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I read in one place that the definition of convergence of a double series is given by:

Let $(a_{mn})_{m, n=1}^{\infty}$ be a double sequence. The corresponding double series is denoted $\sum_{m, n =1}^{\infty} a_{mn}$. The corresponding double sequence of partial sums is denoted $(s_{mn})_{m, n=1}^{\infty}$ and is defined by $s_{mn} = \sum_{i=1}^{m} \sum_{j=1}^{n} a_{ij} = \sum_{j=1}^{n} \sum_{i=1}^{m} a_{ij}$. We say that the double series $\sum_{m, n =1}^{\infty} a_{mn}$ converges to $A \in \mathbf{R}$ if the corresponding double sequence of partial sums $(s_{mn})_{m, n=1}^{\infty}$ converges to $A$. We denote this as \begin{align*} \sum_{m, n =1}^{\infty} a_{mn} = \lim_{m,n \rightarrow \infty} s_{mn} = \lim_{m,n \rightarrow \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} a_{ij} = \lim_{m,n \rightarrow \infty} \sum_{j=1}^{n} \sum_{i=1}^{m} a_{ij} = A. \end{align*}

However, in another place, the definition is as follows:

If the sequence $(s_{nn})$ converges, then we define $\sum_{m, n =1}^{\infty} a_{mn} = \lim_{n \rightarrow \infty} s_{nn}$

My question is: are these two definitions equivalent? For these two definitions to be equivalent, wouldn't we need the condition that $(s_{mn})_{m, n=1}^{\infty}$ converges if and only if $(s_{nn})$ converges and $\lim_{m,n \rightarrow \infty} s_{mn} =\lim_{n \rightarrow \infty} s_{nn}$?

If the definitions are not equivalent, then which one do I follow?

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  • $\begingroup$ What about the terms of $s_{m,n}$ where $m\ne n$? What of $s_{m,n}=m-n$? $\endgroup$ May 31 '17 at 15:16
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Consider $s_{mn} = mn/(m+n)^2$. Then $s_{nn} \to 1/4$ but $\lim_{m,n} s_{mn}$ does not exist. There is no unique limit -- for example, with $m = 2n$ we have $s_{2n,n} \to 2/9 \neq 1/4$.

Convergence of $\lim_{m,n} s_{mn}$ is a stronger condition and implies the convergence of $\lim_{n} s_{nn}$

For double series $\sum_{m,n} a_{mn}$, absolute convergence guarantees that all modes of convergence -- by rows, columns, squares, diagonals, etc. are equivalent:

$$\lim_{M,N}\sum_{m=1}^M \sum_{n=1}^N a_{mn} = \sum_{m=1}^\infty\sum_{n=1}^\infty a_{mn} = \sum_{n=1}^\infty\sum_{m=1}^\infty a_{mn} = \lim_{N \to \infty}\sum_{n=1}^N\sum_{m=1}^N a_{mn}$$

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  • $\begingroup$ Thanks, this a great answer. One last question, is it true that for a double sequence $(a_{mn})_{m, n=1}^{\infty}$, $\lim_{m, n \rightarrow \infty} a_{mn} = A$ if and only if $\lim_{m \rightarrow \infty}\left(\lim_{n \rightarrow \infty} a_{mn}\right) = \lim_{n \rightarrow \infty}\left(\lim_{m \rightarrow \infty} a_{mn}\right) =A$? $\endgroup$
    – SwiftMo
    Jun 1 '17 at 3:28
  • $\begingroup$ If $\lim_{m,n \to \infty}a_{mn} = A$ and $\lim_{n \to \infty}a_{mn}$ exists for all $m$, then $\lim_{m \to \infty} \lim_{n \to \infty}a_{mn} = A$. Similarly if $\lim_{m \to \infty}a_{mn}$ exists for all $n$ then $\lim_{n \to \infty} \lim_{m \to \infty}a_{mn} = A$. The converse is not necessarily true. There is a converse if one of the inner iterated limits is uniformly convergent. $\endgroup$
    – RRL
    Jun 1 '17 at 3:41
  • $\begingroup$ If $\lim_{n \to \infty} a_{mn} = y_m$ with uniform convergence for all $m$ and $\lim_{m \to \infty} a_{mn} = z_n$ exists (not necessarily uniform) then the iterated limits and the double limit exist and are the same: $\lim_{m,n \to \infty} a_{mn} = \lim_{m \to \infty} y_m = \lim_{n \to \infty} z_n$. $\endgroup$
    – RRL
    Jun 1 '17 at 3:45
  • $\begingroup$ Do you have access to the book The Elements of Real Analysis by Bartle? This is nicely covered in Section 19 (second edition). $\endgroup$
    – RRL
    Jun 1 '17 at 3:46
  • $\begingroup$ Thanks for the reference, I'll definitely check it out. I also have one more question on absolute convergence of double series, would be great if you could share your insights: math.stackexchange.com/questions/2305138/… $\endgroup$
    – SwiftMo
    Jun 1 '17 at 4:38

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