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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?

tatha
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    The series converges to one number (not "any possible number"). A rearrangement of the series will be a different series, which may converge to a different number. – Angina Seng Aug 29 '18 at 19:18
  • Hmmm... it can be argued, but numbers are just same and since it is addition so why should order matter? – tatha Aug 29 '18 at 19:20
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    It's not just addition. There is a limit being taken. – Paul Aug 29 '18 at 19:29
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    Be very careful. You are falling into the trap of assuming that things you know to work for the finite case should also work for the infinite case. In sums of finitely many terms you may rearrange them with no consequence (as per the commutative property). This is no longer true when we talk about infinite sums! (as shown explicitly by exactly the theorem you are referring to) – JMoravitz Aug 29 '18 at 19:30
  • Related. – JMoravitz Aug 29 '18 at 19:34
  • Thanks for the link, I see how infinity changes everything. Also I just read that such series is $ \infty - \infty$ so order will make a difference. I guess it that makes sense, – tatha Aug 29 '18 at 19:45

2 Answers2

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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... $\frac{1}{2}\ln 2$.

postmortes
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  • Thanks for detailed explanation, but how do you prove that series can converge to R in any general case? – tatha Aug 29 '18 at 19:45
  • It's called the Riemann series theorem, and it's a bit long for me to post here, so take a look at the wikipedia (sorry) page for it. https://en.wikipedia.org/wiki/Riemann_series_theorem – postmortes Aug 29 '18 at 19:48
  • Thanks for the link. Great help. Can you explain the last part of proof which says to repeat the process of getting p terms for $a^+$ and q terms for $a^-$. ? – tatha Aug 29 '18 at 19:57
  • It just means to repeat the process already described for picking the next p terms (positive ones) and then the next q terms (negative ones). What it says is that you can keep picking out the positive and negative terms selectively to keep stepping just over and just under $M$. You can do this because the positive terms tend to $+\infty$ so there are 'enough' of them, and likewise with the negative terms. Try it out with the alternating series above: pick a number and look at how to get there (choose a number near 1, the harmonic series requires 30,000 or so terms to reach 10) – postmortes Aug 29 '18 at 20:07
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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.

DanielWainfleet
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