Cauchy convergence test for double series

Question

I've known that the original Cauchy convergence test for complex number series. I'm now wondering that if we can generate this critierion to double series.

More formally, consider the double series $\sum_{i,k=1}^{\infty} a_i^{(k)}$. We say this series converges to the complex number $A$ iff $\forall \epsilon >0\exists N\in \mathbb{N}\forall I\ge N\forall K\ge N (|\sum_{i=1}^I\sum_{k=1}^K a_i^{(k)}-A|<\epsilon$.

So what's the generated Cauchy convergence test should be now? My original idea is:

$\forall \epsilon >0\exists N\forall I,K,S,T\ge N (|\sum_{i=1}^I\sum_{k-1}^ka_i^{(k)}-\sum_{s=1}^S\sum_{t=1}^{T}a_s^{(t)}|<\epsilon$ (1)

But I have difficulty in proving this criterion can imply convergence; moreover, I've read Notes on mathematical analysis which states the Cauchy test should be like:

$\forall \epsilon >0\exists N ((\max(p,q)\ge N)\wedge (P>p)\wedge (Q>q)\Rightarrow |\sum_{p\le i\le P,q\le k\le Q} a_i^{(k)}|<\epsilon $ (2)

I've noticed that (2) is weaker than (1) (if write $a_i^{(k)}$ in matrix, (1) is a bigger rectangle minus a smaller one, while (2) just consider the rectangle "diagonally" to the smaller one), but I still can't prove this implies convergence.

In addition, if (2) is the actual criterion, then there should be a counterexample that (1) is true but the original sequence doesn't convergent. What this example can be?

Thanks in advance for any help.

Answer 1

Your condition (2) is what is known as the Cauchy criterion for a double series. I will prove that this condition is sufficient for convergence.

Consider the double series $\sum_i\sum_k a_{ik}.$ You correctly stated the so-called Pringsheim definition of convergence:

The double series converges to $A$ if, given the double sequence of partial sums

$$S_{mn}=\sum_{i=1}^m\sum_{k=1}^n a_{ik},$$

we have for any $\epsilon >0$ a positive integer $N$ such that if $ m,n \geqslant N$ then $|S_{mn} - A| < \epsilon.$

Suppose that the Cauchy criterion is satisfied. For any $\epsilon > 0$ there exists a positive integer $M$ such that if $p \geqslant m \geqslant M$ and $q \geqslant n \geqslant M,$ then we have $|S_{pq}-S_{mn}| < \epsilon.$

If $m =n$ and $p = q,$ then the sequence $T_n = S_{nn}$ satisfies the usual Cauchy criterion for single sequences, where $|T_q - T_n| < \epsilon$ when $q \geqslant n \geqslant N$. Thus the sequence $T_n$ converges to a limit $A$.

Now we can show that the double series also converges to $A$.

By convergence of $T_n$, there exists a positive integer $N_1$ such that if $n \geqslant N_1$ then $|S_{nn} - A| < \epsilon/2.$

By the Cauchy criterion for double series, there exists a positive integer $N_2$ such that $p \geqslant m \geqslant N_2$ and $q \geqslant n \geqslant N_2$ implies $|S_{pq} - S_{mn}| < \epsilon/2.$ In particular, with $m = n$ we have $|S_{pq} - S_{nn}| < \epsilon/2.$

Hence, if $p,q \geqslant \max(N_1,N_2)$ we have

$$|S_{pq} - A| \leqslant |S_{pq} - S_{nn}| + | S_{nn} - A| < \epsilon.$$