Suppose you have a double sequence anm. What are sufficient conditions for you to be able to say that lim? Bonus points for necessary and sufficient conditions.

For an example of a sequence where this is not the case, consider \displaystyle a_{nm}=\left(\frac{1}{n}\right)^{\frac{1}{m}}. \displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}}=\lim_{n\to \infty}{\left(\frac{1}{n}\right)^0}=\lim_{n\to \infty}{1}=1, but \displaystyle \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}=\lim_{m\to \infty}{0^{\frac{1}{m}}}=\lim_{m\to \infty}{0}=0.

**Answer**

If you want to avoid hypotheses that involve uniform convergence, you can always cheat and use the counting measure on {0, 1, 2,…} and then use either the Monotone or Dominated Convergence Theorem from integration theory.

For instance, using the Monotone Convergence Theorem, we get the following (perhaps silly) sufficient criterion:

**Proposition:** If a_{mn} is monotonically increasing in m, and is such that c_{mn} = a_{mn} – a_{m-1,n} is monotonically increasing in n, then \lim\limits_{m \to \infty} \lim\limits_{n \to \infty} a_{mn} = \lim\limits_{n\to\infty}\lim\limits_{m\to\infty} a_{mn}.

*Proof:* Our two hypotheses really just amount to saying that each c_{mn} \geq 0 and that the c_{mn} are monotonically increasing in n. So we can use the Monotone Convergence Theorem with respect to the counting measure:

\lim_{n\to \infty} \int c_{mn} = \int \lim_{n\to \infty} c_{mn}

But really, these integrals are sums, so:

\lim_{n\to \infty} \sum_{m=1}^\infty c_{mn} = \sum_{m=1}^\infty \lim_{n\to \infty} c_{mn}

Since infinite sums are just limits of partial sums, we have:

\lim_{n\to \infty} \lim_{M \to \infty} \sum_{m=1}^M c_{mn} = \lim_{M\to\infty}\lim_{n\to \infty} \sum_{m=1}^M c_{mn}

By our construction of c_{mn}, the left side is \lim\limits_{n\to\infty} \lim\limits_{M\to\infty} a_{Mn}, and the right side is \lim\limits_{M\to\infty} \lim\limits_{n\to\infty}a_{Mn}. \lozenge

**Edit:** In the above, our index ranges are m,n \geq 1, and we make the convention that a_{0n} = 0.

Using the Dominated Convergence Theorem, we could probably get something slightly more useful.

**Attribution***Source : Link , Question Author : asmeurer , Answer Author : Jesse Madnick*