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

## Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin