Divergence Tests for Double Sequences of Real Numbers

# Divergence Tests for Double Sequences of Real Numbers

Recall from the Double Limits and Iterated Limits of Double Sequences of Real Numbers page that a double sequence $(a_{mn})_{m, n=1}^{\infty}$ is said to converge and have a double limit $A \in \mathbb{R}$ if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:

(1)
\begin{align} \quad \mid a_{m, n} - A \mid < \epsilon \end{align}

In such cases we write $\lim_{m, n \to \infty} a_{m,n} = A$.

Furthermore, we defined another type of limit for a double sequence called the iterated limits of the double sequence which are defined to be:

(2)
\begin{align} \quad \lim_{m \to \infty} \left ( \lim_{n \to \infty} a_{mn} \right ) \end{align}
(3)
\begin{align} \quad \lim_{n \to \infty} \left ( \lim_{m \to \infty} a_{mn} \right ) \end{align}

We will now look at some criteria that can tell us if a double sequence diverges.

 Theorem 1: Let $(a_{mn})_{m, n=1}^{\infty}$ be a double sequence of real numbers that converges to $A \in \mathbb{R}$. Then: a) If $\displaystyle{\lim_{n \to \infty} a_{mn}}$ exists for every $m \in \mathbb{N}$ then we have that $\displaystyle{\lim_{m \to \infty} \left ( \lim_{n \to \infty} a_{mn} \right ) = A}$. b) If $\displaystyle{\lim_{m \to \infty} a_{mn}}$ exists for every $n \in \mathbb{N}$ then we have that $\displaystyle{\lim_{n \to \infty} \left ( \lim_{m \to \infty} a_{mn} \right ) = A}$.
• Proof of a) Let $\epsilon > 0$ be given. Define a function $F : \mathbb{N} \to \mathbb{R}$ by:
(4)
\begin{align} \quad F(m) = \lim_{n \to \infty} a_{mn} \end{align}
• Then $F$ is well-defined since $\displaystyle{\lim_{n \to \infty} a_{mn}}$ exists for every $m \in \mathbb{N}$. We wish to show that $\displaystyle{\lim_{m \to \infty} F(m) = A}$.
• Since $(a_{mn})_{m, n=1}^{\infty}$ converges to $A$ we have that for all $\epsilon > 0$ there exists an $N^* \in \mathbb{N}$ such that if $m, n \geq N^*$ then:
(5)
\begin{align} \quad \mid a_{mn} - A \mid < \epsilon \quad \Leftrightarrow \quad A - \epsilon < a_{mn} < A + \epsilon \end{align}
• Now take the limit of both sides as $n \to \infty$. Using the Squeeze theorem we see that for all $m \geq N$ that:
(6)
\begin{align} \quad \lim_{n \to \infty} (A - \epsilon) & < \lim_{n \to \infty} a_{mn} < \lim_{n \to \infty} (A + \epsilon) \\ \quad A - \epsilon & < F(m) < A + \epsilon \end{align}
• So for all $m \geq N$ we have that $\mid F(m) - A \mid < \epsilon$ which implies that:
(7)
\begin{align} \quad \lim_{m \to \infty} F(m) = \lim_{m \to \infty} \left ( \lim_{n \to \infty} a_{mn} \right ) = A \quad \blacksquare \end{align}
• Proof of b) Analogous to that of (a).
 Corollary 1: If $\displaystyle{\lim_{m \to \infty} \left ( \lim_{n \to \infty} a_{mn} \right ) = A}$ and $\displaystyle{\lim_{n \to \infty} \left ( \lim_{m \to \infty} a_{mn} \right ) = B}$ where $A \neq B$ then the double sequence $(a_{mn})_{m,n=1}^{\infty}$ diverges.

Corollary 1 is simply the contrapositive of Theorem 1.

 Theorem 2: Let $(a_{mn})_{m,n=1}^{\infty}$ be a double sequence of real numbers that converges to $A \in \mathbb{R}$ and let $f, g : \mathbb{N} \to \mathbb{R}$ be two functions such that $\displaystyle{\lim_{t \to \infty} f(t) = \infty}$ and $\displaystyle{\lim_{t \to \infty} g(t) = \infty}$. Then $\displaystyle{\lim_{t \to \infty} a_{f(t), g(t)} = A}$.
• Proof: Let $\epsilon > 0$. Since $(a_{mn})_{m, n=1}^{\infty}$ converges to $A$ we have that for all $\epsilon > 0$ there exists an $N^* \in \mathbb{N}$ such that if $n \geq N$ then:
(8)
\begin{align} \quad \mid a_{mn} - A \mid < \epsilon \quad (*) \end{align}
• Since $\displaystyle{\lim_{t \to \infty} f(t) = \infty}$ there exists an $N_1 \in \mathbb{N}$ such that if $n \geq N_1$ then:
(9)
• Similarly, since $\displaystyle{\lim_{t \to \infty} g(t) = \infty}$ there exists an $N_2 \in \mathbb{N}$ such that if $n \geq N_2$ then:
(10)
• Let $N = \max \{ N_1, N_2 \}$. Then $f(t), g(t) \geq N^*$, and so by $(*)$ we have that:
(11)
\begin{align} \quad \mid a_{f(t), g(t)} - A \mid < \epsilon \end{align}
• Therefore $\displaystyle{\lim_{t \to \infty} a_{f(t), g(t)} = A}$. $\blacksquare$
 Corollary 2: If there exists functions $f_1, g_1, f_2, g_2 : \mathbb{N} \to \mathbb{R}$ where $\displaystyle{\lim_{t \to \infty} f_1(t) = \infty}$, $\displaystyle{\lim_{t \to \infty} g_1(x) = \infty}$, $\displaystyle{\lim_{t \to \infty} f_2(t) = \infty}$, and $\displaystyle{\lim_{t \to \infty} g_2(x) = \infty}$ and such that $\displaystyle{\lim_{t \to \infty} a_{f_1(t), g_1(t)} = A}$ $\displaystyle{\lim_{t \to \infty} a_{f_2(t), g_2(t)} = B}$ where $A \neq B$ then the double sequence $(a_{mn})_{m,n=1}^{\infty}$ diverges.

Corollary 2 is simply the contrapositive of Theorem 2.