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:
\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:
\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:
\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:
\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:
\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:
\begin{align} \quad N^* \leq f(t) \quad (**) \end{align}
- 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:
\begin{align} \quad N^* \leq g(t) \quad (***) \end{align}
- Let $N = \max \{ N_1, N_2 \}$. Then $f(t), g(t) \geq N^*$, and so by $(*)$ we have that:
\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.