Double sequence limit

[Solunac]

Hi! Let say that we have double sequence $\{a_{mn}\}_{\mathbb{N}\times\mathbb{N}}:$

$a_{11} ~~~a_{12} ~~~\dots ~~~a_{1n} ~~~\dots ~\to \infty\\$
$a_{21} ~~~a_{22} ~~~\dots ~~~a_{2n} ~~~\dots ~\to \infty\\$
$\vdots ~~~~~~~\vdots ~~~~~~\ldots ~~~\vdots ~~~~~~\ldots ~\to \infty\\$
$a_{m1} ~~a_{m2} ~~\dots ~~~a_{mn} ~~\dots ~\to \infty\\$
$\vdots ~~~~~~~\vdots ~~~~~~\ldots ~~~\vdots \\$
$\downarrow ~~~~~~\downarrow ~~~~~\dots ~~~\downarrow ~~~~~\ddots$
$0 ~~~~~~0 ~~~~~\dots ~~~~0 ~~~~~~~~~~~~\textbf{?}\\$

Where every $a_{mn}\neq\infty$ and $a_{mn}\neq 0$.

What can we say about it when $m\to\infty$ and $n\to\infty$, ie. about the $a_{\infty\infty}$? I was wondering if it is always $\neq\infty$? Under what conditions it could be 0 (if it could be)?


Спасибо.

[bot]

Take any sequence $q_i$ and consider $a_{ij}=q_i2^{-i+j}$
So ? may be as you want including nonexisting
Подсмотрел в профиле - топик стартер понимает по-русски.

Домножение на $q_i$ даёт всё, что угодно, но это "что угодно" будет одинаковым для всех диагоналей.
Домножая на подходящие $q_i$ можно эти "что угодно" сделать независимыми друг от друга.

[Профессор Снэйп]

Variant I.

$$ a_{i,j} = \begin{cases} 1/i, &j \leqslant i \\ j, &j > i \end{cases} $$

$a_{\infty,\infty} = 0$ because $a_{i,i}=1/i$ for all $i$.

Variant II.

$$ a_{i,j} = \begin{cases} 1/i, &j < i \\ j, &j \geqslant i \end{cases} $$

$a_{\infty,\infty} = +\infty$ because $a_{i,i}=i$ for all $i$.

Variant III.

$$ a_{i,j} = \begin{cases} 1/i, &j < i \\ -j, &j = i \\ j, &j > i \end{cases} $$

$a_{\infty,\infty} = -\infty$ because $a_{i,i}=-i$ for all $i$.

[Solunac]

Thanks. So, can we always find subsequence, where we pick just one element of a sequence from every coloumn and every row (we don't jump over some columns or rows), wich convergence to $\infty$? No matter how the sequence is defined...