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So, as a counterexample, let <math>\Sigma_{j=1}^{\infty} (ai)_j = i + (-1) + (-1) + \cdots</math> i times <math> \cdots + (-1) + 0 + 0 + \cdots</math> And note that the partial sums are like the <math>(pi)_j</math> of the problem above, and therefore that this counterexample should follow in the same way. | So, as a counterexample, let <math>\Sigma_{j=1}^{\infty} (ai)_j = i + (-1) + (-1) + \cdots</math> i times <math> \cdots + (-1) + 0 + 0 + \cdots</math> And note that the partial sums are like the <math>(pi)_j</math> of the problem above, and therefore that this counterexample should follow in the same way. | ||
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| + | ---- | ||
| + | '''Actually it holds for the series... part of the hypothesis was that the <math>a_{i,j}\geq 0</math>. Just use comparison with the <math>a_i</math> for convergence of the <math>a_{i,i}</math> series. | ||
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| + | -Coach''' | ||
Latest revision as of 08:41, 22 October 2008
1)
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Why the proof fails
Let $ \{ (pi)_j \}, \{ p_n \} \subseteq \Re; p \in \Re, \forall i,j,n $
And let $ (p1)_j \rightarrow p_1, (p2)_j \rightarrow p_2, (pi)_j \rightarrow p_i $ as $ j \rightarrow \infty $, and $ p_n \rightarrow p $
We show that $ \{ (pi)_j \} $ converges.
Fix $ \epsilon > 0 $, then $ \exists N_1, N_2 \ni \forall n \geq N_1, N_2 $:
Stop here, I'd need to choose more than one $ N_1 $, one for each sequence for what I'm trying to do, and I'm not guaranteed that the $ \sup N_i $ will be finite.
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From this I'll use the following counterexample:
Let $ p_n = 0 \forall n $, and $ (pi)_j = \{ i, i-1, i-2, \cdots, 1, 0, 0, 0, \cdots \} $
2) Ditto with series.
The partial sums of series are sequences, so the same result should hold.
So, as a counterexample, let $ \Sigma_{j=1}^{\infty} (ai)_j = i + (-1) + (-1) + \cdots $ i times $ \cdots + (-1) + 0 + 0 + \cdots $ And note that the partial sums are like the $ (pi)_j $ of the problem above, and therefore that this counterexample should follow in the same way.
Actually it holds for the series... part of the hypothesis was that the $ a_{i,j}\geq 0 $. Just use comparison with the $ a_i $ for convergence of the $ a_{i,i} $ series.
-Coach
