How do we define the sum of a bi-infinite series?

Question

If we have a (usual) series of the form $S = \sum_{n=1}^{\infty} a_n$ then we define $S$ to be the limit as $N$ goes to infinity of the partial sums $S_N = \sum_{n=1}^{N} a_n,$ provided the limit exists. If we instead have a bi-infinite series of the form $$ S = \sum_{n=-\infty}^{\infty} a_n $$ then how do we define this sum? Is it $$ \lim_{N \to \infty} \sum_{n=-N}^N a_n $$ or $$ \lim_{N \to \infty} \lim_{M \to \infty} \sum_{n = -M}^N a_n $$ or something else? Can you also refer me to any standard textbook that deals with bi-infinite series? Thanks!

Answer 1

Bi-infinite summation is summation over $\Bbb Z$. In general (I think that I read about this in Professor Tao's Analysis book), one defines a sum over an arbitrary set $X$ as follows.

Let $X$ be any set and $F(X)$ be the set of all finite subsets of $X$. Let $f: X \to \Bbb R$ be nonnegative.

$$\sum_{x \in X} f(x) := \sup\{\sum_{x \in A} f(x): A \in F(X)\}$$

Now if $f$ is not necessarily nonnegative, then in case $\sum_{x \in X} |f(x)|$ is finite,

$$\sum_{x \in X} f(x) := \sum_{x \in X} f^+(x) - \sum_{x \in X} f^-(x)$$

Where $f^+(x)= \max\{f(x), 0\}$ and $f^-(x) = \max\{-f(x), 0\}$.

Otherwise $\sum_{x \in X} f(x)$ is undefined.

In the case $X = \Bbb Z$, it's easy to check that when $\sum_{x \in \Bbb Z} f(x)$ is defined, we have that the limit:

$$\lim_{N \to \infty} \sum_{n=-N}^N f(n)$$

exists and equals that thing.