# Definition of convergence of $\sum_{i=-\infty}^\infty a_i$

This is a really basic question, but I'm unsure about the definition for convergence of $$\sum_{i=-\infty}^\infty a_i$$ The definition $$\sum_{i=-\infty}^\infty a_i=\lim_{n\to \infty} \sum_{i=-n}^n a_i$$ seems too loose to me, but the strongest definition I can think of: $$\sum_{i=-\infty}^\infty a_i=\lim_{\|A\|\to \infty}\sup_{A\subset\mathbb{Z}}\sum_{i\in A}a_i$$ seems too strong. There are also others like $$\sum_{i=-\infty}^\infty a_i=\lim_{n\to\infty}\sum_{i=-n}^{0}a_i+\lim_{n\to\infty}\sum_{i=1}^na_i$$ but this seems a bit arbitrary.

More specifically: if I'm asked to show that a certain series of functions $$\sum_{i=-\infty}^\infty f_i$$ converges, then which sequence of partial sums should I consider?

edit: yet another possibility is to consider these types of series only if they converge absolutely, in which case these considerations are irrelevant.

edit2: The problem with the splitting definition is that it does not seem to work when considering different modes of convergence. For example, if I want to show that $$\sum_{i=-\infty}^\infty f_i$$ converges uniformly, I would need for a given $$\epsilon$$ need to find a $$\delta(\epsilon)$$ s.t. for $$N>n$$ the $$N$$th partial sum close enough to the limit function for all $$x$$ in my domain. However, here we have no well defined $$N$$th partial sum, since we have split our sum in the definition of the limit. So how do we define uniform convergence in this case?

• The last, the splitting in the series with positive indices and the series with negative indices is the usual definition. Just like an improper Riemann integral $\int_{-\infty}^{\infty} f(x)\,dx$ is said to converge if both $\int_0^{\infty} f(x)\,dx$ and $\int_{-\infty}^0 f(x)\,dx$ converge. Commented Mar 6, 2016 at 18:45
• @DanielFischer thank you for your answer. Can you take a look at my edit? Commented Mar 6, 2016 at 18:55
• You can phrase the splitting definition as "for all $\varepsilon > 0$ there are $M,N \in \mathbb{N}$ such that for all $m \geqslant M$ and $n \geqslant N$ one has $\biggl\lvert \sum\limits_{i = -m}^n a_i - S\biggr\rvert < \varepsilon$". It's easy to get the formulation for uniform convergence from that. Of course absolute convergence is most convenient, but sometimes one also needs to deal with conditional convergence. Commented Mar 6, 2016 at 19:02
• @DanielFischer: Those comments look like an answer to me. Commented Mar 6, 2016 at 20:09
• @joriki Yep, but I haven't yet found the time to write a full answer. Unless something intervenes, I'm going to start writing soon, however. Commented Mar 6, 2016 at 20:14

The usual definition of convergence for doubly infinite series or $\mathbb{Z}$-indexed series is that

$$\sum_{i = -\infty}^{\infty} a_i\tag{1}$$

is defined as convergent if the series

$$\sum_{i = 0}^{\infty} a_i\quad \text{and}\quad \sum_{k = 1}^{\infty} a_{-k}$$

both converge, and the value of the doubly infinite series is the sum of the values of these two series. For doubly infinite series of functions one then has uniform convergence if the series with nonnegative indices and the series with negative indices both converge uniformly.

We can also formulate the criterion without splitting the series, using the product partial order on $\mathbb{N}^2$ to make it a directed set, and say that $(1)$ converges if

$$\lim_{(m,n) \to (\infty,\infty)} \sum_{i = -m}^n a_i\tag{2}$$

exists. For a doubly infinite series of functions, uniform convergence again means uniform convergence of the net

$$S_{m,n} := \sum_{i = -m}^n a_i.$$

There is an exception, however. For Fourier series

$$\sum_{n = -\infty}^{\infty} c_n e^{inx},$$

when one is interested in pointwise (or uniform) convergence, one usually only considers the symmetric partial sums

$$\sum_{n = -N}^N c_n e^{inx}$$

and calls the Fourier series convergent (at $x$) if the limit of the symmetric partial sums at $x$ exists.

The convergence of a doubly infinite series as defined above evidently implies the convergence of the sequence of symmetric partial sums, but the symmetric partial sums can converge when the doubly infinite series doesn't converge, the limit of the symmetric partial sums is then (often) called the principal value of the divergent doubly infinite series. This is all analogous to the situation of improper Riemann integrals.

If all terms are in $\mathbb R$, the we can proceed as follows. Let \begin{align} I & = \{i : a_i\ge0\}, \\ J & = \{i : a_i < 0\}. \end{align} Then let \begin{align} \sum_{i\in I} a_i & = \sup\left\{ \sum_{i\in I_0} a_i : I_0 \text{ is a finite subset of }I \right\} \tag 1 \\ \sum_{i\in J} a_i & = -\sup\left\{ \sum_{i\in I_0} -a_i : I_0 \text{ is a finite subset of }J \right\} \tag 2 \end{align} and finally $$\sum_{i=-\infty}^\infty a_i = \sum_{i\in I} a_i + \sum_{i\in J} a_i. \tag 3$$ This defines the sum except when both $(1)$ and $(2)$ are infinite. The series converges absolutely if both are finite.

One can write $$\sum_{i=-\infty}^\infty a_i = \lim_{n\to\infty} \sum_{i=-n}^n a_i \tag 4$$ and in some cases that converges even when $(1)$ and $(2)$ are infinite. But in that case "rearrangements" of the sum, such as $$\lim_{n\to\infty} \sum_{i=-n}^{2n} a_i$$ sometimes have values differing from that of $(4)$. However, $(4)$ agrees with $(3)$ in all cases where at least one of $(1)$ and $(2)$ is finite.

If $a_i\in\mathbb C$, we can apply all of the above to real and imaginary parts separately in every term, getting the real and imaginary parts of the entire sum.