Proving Boundedness of Double Index Array | Math Problem

[linuxux]

I'm just don't get this question.

from my text:

$$define\ t_{mn}\ =\ \sum^{m}_{i=1}\sum^{n}_{j=1}|a_{ij}|,$$
$$[a_{ij}\ is\ a\ doubly\ indexed\ array\ of\ real\ numbers.]$$
$$(a)\ prove\ that\ the\ set\ \{t_{mn}\ :\ m,n\ \in\ N\}\ is\ bounded\ above.$$
$$(b)\ use\ (a)\ to\ conclude\ that\ (t_{nn})\ converges.$$

This has been confusing me for a couple of days, I've got the rest of the proof, but this part doesn't make any sense. The theorem I'm trying to prove states "IF the doubly indexed array converges absolutely," but if I "prove" {t_mn} is bounded, it suggest then that every doubly indexed array would converge absolutely, so I'm not sure how i can possibly answer (a).

From what i understand, the set {t_mn} is all the possible outcomes of the double summation, for instance, the first member would be a_11, the second might be a_11 + a_12 or a_11 + a_21, etc. But that set has no bound. It can grow as much as it wants, i think.

i'm not sure where I'm going wrong here. please help, thanks.

 

[ozymandias]

You're not given ANY information about the $$a_{ij}$$s?
That's odd. Take $$a_{ij}=1$$ for all i, j. Then
$$t_{mn} = m \times n$$ which is of course NOT bounded above - correct me if I'm mistaken.

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Assaf
http://www.physicallyincorrect.com/"

 

[linuxux]

You're not given ANY information about the $$a_{ij}$$s?
That's odd. Take $$a_{ij}=1$$ for all i, j. Then
$$t_{mn} = m \times n$$ which is of course NOT bounded above - correct me if I'm mistaken.

--------
Assaf
http://www.physicallyincorrect.com/"

Well, the book was introducing double sums, and at one point it defined the "rectangular partial sums":
$$s_{mn}\ =\ \sum^{m}_{i=1}\sum^{n}_{j=1}a_{ij},\ for\ m,n\ \in\ N.$$
And before it defined t_mn, it said "in the same way that we define the 'rectangular partial sums' s_mn above, define t_mn (as indicated above)"

however, it did not specifiy in any way that s_mn converged, all it said was that "in this case the order of the sum doesn't matter since the sum is finite."

 

[ozymandias]

Still, without any information about the a_ijs, we cannot prove the assertion.

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Assaf
http://www.physicallyincorrect.com/"