Suppose you have a double sequence $\displaystyle a_{nm}$. What are sufficient conditions for you to be able to say that $\displaystyle \lim_{n\to \infty}\,\lim_{m\to \infty}{a_{nm}} = \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}$? 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 $\displaystyle \lim_{m\to \infty}\,\lim_{n\to \infty}{a_{nm}}=\lim_{m\to \infty}{0^{\frac{1}{m}}}=\lim_{m\to \infty}{0}=0$.

  • $\begingroup$ $ \displaystyle \lim_{a \to 0}{0^a}=1$ eg. Using calculator $0.0000001^{0.0000001}=0.999998$ $\endgroup$
    – user235005
    Commented Nov 5, 2021 at 17:32
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    $\begingroup$ @user235005: no, your calculator result there is suggesting that $\lim_{a\rightarrow 0} a^a = 1$, which is not the same. $\endgroup$ Commented Nov 23, 2021 at 18:33

6 Answers 6


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.

  • 4
    $\begingroup$ Interesting trick. I have never heard of the counting measure until now (though I am surprised, since it is a nice example of a non-Lebesgue measure). Actually, I thought I remembered in the measure theory that I learned that every countable set was measure zero for any measure, but I could be remembering it wrong. $\endgroup$
    – asmeurer
    Commented Dec 23, 2010 at 5:04
  • $\begingroup$ Perhaps countable sets have measure zero for every Borel measure? Is that right? $\endgroup$ Commented Dec 23, 2010 at 5:07
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    $\begingroup$ But don't call it "cheating". Call it "an advanced method". $\endgroup$
    – GEdgar
    Commented Jun 14, 2011 at 17:40
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    $\begingroup$ Why not play this game again, define $d_{nm}:=c_{m,n}-c_{m,n-1}$, and apply Fubini's theorem on $\sum_{n,m} d_{n,m}$? $\endgroup$
    – PPR
    Commented May 2, 2016 at 12:53
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    $\begingroup$ @LuckyChouhan: To apply the Proposition, one also needs the sequences $c_{mn} = a_{mn} - a_{m-1,n}$ to be monotonically increasing in $n$. I don't believe that your example satisfies that assumption. $\endgroup$ Commented Jul 30, 2023 at 17:06

Here's some results.

Let us say that a sequence $a_{nm}$ of real numbers indexed by pairs of positive integers $(n,m)$ converges to $L$ if and only if for every $\epsilon\gt 0$ there exists $N\gt 0$ such that for all $n,m\geq N$, $|a_{nm}-L|\lt \epsilon$.

This is a fairly strong condition. We have:

Theorem. Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}$ exists and equals $L$. Then the following are equivalent:

  1. For each (sufficiently large) $n_0$, $\lim\limits_{m\to\infty}a_{n_0m}$ exists;
  2. $\lim\limits_{n\to\infty}\lim\limits_{m\to\infty}a_{nm} = L$.

Proof. If 2 holds, then we must have 1 (otherwise the expression in 2 does not even make sense). Now assume that 1 holds, and let $\lim\limits_{m\to\infty}a_{nm} = L_{n}$. We want to prove that $\lim\limits_{n\to\infty}L_n=L$.

Let $\epsilon\gt 0$. Then there exists $N\gt 0$ such that for all $n,m\geq N$, $|a_{nm}-L| \lt \epsilon$. Let $M_N\gt N$ be such that for all $m\geq M_N$, $|a_{M_Nm}-L_{M_N}|\lt \epsilon$. Since $M_n\gt N$, we have $$|L_{M_N} - L| \leq |L_{M_N}-a_{M_Nm}| + |a_{M_Nm}-L| \leq 2\epsilon,$$ so this proves that $L_n\to L$ as $n\to\infty$. In particular, we have for the iterated limit $$\lim\limits_{n\to\infty}\lim\limits_{m\to\infty} a_{nm} = \lim_{n\to\infty}L_n = L.\ \Box$$

A symmetric argument shows that

Theorem. Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}$ exists and equals $L$. Then the following are equivalent:

  1. For each (sufficiently large) $m_0$, $\lim\limits_{n\to\infty}a_{nm_0}$ exists;
  2. $\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}a_{nm} = L$.

However, $\lim\limits_{n,m\to\infty}a_{nm}=L$ does not imply the existence of the iterated limits; in fact, existence of the double limit and the existence of one iterated limit does not suffice to imply that the other iterated limits exists. Take $a_{nm}=\frac{(-1)^n}{m}$. Then $\lim\limits_{n,m\to\infty}a_{nm}=0$ (given $\epsilon\gt 0$, pick $N$ such that $\frac{1}{N}\lt\epsilon$), and the limit as $m\to\infty$ of $\frac{(-1)^n}{m}$ exists for each fixed $n$, but $\lim\limits_{n\to\infty}\frac{(-1)^n}{m}$ does not exist for any $m$. (By taking $a_{nm} = \frac{(-1)^n}{m} + \frac{(-1)^m}{n}$ you get one in which neither iterated limit exists.)

