Analysis: Prove lim sup (an * bn)=lim sup (an)

Question

I have troubles with the following exercise, especially because you don't know anything about the sequence $a_n$:

$b_n$ and $a_n$ are both real sequences and $b_n \to 1$

Prove: $\lim \sup (a_n*b_n)=\lim \sup (a_n)$, and $\lim \sup (a_n^{b_n}))=\lim \sup (a_n)$

One of your question is answered here: If $\lim_{n\to \infty}a_n = a\in \mathbb{R}$ . Prove that $\limsup_{n\to \infty}a_n x_n=a\limsup_{n\to \infty}x_n$ . — Martin Sleziak, Jan 11 '17 at 16:07

score 0 · Accepted Answer · edited Jan 11 '17 at 16:08

You know that: $\ \ \limsup a_n = \begin{cases} +\infty, & \text{if $a_n$ is unbounded above} \\ sup\ V, & \text{if $a_n$ is bounded above and V is not empty}\\ -\infty, & \text{if $a_n$ $\to -\infty$} \end{cases}$

What if you split the question in three parts.

If you state that $\limsup a_n=+\infty$, then you know that $a_n$ is unbounded above, wich means that there is a subsequence of $a_n\ ({a_n}_k)$ that goes to $+\infty$.

Because you know that $b_n \to 1$, you also know that $a_{n_k} {b_n}_k \to +\infty$. This means that $a_n b_n$ is unbounded above. This corresponds to the first situation for $\limsup a_n b_n$. So, $\limsup (a_n b_n) = +\infty$ and therefore $\limsup\ (a_n b_n)=\limsup a_n$.

Hope this helps you out with the other parts.

Analysis: Prove lim sup (an * bn)=lim sup (an)

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