# Thread: Is properly divergent series still properly divergent under rearrangement?

1. ## Is properly divergent series still properly divergent under rearrangement?

For series $\sum\limits_{n = 1}^\infty a_n$, it is said to be properly divergent if its partial sum $\sum\limits_{i = 1}^n a_i$ approaches $+\infty$ (or $-\infty$) as $n\to \infty$. Then, is it true that any properly divergent series is still properly divergent under arbitrary rearrangement? Could you please given a proof if it is true or a counterexample if it does not always hold? Thanks!

2. Originally Posted by zzzhhh
For series $\sum\limits_{n = 1}^\infty a_n$, it is said to be properly divergent if its partial sum $\sum\limits_{i = 1}^n a_i$ approaches $+\infty$ (or $-\infty$) as $n\to \infty$. Then, is it true that any properly divergent series is still properly divergent under arbitrary rearrangement? Could you please given a proof if it is true or a counterexample if it does not always hold? Thanks!
Let's consider the two series...

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

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

... each of them can be derived from the other rearranging the terms. The series (1) is convergent according to the Leibnitz's rule. Now we indicate with $\sigma_{2n}$ the sum of the first 2n terms of (1) and with $\sigma_{3n}$ the sum of the first 3n of (2), so that is...

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

... and from (3) we derive that is...

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

But is...

$\displaystyle \lim_{n \rightarrow \infty} \frac{n}{\sqrt{4n-1}} = \infty$ (5)

... so that the series (2) diverges...

Kind regards

$\chi$ $\sigma$

3. An ingenious answer! thank you very much, chisigma! :-)
This also shows that the countable additivity condition of signed measure is really a stringent requirement.

4. Although the question has been solved, I'd like to comment that for any $S\in [-\infty,+\infty]$ and for any conditionally conergent series there exists an arrangement whose sum is $S$ (Riemann's arrangement theorem) .

Fernando Revilla