Double Sequences and Iterated Limits

This scene is about double sequences. Watch the values

a_{m,n}

on a grid, and see how the row-wise and column-wise limits split apart. The double limit lets both indices grow together. An iterated limit sends one index first, then the other. Swapping the order can change the answer. And even when both iterated limits agree, the double limit may not exist. For example, reading a 📋 table row by row is not the same as reading it column by column. Change the reading order, and you can land on a completely different conclusion.

Examples & Exercises

For the double sequence $a_{m,n} = \frac{m}{m+n}$ , prove completely: (i) For each fixed $m$ , $\lim_{n\to\infty} a_{m,n} = 0$ , so the iterated limit $\lim_{m\to\infty}\lim_{n\to\infty} a_{m,n} = 0$ . (ii) For each fixed $n$ , $\lim_{m\to\infty} a_{m,n} = 1$ , so the iterated limit $\lim_{n\to\infty}\lim_{m\to\infty} a_{m,n} = 1$ . (iii) Nevertheless the double limit $\lim_{m,n\to\infty} a_{m,n}$ (the Pringsheim limit) does not exist.

\lim_{n\to\infty} \frac{m}{m+n} = 0 \quad (m \text{ fixed}), \qquad \lim_{m\to\infty} \frac{m}{m+n} = 1 \quad (n \text{ fixed})

a_{k,k} = \frac{1}{2}, \qquad a_{k,k^2} = \frac{1}{1+k} \to 0

a_{m,n} = \frac{m}{m+n}

: every row limit is

0

, every column limit is

1

— and the diagonal sits at

\frac{1}{2}

forever.

Show proof

(i) Fix $m$ . Since $m + n > n$ , $0 < a_{m,n} = \frac{m}{m+n} < \frac{m}{n}$ . Given $\varepsilon > 0$ , take $N > \frac{m}{\varepsilon}$ . For $n > N$ , $\frac{m}{n} < \varepsilon$ , so $|a_{m,n} - 0| < \varepsilon$ . Hence $\lim_{n\to\infty} a_{m,n} = 0$ for every $m$ , and since the inner limit is the constant $0$ , $\lim_{m\to\infty}\lim_{n\to\infty} a_{m,n} = \lim_{m\to\infty} 0 = 0$ .
(ii) Fix $n$ . Since $a_{m,n} = \frac{m}{m+n} = 1 - \frac{n}{m+n}$ , $|a_{m,n} - 1| = \frac{n}{m+n} < \frac{n}{m}$ . Given $\varepsilon > 0$ , take $N > \frac{n}{\varepsilon}$ . For $m > N$ , $\frac{n}{m} < \varepsilon$ , so $|a_{m,n} - 1| < \varepsilon$ . Hence $\lim_{m\to\infty} a_{m,n} = 1$ for every $n$ , and $\lim_{n\to\infty}\lim_{m\to\infty} a_{m,n} = \lim_{n\to\infty} 1 = 1$ .
(iii) Suppose a double limit $L$ existed. For $\varepsilon = \frac{1}{8}$ there is $N$ with $|a_{m,n} - L| < \frac{1}{8}$ whenever $m, n > N$ . At the diagonal points $(k, k)$ with $k > N$ , $a_{k,k} = \frac{k}{k+k} = \frac{1}{2}$ , so $\left|\frac{1}{2} - L\right| < \frac{1}{8}$ , giving $L > \frac{3}{8}$ . At the points $(k, k^2)$ with $k > \max(N, 8)$ , $a_{k,k^2} = \frac{k}{k+k^2} = \frac{1}{1+k} < \frac{1}{9} < \frac{1}{8}$ , and $|a_{k,k^2} - L| < \frac{1}{8}$ , giving $L < \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$ . Since $\frac{1}{4} < \frac{3}{8}$ , the two conclusions collide. Hence no double limit exists. Remark: the mere fact that the iterated limits in (i) and (ii) differ already forces this, by the contrapositive of the theorem in Problem 2.

Prove the following theorem completely. If a double sequence $a_{m,n}$ converges to $L$ in the Pringsheim sense (for every $\varepsilon > 0$ there is $N$ with $|a_{m,n} - L| < \varepsilon$ whenever $m, n > N$ ), and for each fixed $m$ the inner limit $b_m = \lim_{n\to\infty} a_{m,n}$ exists, then $\lim_{m\to\infty} b_m = L$ . Conclude: when the double limit exists and both iterated limits exist, both iterated limits equal the double limit.

m, n > N \implies |a_{m,n} - L| < \frac{\varepsilon}{2}, \qquad |b_m - L| \le \frac{\varepsilon}{2} < \varepsilon \quad (m > N)

a_{m,n} = \frac{(-1)^n}{m}

: the double limit is

0

, and in the order where the inner limits exist the iterated limit is

0

too — exactly as the theorem predicts.

