Is there an ambiguity on $\sum\limits^{\infty}_{i=0}a_i$?

Question

Sorry if this is basic but I saw the theorem that says if $\sum\limits^{\infty}_{i=0}a_i$ is conditionally convergent then this series can be any number up to rearrangement. Then does it imply that $\sum\limits^{\infty}_{i=0}a_i$ can be any number? How about $\sum\limits^{\infty}_{i=-\infty}a_i$

Answer 1

$\sum\limits_{i=0}^{\infty} a_i$ means one thing. It means the limit of the sequence $\lim\limits_{n \to \infty} S_n$, where

$$S_n = a_0 + a_1 + a_2 + \cdots + a_n.$$

However, $$\sum\limits_{i=-\infty}^{\infty}a_i$$

is ambiguous when it is only conditionally convergent. It is not obvious in what order the terms are meant to be added here, and as you point out adding the terms in a different ways produces different results.

Answer 2

No, it cannot be any number. If the limit $\lim_{n\to\infty}\sum_{i=0}^na_i$ exists and it is a real number $a$, then (and only then)$$\sum_{i=0}^\infty a_i=a.$$So, since any sequence of real numbers has (at most) one limit, the expression $\sum_{i=0}^\infty a_i$ can be at most one number.

What you were told about rearrangements is that if $\sum_{i=0}^\infty a_i$ is conditionally convergent, and if $a\in\Bbb R$, then there is some bijection $b\colon\Bbb Z^+\longrightarrow\Bbb Z^+$ such that$$\sum_{i=0}^\infty a_{b(i)}=a.$$

Answer 3

To answer the second question:

Indeed, the issue with conditional convergence is more pronounced when trying to sum $\sum_{i=-\infty}^\infty a_i$, for example $F(M,N)=\sum_{i=-M}^N\frac{1}{2i+1}$ will tend to different limits or not converge depending on which path of $\lim_{(M,N)\to(\infty,\infty)}$ you're taking.

However the ambituity goes away when both directional partial sums' limits $\lim_{N\to\infty}\sum_{i=0}^N a_i$ and $\lim_{M\to\infty}\sum_{i=-M}^0 a_i$ exist, in which case $$\sum_{i=-\infty}^\infty a_i=\lim_{M\to\infty}\sum_{i=-M}^0 a_i+\lim_{N\to\infty}\sum_{i=1}^N a_i=\lim_{(M,N)\to(\infty,\infty)}\sum_{i=-M}^N a_i$$ will not depend on how $(M,N)\to(\infty,\infty)$.

In some other cases when $a_i\to 0$ as $i\to\pm\infty$, you may see the series $\sum_{i=-\infty}^\infty a_i$ being implicitly summed as $a_0+\sum_{i=1}^\infty \left(a_i+a_{-i}\right)$, even when it's conditionally convergent. For example recall the pole exansion series $$\pi\cot(\pi x)=\sum_{n=-\infty}^\infty\frac{1}{x+n}:=\lim_{N\to\infty}\sum_{n=-N}^N\frac{1}{x+n}$$