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. $\blacksquare$