So, a sufficient conditions for the iterated limits to exist is:

Corollary. Suppose that $\lim\limits_{(n,m)\to\infty}a_{nm}=L$. Then the iterated limits $$\lim_{n\to\infty}\lim_{m\to\infty}a_{nm}\text{ and }\lim_{m\to\infty}\lim_{n\to\infty}a_{nm}$$ both exist and are equal to $L$ if and only if $\lim\limits_{n\to\infty}a_{nm}$ exists for almost all $m$ and $\lim\limits_{m\to\infty}a_{nm}$ exists for almost all $n$.

(Here, "almost all" means "all except perhaps for a finite number").

As I said, the condition above is pretty strong. You can have both iterated limits exist and be equal and yet for the double limit not to exist. To adapt the standard two-variable example, take $a_{nm}=\frac{nm}{n^2+m^2}$. The iterated limits both exist and are equal to $0$, but the double limit does not exist (for any $N\gt 0$ there exist $n,m\geq N$ such that $a_{nm}=\frac{1}{2}$ and there exist $n,m\geq N$ such that $a_{nm}=\frac{2}{5}$; just take $n=m=N$ for the first, and $n=2m=2N$ for the second).

You can get that the double limit exists and is equal to (one of) the iterated limits if you have some uniformness conditions.

I don't know of any necessary and sufficient conditions, and I suspect there won't generally be without some other overarching conditions. This is essentially the same problem as the problem of iterated limits in functions of two variables (just as the problem of finding the limit of a real function of real variable is closely connected to the problem of finding limits of sequences of real numbers). They are connected to the double limits, and you often have conditions based on uniform convergence that guarantee good things happen, but to state some general, simple condition for the iterated limits to exist and be equal seems difficult in general.

  • 1
    $\begingroup$ I suspected that something like that might hold, though as you said, it is pretty strong. Very nice proof and examples, though. $\endgroup$
    – asmeurer
    Commented Dec 22, 2010 at 23:35
  • 1
    $\begingroup$ And now you've got me wondering what the set of numbers that $a_{nm}=\frac{nm}{n^2 + m^2}$ can converge to is. $\endgroup$
    – asmeurer
    Commented Dec 22, 2010 at 23:37
  • 2
    $\begingroup$ @asmeurer: Certainly, any number of the form $\frac{k}{k^2+1}$ with $k$ an integer; also $0$ (take $n=m^2$); if $n=\lfloor tm\rfloor$, you might be able to get any $\frac{t}{t^2+1}$. $\endgroup$ Commented Dec 22, 2010 at 23:45
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    $\begingroup$ I believe that the convergence of double sequences introduced in the beginning of this post is sometimes called Prigsheim convergence. $\endgroup$ Commented Jun 14, 2011 at 15:59
  • $\begingroup$ @Arturo: A question about Pringsheim convergence has been posted which, if I correctly understood OP's comment, is asking about a reference for the results mentioned in this answer. I thought it might be good to ping you about this. $\endgroup$ Commented Dec 24, 2011 at 9:28

This result is similar to the result from Jesse Madnick's post.

Suppose that $a_{km}$ is non-decreasing in both variables, i.e., $a_{km}\le a_{k,m+1}$ and $a_{km}\le a_{k+1,m}$ for each $k$ and $m$.

Then $\lim\limits_{k\to\infty} \lim\limits_{m\to\infty} a_{km} = \lim\limits_{m\to\infty} \lim\limits_{k\to\infty} a_{km}$.

Proof: Denote $b_k=\lim\limits_{m\to\infty} a_{km}$ and $c_m=\lim\limits_{k\to\infty} a_{km}$. Monotonicity implies that these limits exist (they might be $+\infty$) and the sequences $(b_k)$, $(c_m)$ are non-decreasing.

Put $b:=\lim\limits_{k\to\infty} b_k$, $c:=\lim\limits_{m\to\infty} c_m$.

We have $a_{km}\le c_m \le c$ $\Rightarrow$ $b_k \le c$ $\Rightarrow$ $b\le c$.

The proof that $c\le b$ is analogous.

  • $\begingroup$ @MartinSleziak, why is $b_k \leq c$? I don't understand how you deduce that step. $\endgroup$
    – Kamil
    Commented Jul 11, 2016 at 8:16
  • 1
    $\begingroup$ @Kamil Using the fact that limits preserve inequality: math.stackexchange.com/questions/1850471/… If $a_{km} \le c_m$ for each $m$, then also $\lim\limits_{m\to\infty} a_{km} \le \lim\limits_{m\to\infty} c_m$ (assuming the limits exist). $\endgroup$ Commented Jul 11, 2016 at 8:24

Here's a fact I use all the time:

Let $a_{mn}$ be any double sequence in the extended real numbers $\overline{\mathbb{R}}$.

Then $$ \sup_m \sup_n \, a_{mn} = \sup_n \sup_m \, a_{mn} = \sup_{m,n} \, a_{mn} $$

Proof. Let $A=\sup_{m,n} \, a_{mn}, \, B = \sup_m \sup_n \, a_{mn}$.

