Math Help - bolzano weirshtrass question

1. bolzano weirshtrass question

http://img135.imageshack.us/img135/7662/60430495yx6.gif

i only know that these series are bounded
so i know that that there is at least one sub sequence which converges to
one of the bounds of each sequence
which gives me nothing because if it were converges
then i could transform lim inf An ==> lim An
because every sub sequence converges to the same limit which is the
main limit of the sequence.

what to do here?

2. Let us prove: $\liminf a_n \cdot \liminf b_n \leq \liminf a_nb_n$.
(If $\{a_n\},\{b_n\}$ are bounded then so is $\{a_nb_n\}$).

Let $a = \liminf a_n$, well this means $a = \lim \bigg( \inf \{ a_k | k\geq n\} \bigg)$.
The sequence, $\inf \{ a_k | k\geq n \}$ is non-increasing therefore the limit, $a$,
is the greatest lower bound, so $a\leq \inf \{ a_k | k\geq n\}$ for all $n\geq 1$.
For $n=1$ we see $a\leq \inf \{ a_1,a_2,... \}$ therefore $a$ is a lower bound for $\{a_1,a_2,...\}$.
This means, $a\leq a_n$ for all $n\geq 1$.

Likewise if $b = \liminf b_n = \lim \bigg( \inf \{ a_k | k\geq n\} \bigg)$ then $b\leq b_n$ for all $n\geq 1$.

We are told that $0\leq a_n,b_n$ which means $0\leq a,b$.
Thus, we can write inequality, $ab \leq a_n b_n$ for all $n\geq 1$.
This means that, $ab \leq \inf\{ a_1b_1,a_2b_2,... \} \leq \inf \{ a_kb_k | k\geq n \}$ for all $n\geq 1$.
Thus, $ab$ is a lower bound for the sequence $\inf \{ a_kb_k | k\geq n\}$.
But this 'inferior sequence' is non-increasing so it has a limit.
Thus, $ab \leq \lim \bigg( \inf \{ a_kb_k| k\geq n\} \bigg) = \liminf (a_nb_n)$.
This completes the proof.

You can try doing the other problems now.

3. regarding this definition

$

a = \lim \bigg( \inf \{ a_k | k\geq n\} \bigg)
$

the sequence
$

\inf \{ a_k | k\geq n \}
$

is non decreasing . its inf gets bigger or not changing in each following sequence
so its limit is its least upper bound
not like you said "greatest lower bound"

am i correct??