Double sequence, two sequences converge, but to different limits?

Question

Let $(a_{m, n})_{m, n}$ be a "double sequence" of real numbers; that is, for every pair $(m, n) \in \mathbb{N} \times \mathbb{N}$, $a_{m, n}$ is a real number. Suppose that, for every $m$, $(a_{m, n})_n$ converges to a limit $x_m$, and that for every $n$ the sequence $(a_{m, n})_m$ converges to a limit $y_n$. What is an example in which $(x_m)_m$ and $(y_n)_n$ both converge, but converge to different limits, i.e.$$\lim_{m \to \infty} \lim_{n \to \infty} a_{m, n} \neq \lim_{n \to \infty} \lim_{m \to \infty} a_{m, n}?$$

Answer 1

A very simple example:

$$a_{m,n}=\begin{cases} 1,&\text{if }m\le n\\ 0,&\text{if }m>n\;. \end{cases}$$

If you write out the double sequence as an infinite array, it’s very easy to see what happens:

$$\begin{array}{ccc} 1&1&1&1&1&\ldots&\to&1\\ 0&1&1&1&1&\ldots&\to&1\\ 0&0&1&1&1&\ldots&\to&1\\ 0&0&0&1&1&\ldots&\to&1\\ 0&0&0&0&1&\ldots&\to&1\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\ \downarrow&\downarrow&\downarrow&\downarrow&\downarrow&&&\vdots\\ 0&0&0&0&0&\ldots \end{array}$$

Answer 2

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

One limit is 0 another is 1.