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Suppose that $\limsup a_n$ is finite and $b_n \rightarrow b>0$ ($b\neq \infty$) as $n \rightarrow \infty$, and prove that $\limsup a_n b_n=(\limsup a_n)b$. Note in this problem $a_n$ can be unbounded below, I have already shown the result if $a_n$ is bounded.

Here is my approach so far, please let me know if I am on the right track. We will show $(\limsup a_n)b=$ sup$E$ where $E$ denotes the set of all subsequential limits of $a_n b_n$ along with $+\infty, -\infty$.

First we show $(\limsup a_n)b$ is an upper bound for $E$. So let $a_{n_k}b_{n_k}$ be a convergent subsequence of $a_n b_n$ with $n_1>n_2>...$ then we have the following inequalties

$$\lim_{k \to\infty}a_{n_k}b_{n_k}\leq \lim_{k \to\infty}\left(\sup_{i\geq k}{a_{n_i}}b_{n_k}\right)\leq \lim_{k\to\infty}\left(\sup_{n\geq k}{a_n}b_{n_k} \right)=\lim_{k\to\infty}\sup_{n\geq k}a_n\lim_{k\to\infty}{b_{n_k}}=b\limsup a_n$$

Hence $(\limsup a_n)b$ is an upper bound for $E$. Now we show that it is the least upper bound.

Suppose there exists an $M<b\limsup a_n$ such that for all convergent subsequences $a_{n_k}b_{n_k}$ of $a_nb_n$ we have that $\lim_{k\to\infty}{a_{n_k}b_{n_k}}\leq M$. Contradiction since by taking $a_n$ to be bounded we have that $\limsup a_nb_n=b\limsup a_n$.

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  • $\begingroup$ yes fixing it now $\endgroup$ – zerothedog Mar 13 '17 at 6:11
  • $\begingroup$ You write that "$a_n$ can be unbounded", but how is that possible given that $\limsup a_n$ is finite? $\endgroup$ – quasi Mar 13 '17 at 6:16
  • $\begingroup$ I meant that it doesn't have to have a lower bound. $\endgroup$ – zerothedog Mar 13 '17 at 6:17
  • $\begingroup$ Yes, of course, my mistake. $\endgroup$ – quasi Mar 13 '17 at 6:17
  • $\begingroup$ For your statement ". . . let $a_{n_k}b_{n_k}$ be a subsequence of . . .", it should be ". . . let $a_{n_k}b_{n_k}$ be a convergent subsequence of . . .". Similarly for your later statement ". . . for all subsequences . . .", it should be ". . . for all convergent subsequences . . .". Other than that, your proof looks OK. $\endgroup$ – quasi Mar 13 '17 at 6:40
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Let $u = \limsup a_n$.

Choose a convergent subsequence $(c_k)$ of $(a_n)$ which converges to $u$, and let $(d_k)$ be the subsequence of $(b_n)$ whose terms correspond to the terms of $(a_n)$ selected for $(c_k)$.

Then $(c_kd_k)$ is a convergent sequence with $\displaystyle{\lim_{k \to\infty}c_kd_k =ub}$.

It follows that $\limsup a_nb_n \ge b(\limsup a_n)$.

Suppose $\limsup a_nb_n > b(\limsup a_n)$.

Then there exists a convergent subsequence $(e_k)$ of $(a_nb_n)$ such that $\displaystyle{\lim_{k \to\infty}e_k = v > bu}$.

For the sequence $(e_k)$, let $(f_k)$ and $(g_k)$ be the corresponding subsequences of $(a_n)$ and $(b_n)$ respectively.

Since $(e_k)$ converges, and $(g_k)$ converges to $b > 0$, it follows that $(f_k)$ converges, and moreover

$$ \lim_{k \to\infty}f_k = \frac{\lim_{k \to\infty}e_k}{\lim_{k \to\infty}g_k} = \frac{v}{b} > \frac{bu}{b} = u $$

But $(f_k)$ is a subsequence of $(a_n)$, so $\,\lim_{k \to\infty}f_k > u\,$ contradicts $\,\limsup a_n = u$.

It follows that $\limsup a_nb_n = b(\limsup a_n)$, as was to be shown.

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