here. Then, via a $\delta$-$\epsilon$ proof, I proved the reverse inequality, limsup $a_nb_n \le \alpha\beta$, and I did not need $\beta$ to be nonzero here either.

So far, I have assumed $\alpha$ to be finite. However, if $\alpha = \infty$, then I do see why the condition is necessary. Can you kindly confirm my understanding that the condition is not necessary if $\alpha$ is finite?

When $\alpha = \infty$: There exists a $N$ such that $ n > N \to b_n > \beta/2$. Let $ M \in \mathbb R^+$ be arbitrary but fixed. By the limit superior of $\{a_n\}$, for each $n > N$, there is a $ k \ge n$ such that $a_k > M(\frac{2}{\beta})$. Therefore, $a_kb_k > M$ for the particular $k$ referenced above. So, limsup $a_nb_n = \infty = \alpha\beta$.

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    $\begingroup$ Yes, if $\alpha$ is finite, it is immaterial whether $\beta = 0$, only for $\alpha = \infty$ is it important (and then $\alpha\beta$ is an indeterminate form if $\beta = 0$). $\endgroup$ – Daniel Fischer Oct 15 '14 at 18:24
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Can you kindly confirm my understanding that the condition is not necessary if $\alpha$ is finite?

I can confirm that. Only if $\alpha = \infty$ the condition $\beta \neq 0$ is relevant. For $\alpha = \infty$ and $\beta = 0$, the product $\alpha\beta$ is an indeterminate form, and indeed $\limsup\limits_{n\to\infty} a_n b_n$ can be any value in $[0,\infty]$ then.

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