(New page: 1) --- Why the proof fails Let <math>\{ (pi)_j \}, \{ p_n \} \subseteq \Re; p \in \Re, \forall i,j,n</math> And let <math>(p1)_j \ra p_1, (p2)_j \ra p_2, (pi)_j \ra p_i</math> as <math...) |
m |
||
| Line 7: | Line 7: | ||
Let <math>\{ (pi)_j \}, \{ p_n \} \subseteq \Re; p \in \Re, \forall i,j,n</math> | Let <math>\{ (pi)_j \}, \{ p_n \} \subseteq \Re; p \in \Re, \forall i,j,n</math> | ||
| − | And let <math>(p1)_j \ | + | And let <math>(p1)_j \rightarrow p_1, (p2)_j \rightarrow p_2, (pi)_j \rightarrow p_i</math> as <math>j \rightarrow \infty</math>, and <math>p_n \rightarrow p</math> |
We show that <math>\{ (pi)_j \}</math> converges. | We show that <math>\{ (pi)_j \}</math> converges. | ||
Revision as of 15:52, 9 October 2008
1)
---
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.
---
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.
