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Let $a_{mn}$ be a sequence in some $\mathbb{R}^k$. We know in advance that

$$\lim_{n} ~a_{nn} = L_1, \qquad \lim_{m}~ \lim_{n} ~a_{mn} = L_2 $$

exist. Is there a sufficient criteria to conclude that $L_1 = L_2$? A simple example where this is not true is when $$a_{mn} = \frac{m}{n}.$$

Please note that I did not say $$ \lim_{n}~ \lim_{m} ~a_{mn} = L_3 $$ exists. If there is an answer available after assuming $L_3$ exists, then I would like to know that. But my original question doesn't assume $L_3$ exists.

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For any values of $L_1$, $L_2$ and $L_3$, whether equal or not, there is a sequence $a_{nm}$ achieving your three limit requirements. The reason is that $\lim_n a_{nn}=L_1$ refers only to the entries on the diagonal, $\lim_n\lim_m a_{nm}=L_2$ depends only on entries above the diagonal, and $\lim_m\lim_n a_{nm}=L_3$ depends only on entries below the diagonal. We could simply let $a_{nm}$ be $L_1$ when $n=m$ and $L_2$ when $n\lt m$ and $L_3$ when $n\gt m$. You may similarly ask independently for any of the three limits not to exist, without affecting the other limits, simply by modifying the values of $a_{nm}$ only in the appropriate region (that is, either on, above, or below the diagonal). So in general, knowing any two of the three limits tells you nothing about the third limit or whether it exists.

But meanwhile, you can get $L_1=L_2$ if you know something about how fast the two limits converge. What you need to know, of course, is that $a_{nn}$ becomes sufficiently close to both limits, and one can easily invent criteria that will ensure this. But if you think about it, such requirements amount merely to a restatement of the equality of the limits, and so they will probably be unsatisfying.

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Well, it is a bit too strong for you, but should be noted that if $a_{nm}$ converges uniformly in $m$, and your $L_2$ exists, then all the three limits exist and are equal. This is a classical result having generalizations to net-convergence, and its first version is attributed to Cauchy.

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