Show proof

Fix $\varepsilon > 0$ . Applying the definition of the double limit with $\frac{\varepsilon}{2}$ gives an $N$ such that $m, n > N \implies |a_{m,n} - L| < \frac{\varepsilon}{2}$ .
Now fix $m > N$ . For every $n > N$ , $|a_{m,n} - L| < \frac{\varepsilon}{2}$ . We show $|b_m - L| \le \frac{\varepsilon}{2}$ . Suppose instead $|b_m - L| > \frac{\varepsilon}{2}$ , and put $\delta = |b_m - L| - \frac{\varepsilon}{2} > 0$ . Since $a_{m,n} \to b_m$ as $n \to \infty$ , for all large $n$ (in particular some $n > N$ ) we have $|a_{m,n} - b_m| < \delta$ . Then $|a_{m,n} - L| \ge |b_m - L| - |a_{m,n} - b_m| > |b_m - L| - \delta = \frac{\varepsilon}{2}$ , contradicting the previous step. Hence $|b_m - L| \le \frac{\varepsilon}{2} < \varepsilon$ .
That is, $|b_m - L| < \varepsilon$ for every $m > N$ . Since $\varepsilon > 0$ was arbitrary, $\lim_{m\to\infty} b_m = L$ : that is, $\lim_{m\to\infty}\lim_{n\to\infty} a_{m,n} = L$ . The symmetric argument shows that if $\lim_{m\to\infty} a_{m,n}$ exists for each $n$ , then $\lim_{n\to\infty}\lim_{m\to\infty} a_{m,n} = L$ . Therefore, when the double limit exists and both iterated limits exist, all three agree.

For the double sequence $a_{m,n} = \frac{(-1)^n}{m}$ , prove completely: (i) The double limit $\lim_{m,n\to\infty} a_{m,n} = 0$ exists. (ii) But for each fixed $m$ , $\lim_{n\to\infty} a_{m,n}$ does not exist, so the iterated limit $\lim_{m\to\infty}\lim_{n\to\infty} a_{m,n}$ does not exist. (iii) The other iterated limit $\lim_{n\to\infty}\lim_{m\to\infty} a_{m,n}$ does exist and equals $0$ .

|a_{m,n}| = \frac{1}{m}, \qquad a_{m,2j} = \frac{1}{m}, \qquad a_{m,2j+1} = -\frac{1}{m}

a_{m,n} = \frac{(-1)^n}{m}

: each row oscillates between

\pm\frac{1}{m}

forever, but the amplitude itself dies to

0

as the row index grows.

Show proof

(i) For all $m, n$ , $|a_{m,n} - 0| = \frac{|(-1)^n|}{m} = \frac{1}{m}$ — no dependence on $n$ at all. Given $\varepsilon > 0$ , take $N > \frac{1}{\varepsilon}$ . If $m > N$ (whatever $n$ is), then $|a_{m,n}| = \frac{1}{m} < \frac{1}{N} < \varepsilon$ . Hence $a_{m,n} \to 0$ in the Pringsheim sense.
(ii) Fix $m$ . At even $n = 2j$ , $a_{m,n} = \frac{1}{m}$ ; at odd $n = 2j+1$ , $a_{m,n} = -\frac{1}{m}$ . Suppose a limit $b = \lim_{n\to\infty} a_{m,n}$ existed. For $\varepsilon_0 = \frac{1}{2m} > 0$ there is $N_0$ with $|a_{m,n} - b| < \varepsilon_0$ for all $n > N_0$ . Choosing an even $n > N_0$ gives $\left|\frac{1}{m} - b\right| < \frac{1}{2m}$ ; choosing an odd $n > N_0$ gives $\left|-\frac{1}{m} - b\right| < \frac{1}{2m}$ . By the triangle inequality, $\frac{2}{m} = \left|\left(\frac{1}{m} - b\right) + \left(b + \frac{1}{m}\right)\right| \le \left|\frac{1}{m} - b\right| + \left|-\frac{1}{m} - b\right| < \frac{1}{2m} + \frac{1}{2m} = \frac{1}{m}$ — the false inequality $\frac{2}{m} < \frac{1}{m}$ . This contradiction shows $\lim_{n\to\infty} a_{m,n}$ does not exist. Since this holds for every $m$ , the iterated limit $\lim_{m\to\infty}\lim_{n\to\infty} a_{m,n}$ does not exist.
(iii) For fixed $n$ , $|a_{m,n}| = \frac{1}{m} \to 0$ as $m \to \infty$ , so $\lim_{m\to\infty} a_{m,n} = 0$ for every $n$ . Hence $\lim_{n\to\infty}\lim_{m\to\infty} a_{m,n} = \lim_{n\to\infty} 0 = 0$ , matching the double limit in (i). There is no conflict with the theorem of Problem 2: it only speaks about an order in which the inner limits exist, and here they exist in just one order.