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14:09
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A: Can we apply squeeze in that way?

Thomas$$\sum_{i=k}^l a_i \le \sum_{i=k}^l b_i \le \sum_{i=k}^l c_i$$ This implies that $$|\sum_{i=k}^l b_i| \le\max \{ |\sum_{i=k}^l a_i|, |\sum_{i=k}^l c_i|\}$$ Now $\sum a_i$, $\sum c_i$ exist, hence and right term becomes arbitrarily small (Cauchy-criterion), if $k,l$ are chosen large enough, h...

mesel
mesel
@thomas: if $A_i\leq B_i\leq C_i$ can we say that $|A_i-A_j|\leq |B_i-B_j|\leq |C_i-C_j|$ ?? or you mean something else?
Thomas
Thomas
@mesel No. But you have a bound from below and above through terms which become small in absolute value. So $|B_i -B_j|$ is less than the $\max$ of the absolute values.
@Ant: yes, that's correct.
mesel
mesel
I am sorry but I did not get the point why $B_i$ is cauchy sequence.
Thomas
Thomas
@mesel I'll edit my answer.
mesel
mesel
@Thomas: right hand side need not be arbitrary small,if you mean $max\{|A_i|,|C_i|\}$ ?
hi
Thomas
Thomas
14:12
Hi, If both sequence $\sumî_k a_i$ and $\sumî_k c_i$ converge, then their partial sums are both Cauchy sequences, sure.
That is, by definition, the statement that the rhs (the one with the max) becomes arbitrarily small.
Just choose $\varepsilon > 0$
Then $K_0$ st $|\sum_l^k a_i| < \varepsilon$
mesel
mesel
You mean that $A_i$ $C_i$ are cauchy ,I agree on that, By capital letter I mean their partial sum.
Thomas
Thomas
Thel $L_0$ such that $|\sum_l^k c_i| < \varepsilon$
Then $N_0 = \max{K_0, L_0\}$
$A_i -B_j = \sum_j^i a_k$
damn. sorry
$A_i -A_j = \sum_j^i a_k$
If $A_j$ converges, this difference becomes small.
mesel
mesel
if a squence cauchy it means that $A_i-A_J$ or $C_i-C_j$ can be arbitrarly small.
Thomas
Thomas
Yes.
In absolute value, that is $|A_i-a_j|§
$|A_i-A_j|$ (sorry)
What exactly is it you do not get?
mesel
mesel
you get $0\leq|B_i|\leq max{A_i,C_i}$,right?
Thomas
Thomas
14:23
Note: the rhs in my answer is equal to $\max \{ |A_l -A_k|, |C_l - C_k|\}$
yes
no
sorry
I wrote down what I get.
note that the sum is from $l$ to $k$. Not from $0$ to $k$
$A_l - A_k = \sum_0^l a_i - \sum_0^k a_i = \sum_k^l a_i$
mesel
mesel
well, it seems to me that there is a small piece of mistake.
Thomas
Thomas
no. Where?
mesel
mesel
let $r$ be a given epsilon
Thomas
Thomas
fine
mesel
mesel
now we we want find i and j s.t
|B_i-B_j| is smaller than $r$.
Thomas
Thomas
14:29
correct
as soon as $j, i> J_0$, say
you want me to continue?
mesel
mesel
then $|B_i-B_j|$ is smaller than max of |A_i-A_j| or |C_i-C_j|.
Thomas
Thomas
yes
mesel
mesel
so for given $r$ we can find i,j for Ai and C_i since they are cauchy.
Thomas
Thomas
this is a bit imprecise.
mesel
mesel
So,you mean the found i,j are enough for us by the inequalty.
Thomas
Thomas
14:34
You find $J_A, J_C$ such that the desired inequality holds for $A_i-A_j$ if $i,j >J_A$, and for $C_i -C_j $ if $i, j >J_C$.
now let $J_B := \max \{J_A, J_C\}$
mesel
mesel
now,it seems to that it was my mistake :)
Thomas
Thomas
then boht $A_i-A_j$ and $C_i-C_j$ are small
both
mesel
mesel
thanks Thomas.
Thomas
Thomas
You're welcome
(have to leave now)
mesel
mesel
no
I am here
are you there?
see you

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