- #1

- 1,462

- 44

- I
- Thread starter Mr Davis 97
- Start date

- #1

- 1,462

- 44

- #2

FactChecker

Science Advisor

Gold Member

- 6,053

- 2,339

- #3

- 1,462

- 44

So this hold only if the the two series are absolutely convergent?

- #4

fresh_42

Mentor

- 14,371

- 11,691

If they do, you can use the algebraic property you've asked for, because the order is irrelevant and we can add the infinite sums.So this hold only if the the two series are absolutely convergent?

Otherwise you need the limit argument:

##(L_{2n}+L_{2n-1})-\varepsilon =L_{2n}-\varepsilon/2 + L_{2n-1}-\varepsilon/2 \leq S_n = S_{2n}+S_{2n-1} \leq L_{2n}+\varepsilon/2 + L_{2n-1}+\varepsilon/2 = (L_{2n}+L_{2n-1})+\varepsilon##

with the partial sums ##S_{*}## and its limits ##L_{*}##. Then ##\lim_{n \to \infty}S_n= L_{2n}+L_{2n-1}##.

- #5

WWGD

Science Advisor

Gold Member

- 5,421

- 3,685

A series can be seen as a sequence of partial sums.

- #6

FactChecker

Science Advisor

Gold Member

- 6,053

- 2,339

If any one of them is NOT absolutely convergent, then either the sum of the negative tems is ##-\infty## or the sum of the positive terms is ##+\infty##, or both. That gives enough terms of one sign to swing the partial sums as far as you want in that direction, over and over again. So that situation will also be true for the combined series. The result depends on how the terms are ordered and you can never count on the partial sums settling down. In general, that makes it impossible to apply algebraic properties.So this hold only if the the two series are absolutely convergent?

If both series are absolutely convergent, then there is a limited ability of the terms of one sign to swing the sum in that direction. That also remains true of the combined series. The partial sums are well behaved and the final limits are the same regardless of the order of the terms. Algebraic properties can be applied.

- #7

fresh_42

Mentor

- 14,371

- 11,691

Can you prove this, please? The minus signs don't have to be distributed in an alternating way.If any one of them is NOT absolutely convergent, then either the sum of the negative terms is ##\infty## or the sum of the positive terms is ##\infty##, or both.

- #8

FactChecker

Science Advisor

Gold Member

- 6,053

- 2,339

I assume that we are talking about real numbers, not complex. Then, sure, I can prove it. If ##\sum \mid a_n \mid = \infty##, then either the sum of the absolute values of the negative terms is infinite or the sum of the positive terms is infinite or both.Can you prove this, please? The minus signs don't have to be distributed in an alternating way.

- #9

fresh_42

Mentor

- 14,371

- 11,691

- #10

FactChecker

Science Advisor

Gold Member

- 6,053

- 2,339

That is the opposite of what I said.

- #11

fresh_42

Mentor

- 14,371

- 11,691

No, it's the negation, not the opposite. ##\sum a_{2n} < \infty \wedge \sum a_{2n-1}<\infty \Longrightarrow \sum a_n<\infty##. You stated the stronger statement that ##\sum |a_n|<\infty## which I do not see. This is equivalent to what you said, namely ##\sum |a_n| = \infty \Longrightarrow \sum a_{2n} = \infty \vee \sum a_{2n-1}=\infty##.That is the opposite of what I said.

I have no counterexample at hand, so I currently cannot rule out, that it is correct, but I doubt it. In any case, it is not obvious. We cannot conclude from convergence to absolute convergence, so your stronger statement needs to be proven. The point is, that convergence + not absolute convergence does not imply, that the distribution of signs goes along ##2n / 2n-1##, which is why I cannot see how you can conclude especially on these two subsequences.

For short: I see that absolute convergence is sufficient. But you claim that it is also necessary, which I do not believe.

- #12

FactChecker

Science Advisor

Gold Member

- 6,053

- 2,339

Oh. I see your point. I meant to say that absolute convergence is the only way it is safe to apply simple rules of algebra as the OP asked. Otherwise, one must consider the order and it gets messy.For short: I see that absolute convergence is sufficient. But you claim that it is also necessary, which I do not believe.

- #13

fresh_42

Mentor

- 14,371

- 11,691

That was what I said in post #4.

- #14

mathwonk

Science Advisor

Homework Helper

2020 Award

- 11,100

- 1,302

just off the top of my head,

how about as odd series: 1 + 1/2 - 1/3 + 1/4 - 1/5 + 1/6 - 1/7 + 1/8 -+...., and as even series: 1 - 1/4 + 1/3 - 1/8 + 1/5 - 1/12 + 1/7 - 1/16 +-......

Then the combined series seems to diverge, but both series seem to satisfy the criterion for conditional convergence. (OOPs! claim retracted below.)

how about as odd series: 1 + 1/2 - 1/3 + 1/4 - 1/5 + 1/6 - 1/7 + 1/8 -+...., and as even series: 1 - 1/4 + 1/3 - 1/8 + 1/5 - 1/12 + 1/7 - 1/16 +-......

Then the combined series seems to diverge, but both series seem to satisfy the criterion for conditional convergence. (OOPs! claim retracted below.)

Last edited:

- #15

fresh_42

Mentor

- 14,371

- 11,691

But if both subseries converge, then by transition to partial sums we can add those and the sum again converges. Was my argument wrong in post #4?just off the top of my head,

how about as odd series: 1 + 1/2 - 1/3 + 1/4 - 1/5 + 1/6 - 1/7 + 1/8 -+...., and as even series: 1 - 1/4 + 1/3 - 1/8 + 1/5 - 1/12 + 1/7 - 1/16 +-......

Then the combined series seems to diverge, but both series seem to satisfy the criterion for conditional convergence.

- #16

FactChecker

Science Advisor

Gold Member

- 6,053

- 2,339

I believe that simple general algebraic properties can only be applied if the series are absolutely convergent. There may be non-general "slick" proofs for the case you asked about, but I think they will depend on the order of the terms and will be about the same as the proofs you say you can already do. I have never studied special properties that depend on specific orders of summation (like the even/odd question asked), so I will leave this to others.So this hold only if the the two series are absolutely convergent?

- #17

mathwonk

Science Advisor

Homework Helper

2020 Award

- 11,100

- 1,302

- Replies
- 11

- Views
- 5K

- Replies
- 6

- Views
- 1K

- Replies
- 6

- Views
- 8K

- Last Post

- Replies
- 3

- Views
- 844

- Last Post

- Replies
- 3

- Views
- 11K

- Last Post

- Replies
- 6

- Views
- 3K

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 3K

- Replies
- 7

- Views
- 2K

- Replies
- 2

- Views
- 2K