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So here is the problem. Given absolute convergence for a double series (infinite sum over $|a_{ij}|$) , show the double series $(a_{ij})$ converges . The proof strategy is: 1) keep one index fixed - so given i is fixed we know the series over j converges absolutely to some $b_i$ i.e $\Sigma_j |a_{ij}| $ converges to $b_i$, so the actual series $\Sigma_j a_{ij}$ must converge to a $b'_i$. Also we know $b'_i \leq b_i$ (because $a_{ij} \leq |a_{ij}|$). Now if we knew $b'_i \geq 0$, then comparison test applies (2.7.4). And we are done (just have to take infinite sum over $b_i$). But - it is not obvious to me why $b'_i \geq 0$ (i know $b_i \geq 0$ for sure as its the limit of the absolute values). I am missing something obvious - HELP!

Based on T(op?) Gunn's response, we have $|b'_i| <b_i$, so $|b'_i| \geq 0$ so comparison test claims $b'_i$ has absolute convergence - hence $b'_i$ has convergence.

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    $\begingroup$ Time to learn some math typesetting. $\endgroup$ – zhw. May 19 '17 at 19:12
  • $\begingroup$ just did! would help if someone can figure this out $\endgroup$ – pythOnometrist May 19 '17 at 19:18
  • $\begingroup$ Your notion of double series convergence is $\lim_{N\to \infty} \sum_{i=1}^{N}\sum_{j=1}^{N} a_{ij}$ exists and is finite? $\endgroup$ – zhw. May 19 '17 at 19:22
  • $\begingroup$ yeah using rectangles/ squares etc. $\endgroup$ – pythOnometrist May 19 '17 at 19:27
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Suppose $ \sum_i \sum_j |a_{ij}|$ converges. Then for each $i$, $\sum_j |a_{ij}|$ converges. Hence $\sum_j a_{ij}$ converges. The key observation from here is the "Infinite Triangle Inequality":

$$ \left\lvert \sum_{j = 1}^\infty a_{ij} \right\rvert \le \sum_{j = 1}^\infty |a_{ij}|. $$

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  • $\begingroup$ Neat - is there a reference I can go over the derivation - the condition i suppose holds when the series in the absolute sign on lhs converges and the rhs series converges. Is that it? $\endgroup$ – pythOnometrist May 19 '17 at 19:28
  • $\begingroup$ Well if we're using sums over squares as our starting point, I'd say you have more work left to do. $\endgroup$ – zhw. May 19 '17 at 19:50
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The comparison test can only be used if the terms of the series are nonnegative.

Suppose the series $\sum_{i,j}|a_{ij}|$ converges. Let $b_{ij} = \max(a_{ij},0$) and $c_{ij} = \max(-a_{ij},0)$.

Since $0 \leqslant b_{ij} \leqslant |a_{ij}|$ and $0 \leqslant c_{ij} \leqslant |a_{ij}|$ we can now apply the comparison test to conclude that

$\sum_{i,j}b_{ij}$ and $\sum_{i,j}c_{ij}$ converge.

Therefore, we have convergence of

$$\sum_{i,j}a_{ij} = \sum_{i,j}(b_{ij} - c_{ij}) = \sum_{i,j}b_{ij} - \sum_{i,j}c_{ij} $$

Also be aware that convergence of a double series (to $S$) in the strictest sense is that for any $\epsilon >0$ there exists a positive integer $N$ such that for all $n,m > N$ we have

$$\left|\sum_{i=1}^m \sum_{j=1}^n a_{ij} - S \right| < \epsilon$$

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  • $\begingroup$ actually T. Gunn's answer addresses the non -negativity requirement. $\endgroup$ – pythOnometrist May 19 '17 at 19:55
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Let $$S_N = \sum_{i,j=1}^{n} a_{ij}, \,\,T_n = \sum_{i,j=1}^{n} |a_{ij}|.$$ For $m<n,$  define $A(m,n) = \{(i,j): 1\le i,j\le n\} \setminus \{(i,j): 1\le i,j\le m\}.$ Then

$$\tag 1 |S_n - S_m| = |\sum_{i,j \in A(m,n)}a_{ij}| \le \sum_{i,j \in A(m,m)}|a_{ij}| = T_n- T_m.$$

Because $T_n$ converges, it is Cauchy. $(1)$ then shows $S_n$ is Cauchy. This implies $S_n$ converges, which is the desired conclusion.

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By assumption for any $i \in \{1, 2, 3, \cdots \}$, $$\sum_{j = 1}^{\infty} |a_{ij}|$$ converges,
i.e. $$ \sum_{j = 1}^{\infty} a_{ij} $$ converges absolutely.

So, $$ -\sum_{j = 1}^{\infty} |a_{ij}| \leq \sum_{j = 1}^{\infty} a_{ij} \leq \sum_{j = 1}^{\infty} |a_{ij}|, \\ |\sum_{j = 1}^{\infty} a_{ij}| \leq \sum_{j = 1}^{\infty} |a_{ij}|. $$

Let $$ b_i := \sum_{j = 1}^{\infty} |a_{ij}|, \\ c_i := \sum_{j = 1}^{\infty} a_{ij}. $$

Then, $$ |c_i| \leq b_i. $$

$$ \sum_{i=1}^{\infty} b_i $$ converges by assumption.
So, $$ \sum_{i=1}^{\infty} |c_i| $$ also converges, i.e. $$ \sum_{i=1}^{\infty} c_i $$ converges absolutely.

$$ \sum_{i=1}^{\infty} c_i = \sum_{i=1}^{\infty} \sum_{j = 1}^{\infty} a_{ij} $$ converges.

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