Absolute and Conditional Convergence of Double Series of Real Nums.

# Absolute and Conditional Convergence of Double Series of Real Numbers

Recall from the Absolute and Conditional Convergence of Series of Real Numbers page that a series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ is said to be absolutely convergent if $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ converges, and conditionally convergent if $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges and $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ diverges.

We will now impose an analogous definition to double series of real numbers.

 Definition: A double series $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ is said to be Absolutely Convergent if $\displaystyle{\sum_{m,n=1}^{\infty} \mid a_{mn} \mid}$ converges. If $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ converges but $\displaystyle{\sum_{m,n=1}^{\infty} \mid a_{mn} \mid}$ diverges then we said that $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ is Conditionally Convergent.

Like with regular series, any strictly positive double series that converges is absolutely convergent.

 Theorem 1: If a double series $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ converges absolutely then $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ converges.
• Proof: Suppose that $\displaystyle{\sum_{m,n=1}^{\infty} a_{mn}}$ converges absolutely. We first note that for all $m, n \in \mathbb{N}$ that:
(1)
\begin{align} \quad - \mid a_{mn} \mid \leq a_{mn} \leq \mid a_{mn} \mid \end{align}
• Therefore we have that:
(2)
\begin{align} \quad 0 \leq a_{mn} + \mid a_{mn} \mid \leq 2 \mid a_{mn} \mid \end{align}
• Taking the double sum from $m, n = 1$ to $\infty$ of all sides gives us that:
(3)
\begin{align} \quad 0 \leq \sum_{m,n=1}^{\infty} [a_{mn} + \mid a_{mn} \mid] \leq 2 \sum_{m,n=1}^{\infty} \mid a_{mn} \mid \end{align}
• By comparison we must have that the double series $\displaystyle{\sum_{m,n=1}^{\infty} [a_{mn} + \mid a_{mn} \mid]}$ converges, and so:
(4)
\begin{align} \quad \sum_{m,n=1}^{\infty} [a_{mn} + \mid a_{mn} \mid] - \sum_{m,n=1}^{\infty} \mid a_{mn} \mid = \sum_{m,n}^{\infty} a_{mn} \end{align}
• Therefore, $\displaystyle{\sum_{m,n}^{\infty} a_{mn}}$ is the sum of two convergent double series and is hence convergent. $\blacksquare$