# Thread: Regrouping and rearranging series

1. ## 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.

2. This is only possible for a conditionally convergent series- that is, a series having both positive and negative terms such that $\sum a_n$ converges but $\sum |a_n|$ does not converge.

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

It is then easy to see that $a_n= b_n- c_n$ and $|a_n|= b_n+ c_n$. (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 $\sum b_n$ and $\sum c_n$? If both series converged, then, since $|a_n|= b_n- c_n$, we would have $\sum |a_n|= \sum b+n+ \sum c_n$ converges which is not true so the cannot both converge.

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

That is, both $\sum b_n$ and $\sum a_n$ 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 $N_1$ such that $\sum_{n=1}^N_1 b_n> M$. We can also find $N_2$ such that $M/2- 1< \sum_{n=1}^N_2< M/2$. 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 $b_n$ and $c_n$, that is, terms from $a_n$ 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 $a_n$ 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!

3. An example of a convergent series that diverges rearranging the terms. Let consider the series...

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

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

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

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

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

... and because is...

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

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

Kind regards

$\chi$ $\sigma$

4. 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?

5. 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

$\chi$ $\sigma$

6. Some years ago I was involved in an unresolved question. Let consider the convergent alternating terms series...

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

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

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

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

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

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

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

My solution: because is...

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

... and...

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

... we conclude that...

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

Kind regards

$\chi$ $\sigma$