1. ## Bolzano-Weirstrauss

Suppose the sequence {a_n} converges to L and limsup{b_n} = M < infiniti
//({b_n} is bounded below and above)// Prove limsup{a_n*b_n} = LM.

b) also show that the statement is false if {b_n} is not bounded below.

My Sketch: I have to show that limsup{a_n*b_n} = Q is finite; then use
subsequences to show Q less than or equal to LM, and that is greater than or equal to LM.

2. Hang on just a minute.

If $a_n=-1$ for all $n$ then $\lim a_n=-1$, i.e. $L=-1$.

If $b_n=\begin{cases}1&\textrm{for odd }n\\-2&\textrm{for even }n\end{cases}$ then $\limsup b_n=1$, i.e. $M=1$. Notice $b_n$ is bounded below.

Then $a_nb_n=\begin{cases}-1&\textrm{for odd }n\\2&\textrm{for even }n\end{cases}$ and so $\limsup a_nb_n=2\neq LM$.

As is well known $\limsup(-a_n)=-\liminf a_n$ when both exist.

The result as stated does not hold, therefore. You might need an extra condition such as $L>0$, but then why would $b_n$ have to be bounded below? I think we need to be told.

3. [incorrect]

4. Here's another way to look at it (though I still don't use that $b_n$ is bounded below. and am no longer sure about needing L>0)

Let $n_k$ be a sequence such that $b_{n_k}\to M$. Then $a_{n_k}b_{n_k}\to\limsup a_nb_n$ since $a_n\to L$ and thus does so along ANY subsequence.