# Regrouping and rearranging series

• Mar 8th 2010, 09:20 PM
zxcv
Regrouping and rearranging series
Hi,

Can someone give me an example showing that a rearrangement of a divergent series that may diverge and converge? And also help me by explaining why the original series diverges and the re-arranged one converges (or diverges) because i am not too sure about this part of the theory and it would be easier if I can see an example with explanation.
Thank you very much.
• Mar 9th 2010, 03:43 AM
HallsofIvy
This is only possible for a conditionally convergent series- that is, a series having both positive and negative terms such that $\displaystyle \sum a_n$ converges but $\displaystyle \sum |a_n|$ does not converge.

We start by separating positive and negative "parts". Specifically, we define $\displaystyle b_n= a_n$ if $\displaystyle a_n\ge 0$, $\displaystyle b_n= 0$ if $\displaystyle a_n< 0$ and then define $\displaystyle c_n= -a_n$ if $\displaystyle a_n\le 0$, $\displaystyle c_n= 0$ if $\displaystyle a_n> 0$. Note that both $\displaystyle b_n$ and $\displaystyle c_n$ are positive sequences.

It is then easy to see that $\displaystyle a_n= b_n- c_n$ and $\displaystyle |a_n|= b_n+ c_n$. (There isn't really a sum there- for all n one of the terms in each of those is 0.)

Now, what about the convergence of $\displaystyle \sum b_n$ and $\displaystyle \sum c_n$? If both series converged, then, since $\displaystyle |a_n|= b_n- c_n$, we would have $\displaystyle \sum |a_n|= \sum b+n+ \sum c_n$ converges which is not true so the cannot both converge.

Suppose one converged and the other didn't? If $\displaystyle \sum b_n$ converges we have that $\displaystyle c_n= b_n- a_n$ so $\displaystyle \sum c_n= \sum b_n- \sum a_n$ converges- but we already know they cannot both converge. Similarly, if we assume $\displaystyle ]sum c_n$ converges, we can show that $\displaystyle \sum b_n$ also converges, which cannot happen.

That is, both $\displaystyle \sum b_n$ and $\displaystyle \sum a_n$ must diverge and, since they are both series of positive numbers must diverge to positive infinity- and that means they are unbounded.

Given any integer M, there is some $\displaystyle N_1$ such that $\displaystyle \sum_{n=1}^N_1 b_n> M$. We can also find $\displaystyle N_2$ such that $\displaystyle M/2- 1< \sum_{n=1}^N_2< M/2$. The sum of those two sums is between M/2- 1 and M/2. But if we remove a finite number of terms from a divergent series, the terms left still diverge. We can do this again. And after that, still again, finding a finite number of terms from $\displaystyle b_n$ and $\displaystyle c_n$, that is, terms from $\displaystyle a_n$ that always converge to a number between M/2-1 and M/2. But that means that we can order all of the numbers in $\displaystyle a_n$ so that the sum becomes larger and larger- it diverges.

In fact, doing basically that same process, we can rearrange the terms in the sequence to converge to any number at all!
• Mar 9th 2010, 05:08 AM
chisigma
An example of a convergent series that diverges rearranging the terms. Let consider the series...

$\displaystyle 1 - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \dots + \frac{1}{\sqrt{2n-1}} - \frac{1}{\sqrt{2n}} + \dots$ (1)

... that for the Leibnitz cvriterion converges, and the series...

$\displaystyle 1 + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{7}} - \frac{1}{\sqrt{4}} + \dots + \frac{1}{\sqrt{4n-3}} + \frac{1}{\sqrt{4n-1}} - \frac{1}{\sqrt{2n}} + \dots$ (2)

... that is the (1) after rearranging of the terms. If we indicate with $\displaystyle S_{2n}$ the sum of the first 2n terms of (1) and with $\displaystyle S_{3n}$ the sum of the first 3n terms of (2) we have...

$\displaystyle S_{3n} = S_{2n} + \frac{1}{\sqrt{2n+1}} + \frac{1}{\sqrt{2n+3}} + \dots + \frac{1}{\sqrt{4n-1}} > S_{2n} + \frac{n}{\sqrt{4n-1}}$ (3)

... and because is...

$\displaystyle \lim_{n \rightarrow \infty} \frac{n}{\sqrt{4n-1}} = \infty$ (4)

... we conclude that the series (2) diverges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Mar 9th 2010, 05:43 PM
zxcv
Since I was questionning about a rearrangement of a divergent series which may converge or diverge, can the following example satisfy the condition?

an= 1/n if n is odd
an= 1/(n^2) if n is even
(-1)^n-1an (where n is subscript of a) ---> divergent series
Would this be correct?
• Mar 9th 2010, 10:08 PM
chisigma
Quote:

Originally Posted by zxcv
Since I was questionning about a rearrangement of a divergent series which may converge or diverge, can the following example satisfy the condition?

an= 1/n if n is odd
an= 1/(n^2) if n is even
(-1)^n-1an (where n is subscript of a) ---> divergent series
Would this be correct?

The series of Your example diverges... but what is the rearrangement of the terms in order to have a convergent series?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Mar 9th 2010, 11:03 PM
chisigma
Some years ago I was involved in an unresolved question. Let consider the convergent alternating terms series...

$\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \ln 2$ (1)

That is a classical conditionally convergent series whose sum is different if we rearrange the terms, as example in that way...

$\displaystyle 1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} + \dots =$

$\displaystyle = \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} + \dots =$

$\displaystyle = \frac{1}{2}\cdot (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots) = \frac{1}{2}\cdot \ln 2$ (2)

Very well!... the discussion occurred after the solution I found to the following integral...

$\displaystyle \int_{0}^{\infty} e^{-x}\cdot \frac{1 - \cos x}{x}\cdot dx$ (3)

My solution: because is...

$\displaystyle \frac{1-\cos x}{x} = \sum_{n=1}^{\infty} (-1)^{n-1}\cdot \frac{x^{2n-1}}{(2n)!}$ (4)

... and...

$\displaystyle \int_{0}^{\infty} x^{n}\cdot e^{-x}\cdot dx = n!$ (5)

... we conclude that...

$\displaystyle \int_{0}^{\infty} e^{-x}\cdot \frac{1 - \cos x}{x}\cdot dx= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2n} = \frac{1}{2}\cdot \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = \frac{\ln 2}{2}$ (5)

Succesive numerical computation extablished that the result found in (5) is correct... so that's all right!... not properly because, as I had care of point out, though the series (4) is inconditionally convergent and its sum doesn't change rearranging in any way the order of terms, the same is not true for the series in (5)... at the time no rigorous clarification for that was proposed and the question was left unresolved (Thinking)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$