Then for all $m$ and $n$, $A\ge a_{mn}$, so that $A \ge B$.

To show the other direction, first consider the case that $A<\infty$. Then $\forall \varepsilon > 0 \exists m^*,n^*$ s.t. $a_{m^*n^*} \ge A-\varepsilon$. It follows that $$ B=\sup_m \sup_n \, a_{mn} \ge \sup_m \, a_{mn^*} \ge a_{m^*n^*} \ge A-\varepsilon. $$ Taking the limit as $\varepsilon\to 0$, one obtains $B\ge A$.

In case $A=\infty$, $\forall M>0\exists m^*,n^*$ s.t. $a_{m^*n^*} \ge M$. Then $$ B=\sup_m \sup_n \, a_{mn} \ge \sup_m \, a_{mn^*} \ge a_{m^*n^*} \ge M. $$ Sending $M$ to $\infty$, one obtains $B=\infty$.

$A=\sup_n \sup_m \, a_{mn}$ follows by a symmetric argument. $\square $

As a corrolary one obtains:

Let $a_{mn}$ be any double sequence in the extended real numbers $\overline{\mathbb{R}}$ which is non-decreasing in $m$ and $n$. Then $$ \lim_{n\to\infty}\lim_{m\to\infty} a_{mn} = \lim_{m\to\infty}\lim_{n\to\infty} a_{mn} $$

This can be used with sums of non-negative numbers, for example.

  • $\begingroup$ To prove the first assertion, we can also use the "universal property" definition of the supremum. Expanding the definition of the supremum, we get that for an arbitrary real $A$, we have $A ≥ \sup_n \sup_m a_{nm}$ if and only if $A≥\sup_n a_{nm}$ for all $m$, if and only if $A≥a_{nm}$ for all $n$ and all $m$. But this is exactly the definition of $\sup_{n,m} a_{nm}$. $\endgroup$ Commented Feb 21, 2019 at 2:09
  • $\begingroup$ In the corollary would it be correct to add $\lim_{m,n\to\infty}a_{mn}$ to the equality? $\endgroup$ Commented Jan 17, 2020 at 8:41
  • $\begingroup$ @MohamedAli First you have to define what is the meaning of $\lim_{m,n\to\infty}a_{mn}$. Note that we have a “direction” when taking limit while we don’t need a “direction” to take supremum. So your notation may not make sense and be defined anyway. $\endgroup$
    – Sam Wong
    Commented Sep 22, 2021 at 9:19

The Moore-Osgood Theorem might also be relevant here.


The following is a minor variation on Arturo Magidin's answer, but according to my experience it is more widely used.

Proposition: Assume $$ \lim_{m \to \infty} \lim_{n \to \infty} a_{mn} = \lim_{m \to \infty} a_{m\times} = a \qquad\text{and}\qquad \lim_{m \to \infty} a_{mn} = a_{\times n} $$ i.e. we have three sequences $(m,n) \mapsto a_{mn}$, $m \mapsto a_{m\times}$ and $n \mapsto a_{\times n}$ such that all of the above limits exist and converge to the indicated values. We further assume that the single-variable limits converge uniformly in the other variable, i.e. for every $\varepsilon>0$ there exists $M > 0$ such that $|a_{mn} - a_{\times n}| < \varepsilon$ for all $m > M$ and all $n$, and likewise for $a_{m\times}$. Then $\displaystyle \lim_{n \to \infty} \lim_{m \to \infty} a_{mn}$ exists and it holds $$ \lim_{n \to \infty} \lim_{m \to \infty} a_{mn} = \lim_{m \to \infty} \lim_{n \to \infty} a_{mn} . $$

Proof: For any $\varepsilon > 0$, we can find $N > 0$ such that $|a_{mn} - a_{m\times}| < \tfrac{\varepsilon}{3}$ for all $n > N$ and all $m$. Similarly, we can find $M>0$ such that both $|a_{\times n} - a_{mn}| < \tfrac{\varepsilon}{3}$ for all $n$ and $|a_{m\times} - a| < \tfrac{\varepsilon}{3}$ holds for all $m > M$. Thus, it holds $$ |a_{\times n} - a| \leq |a_{\times n} - a_{mn}| + |a_{mn} - a_{m\times}| + |a_{m\times} - a| < \varepsilon $$ for all $n > N$ and $m > M$.

  • $\begingroup$ It seems to me that the notation $\lim_{m\rightarrow\infty}a_{mn}=a_n$ is problematic here since we already have $\lim_{n\rightarrow\infty}a_{mn}=a_m$. In other words, the same sequence cannot be used for the limit where the first (resp. second) index is kept constant. This however doesn't seem to invalidate the proof which looks correct after using adequate notations. $\endgroup$ Commented Jan 31 at 12:59
  • $\begingroup$ You are right, that was potentially confusing. I've updated the notation. $\endgroup$
    – gTcV
    Commented Feb 1 at 3:40

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