Double Limits and Iterated Limits of Double Seqs. of Real Numbers

Double Limits and Iterated Limits of Double Sequences of Real Numbers

Recall from the Double Sequences of Real Numbers page that a function $f : \mathbb{N} \times \mathbb{N} \to \mathbb{R}$ is called a double sequence denoted $(a_{mn})_{m, n=1}^{\infty}$ where $f(m, n) = a_{mn}$ for all $m, n \in \mathbb{N}$.

Like with regular sequences of real numbers, we can define when a double sequence converges or not. We first define a double limit of a double sequence.

Definition: The double sequence $(a_{m, n})_{m, n = 1}^{\infty}$ is said to Converge to the real number $A \in \mathbb{R}$ if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\mid a_{m, n} - A \mid < \epsilon$ and we say $A$ is the Double Limit of this double sequence written $\displaystyle{\lim_{m, n \to \infty} a_{m, n} = A}$. If no such $A \in \mathbb{R}$ satisfies this, then we say that the the double sequence $(a_{m, n})_{m, n = 1}^{\infty}$ diverges.

In general, it may be quite a bit more cumbersome to show that a double sequence converges to a particular limit $A$, and even if a double sequence does converge, it may be difficult to guess the value $A$ that it converges to.

Before we move on, we will also need to describe another type of limit of a double sequence known as an iterated limit.

Definition: Let $(a_{m, n})_{m, n=1}^{\infty}$ be a double sequence. Then the Iterated Limits of this double sequence are defined as $\displaystyle{\lim_{m \to \infty} \left ( \lim_{n \to \infty} a_{m, n} \right )}$ and $\displaystyle{\lim_{n \to \infty} \left ( \lim_{m \to \infty} a_{m, n} \right )}$.

It is very important to note the distinction between the double limit and the iterated limits of a double sequence. In general, these three limits may be different.

For example, consider the following double sequence:

\begin{align} \quad (a_{m,n})_{m, n=1}^{\infty} = \left ( \frac{m^2 + 2n^2}{n^2} \right )_{m, n =1}^{\infty} \end{align}

The two iterated limits for this double sequence are:

\begin{align} \quad \lim_{m \to \infty} \left ( \lim_{n \to \infty} \frac{m^2 + 2n^2}{n^2} \right ) = \lim_{m \to \infty} \left ( \lim_{n \to \infty} \frac{m^2}{n^2} + 2 \right ) = \lim_{m \to \infty} 2 = 0 \end{align}
\begin{align} \quad \lim_{n \to \infty} \left ( \lim_{m \to \infty} \frac{m^2 + 2n^2}{n^2} \right ) = \infty \end{align}
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