Rearrangement of conditional convergent series

Question

Given - 'For given conditional convergent series, it can converges to any number R with suitable rearrangement'

So how can we say that series is convergent, as it can be made to converge to any possible number? Doesn't that defeat purpose of convergence?

Also, how can we prove that it converges to any number R (say 1 or 10) in general case without any specific series?

Answer 1

To answer your comment first: if you have a finite set of numbers that you wish to sum, then the order doesn't matter. You will always get the same answer however you arrange them. However, for infinite sets of numbers things can get tricky. Consider, please:

$$ 1-x+x^2-x^3+\cdots = \frac{1}{1+x} $$ and $$ 1-x+x^3-x^4+x^6 - \cdots = \frac{1}{1+x+x^2} $$ and $$ 1-x^2+x^3-x^5+x^6 - x^8 +\cdots = \frac{1+x}{1+x+x^2}$$ and now set $x=1$ in each one. The LHS for each sum is $$1 -1 +1 -1 +1 -1 + \cdots $$ and yet the right-hand side tells us that this (seemingly identical) series has three possible sums: $1/2$, $1/3$ and $2/3$.

These series are divergent, yet have three different answers depending on you believe the series is generated. The same thing happens with conditionally convergent series: by rearranging the series you can obtain same set of numbers, but the series sum to different values.

Classically, the alternating series is given by $$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \sum_{k=1}^\infty \frac{1}{k}$$ Rearrange the terms: $$ 1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} + \frac{1}{6} - \cdots = \sum_{k=1}^\infty \left( \frac{1}{2k-1} - \frac{1}{2(2k-1)} - \frac{1}{4k} \right) $$ The first series sums to $\ln 2$ and the second to... $\frac{1}{2}\ln 2$.

Answer 2

To define what a series is, we must first define what a sequence $(A_n)_{n\in \Bbb N}$ is.

A sequence $(A_n)_{n\in \Bbb N}$ is a function $f$ with domain $\Bbb N$ such that $f(n)=A_n$ for each $n\in \Bbb N.$ So $(A_n)_{n\in \Bbb N}=(f(n))_{n\in \Bbb N}.$

And we define $\sum_{n=1}^{\infty}A_n=S$ to mean that $\lim_{n\to \infty}g(n)=S,$ where $g(n)=\sum_{j=1}^nA_j $for each $n\in \Bbb N.$

" Re-arranging the series" means taking a bijection $\psi:\Bbb N\to \Bbb N$ and defining a new sequence $(A_{\psi (n)})_{n\in \Bbb N}=(f(\psi(n)))_{n\in \Bbb N}.$

And then $\sum_{n=1}^{\infty}A_{\psi(n)}=T$ means that $\lim_{n\to \infty}h(n)=T,$ where $h(n)=\sum_{j=1}^nA_{\psi(j)}$ for each $n\in \Bbb N.$

The sequence $(A_n)_{n\in \Bbb N}$ is usually NOT the same sequence as $(A_{\psi(n)})_{n\in \Bbb N}.$

Note that a sequence is NOT the same thing as the set of all the terms of the sequence.

And note that unless $A_{\phi (n)}=A_n$ for every $n$ then we do NOT say that $\sum_n A_n$ and $\sum_n A_{\phi(n)}$ are the same series. That is the re-arranged series is generally a $different$ series.