1
$\begingroup$

What is the domain of convergence of the double sum $\sum_{i,j \in \mathbb N}\frac{(i+j)!}{i!j!}z^iw^j$, where $z,w$ are complex numbers? How do you define the partial sum of the double sum really? I cannot find any reference on this matter.

  • 0
    Have you spotted particular values of $z$ and/or $w$ where you can assert this convergence ? – 2017-01-28
  • 0
    The set $\mathbb{N}\times\mathbb{N}$ is countable, so enumerate it as you wish and sum in this order. By definition, the sum converges if the result exists and doesn't depend on this order. – 2017-01-28
  • 0
    The simplest values coming into mind are $B_1(0) \times {0}$ and ${0} \times B_1(0)$. – 2017-01-28
  • 0
    Do you have any reference on this matter? @Wolfram – 2017-01-28
  • 0
    I don't know this theme, but probably you can find some useful information in IVth chapter of Cartan book on complex analysis. (Elementary Theory of Analytic Functions of One or Several Complex Variables) – 2017-01-28

1 Answers 1

1

Write this as

$$\sum_{i=0}^\infty \sum_{j=0}^\infty \frac{(i+j)!}{i!j!}z^iw^j- 1 - \sum_{i=0}^\infty z^i- \sum_{j=0}^\infty w^j,$$

where the single series are easily handled.

Consider the double series. The partial sum is defined as

$$S_{mn}(z,w) =\sum_{i=0}^m \sum_{j=0}^n \frac{(i+j)!}{i!j!}z^iw^j.$$

This converges in the strongest sense to $S(z,w)$ if for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that if $n,m > N$ we have $\left|S_{mn}(z,w) - S(z,w)\right| < \epsilon. $

There are weaker forms of convergence, such as summing by rows, columns, and diagonals. In general, convergence in these cases does not guarantee convergence in the strongest sense.

To sum along diagonals, we consider convergence of partial sums of the form $S_{m}(z,w)$ where

$$S_m(z,w) = \sum_{p=0}^m \sum_{q=0}^p \frac{p!}{q!(p-q)!}z^qw^{p-q}. $$

Using the binomial theorem we get

$$S_m(z,w) = \sum_{p=0}^m (z+w)^p, $$

which converges as $m \to \infty$ if $|z + w| < 1$.

Addendum

We can show that this double series is absolutely convergent in the strongest sense when $|z| + |w| < 1.$ This is, of course, a stronger condition than is necessary for diagonal convergence since $|z + w| \leqslant |z| + |w|$.

Note that

$$\sum_{i=0}^m \sum_{j=0}^n \frac{(i+j)!}{i!j!}|z|^i|w|^j \leqslant \sum_{p=0}^{m+n} \sum_{q=0}^p \frac{p!}{q!(p-q)!}|z|^q|w|^{p-q} = \sum_{p=0}^{m+n} (|z| + |w|)^p. $$

The double series converges as $m,n \to \infty$ if the geometric series on the RHS converges. This is the case when $|z| + |w| < 1$.

  • 0
    How do you conclude that it converges in the strongest sense when $|z|+|w| < 1$ after you subtract those terms? – 2017-01-28
  • 0
    I see, thank you so much, but I I am still interested in how to find its domain of convergence under the strongest sense. Can you help? – 2017-01-28
  • 0
    One can also look at iterated sums $\lim_{m \to \infty} \sum_{i=1}^m (\sum_{j=1}^\infty a_{ij})$. Convergence here is does not always guarantee the double sum convergence in the strict sense, but it does if the terms are nonnegative. I don't know how much you want to go into this topic. – 2017-01-28
  • 0
    I am so sorry, and I have known that there are different kinds of summation from your answer. I am just curious about how to find its domain of convergence under the strongest sense you mention, if you can recall the proof. But if you cannot, I can certainly accept this answer now. – 2017-01-28
  • 0
    @koch: I added a proof of absolute convergence when $|z| + |w| < 1$. – 2017-01-28
  • 0
    Is it divergent if $|z| + |w| \geqslant 1$? – 2017-01-29
  • 0
    Yes it converges absolutely if and only if $|z| +|w| < 1$. To see divergence if $|z| +|w| \geqslant 1$ use the comparison $S_{nn}(z,w) \geqslant \sum_{p=0}^{n} \sum_{q=0}^p \frac{p!}{q!(p-q)!}|z|^q|w|^{p-q} = \sum_{p=0}^{n} (|z| + |w|)^p$. So this particular subsequence of partial sums diverges with the geometric series on the RHS when $|z| + |w| \geqslant 1.$ – 2017-01-29
  • 0
    Sorry for making this question lasting so long, but do we care about conditional convergence? – 2017-01-29
  • 0
    But there can be some form of convergence (but not absolute) as you see with the diagonal sums when $|z + w| < 1 \leqslant |z| + |w|$. – 2017-01-29
  • 0
    I would have to think more about conditional convergence as $m,n \to \infty$ (in what I called the strongest sense also known as the Pringsheim sense) with a complex term. Have you looked for any references? – 2017-01-29