bolzano weirshtrass question

• Jan 7th 2009, 11:58 AM
transgalactic
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?
• Jan 7th 2009, 07:54 PM
ThePerfectHacker
Let us prove: $\displaystyle \liminf a_n \cdot \liminf b_n \leq \liminf a_nb_n$.
(If $\displaystyle \{a_n\},\{b_n\}$ are bounded then so is $\displaystyle \{a_nb_n\}$).

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

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

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

You can try doing the other problems now.
• Jan 9th 2009, 02:58 AM
transgalactic
regarding this definition

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

the sequence
$\displaystyle \inf \{ a_k | k\geq n \}$

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??