What is the sum of this infinite series?

[PeroK]

S=0=...+1-1+1-1...

Okay, what about?

T = -1 + 1 - 1 + 1 ...

 

[PeroK]

I've got to go, so let me make my point:

a) T = 0 by the same argument you used for S. Just group the terms.

But:

b) S = 1 + T. So, if T = 0, then S must be 1.

So, something must have gone wrong.

 

[jonjacson]

I've got to go, so let me make my point:

a) T = 0 by the same argument you used for S. Just group the terms.

But:

b) S = 1 + T. So, if T = 0, then S must be 1.

So, something must have gone wrong.

We can continue talking another day, no problem.
a) T=1 because you forget that S contains a -1 for every +1, and that proportion holds no matter the size of the set. If you add +1 to create T there is no -1 as a counterparty, that is why T sums +1 and S sums 0, because you added an extra +1 to S but you did NOT add the corresponding -1 that would maintain the same proportion of +1 and -1.

 

[PeroK]

We can continue talking another day, no problem.
a) T=1 because you forget that S contains a -1 for every +1, and that proportion holds no matter the size of the set. If you add +1 to create T there is no -1 as a counterparty, that is why T sums +1 and S sums 0, BECAUSE YOU ADDED AN EXTRA 1 that did not have his corresponding -1 in the series.

How do you know that S does not have an extra 1?

T = -1 + 1 -1 +1 ... = (-1 + 1) + (-1 + 1) ... = 0 + 0 + 0 ... = 0

So, T has just as much right to be 0 as S has. Why prefer S? Just because it starts with a positive number? I might prefer T = 0 and S = 1.

Also, why isn't T = -S?

Surely if you mutiply S by -1 you get T? So, how can S = 0 and T = -1?

 

[jonjacson]

How do you know that S does not have an extra 1?

T = -1 + 1 -1 +1 ... = (-1 + 1) + (-1 + 1) ... = 0 + 0 + 0 ... = 0

So, T has just as much right to be 0 as S has. Why prefer S? Just because it starts with a positive number? I might prefer T = 0 and S = 1.

Also, why isn't T = -S?

Surely if you mutiply S by -1 you get T? So, how can S = 0 and T = -1?

The first is a good question. Because that would imply a kind of assimetry, if there is an extra 1 it means the amount of -1 is less than the amount of +1. But why isn't it an extra -1 instead of a +1?

It doesn't make sense, the only consistent answer to that question is that the amount of +1 in the series is equal to the amount of -1 in it.

In other words, for every +1 in the series we can find its corresponding -1, and that is exactly what we see if we look at the terms in that series, you can take 10, 100, or a million terms and you will always see there is a -1 for every +1.

Every time you represent T you forget to show you added an extra +1 to S, let me show that extra +1 and all your confusions will dissapear:

T= ...-1+1-1+1+1-1+1-1+1
S=...-1+1-1+1-1+1-1+1-1

As you see there is an "anomaly" in T and is precisely the extra +1 you added to S. With that representation all the answers to your questions are simple:

T=S+1=1

I see you insist in multiplying -1 by S to get -S.

Let's do the same for T, we get:

-T=...+1-1+1-1-1+1-1+1-1... =-1 =-s-1 or in other words, if you add an extra -1 to -S you get -T which sums -1 instead of 0

 

[FactChecker]

Why? Why must I look one by one? Why can't I just put two terms together?

I mean, let's forget series, Isn't this +1-1 equal to (+1-1)?

Yes, they are equal.
Suppose we allow -1+1-1+1... to be a valid infinite summation with a single answer.
On one hand, we have -1 +(+1-1) + (+1-1) +(+1-1) + ... = -1 +0 +0+0...= -1
On the other hand we have (-1+1) + (-1+1) + (-1+1) + ... = 0+0+0... = 0
Which one is correct? This violates the associative property of addition, which is fundamental to all mathematics.

In math, a summation has a valid total only if changing the order or clustering of terms does not change the number that the partial sums eventually converge to. The associative and commutative properties of addition are not violated by the math definition of an infinite sum.

We either have to give up on -1+1-1+1... having a well-defined total, or we have to give up the associative and commutative properties of addition. The choice is clear. You will see eventually that "summations" like -1+1-1+1... are not very interesting and we haven't given up much.

 

[jonjacson]

Yes, they are equal.
Suppose we allow -1+1-1+1... to be a valid infinite summation with a single answer.
On one hand, we have -1 +(+1-1) + (+1-1) +(+1-1) = -1 +0 +0+0...= -1
On the other hand we have (-1+1) + (-1+1) + (-1+1) + ... = 0+0+0... = 0
Which one is correct? This violates the associative property of addition, which is fundamental to all mathematics.

In math, a summation has a valid total only if changing the order or clustering of terms does not change the number that the partial sums eventually converge to. The associative and commutative properties of addition are not violated by the math definition of an infinite sum.

We either have to give up on -1+1-1+1... having a well-defined total, or we have to give up the associative and commutative properties of addition. The choice is clear. You will see eventually that "summations" like -1+1-1+1... are not very interesting and we haven't given up much.

The second one is correct and it is equal to 0.

Since in the first case you are telling us there are extra +1 that do not have their corresponding -1. And my question is, Why only one? Why don't you add magically 25 extra +1? To me you are doing this mistake:

+1+1+1+1+1+1+1 +(+1-1) + (+1-1) + (+1 - 1) = 7 +0+0= 7 ; what I have been telling since the first post is that for those seven +1 there must be in the series their corresponding seven -1, what confuses you is you don't write them, so it looks like they don't exist and the result is absurd.

What I think is accepting the sum is = 0 does not generate any inconsistency nor absurd result and I would like you to demonstrate where I am wrong since I am not even rearranging terms. I have a question for you, assuming the reesult is 0, Does it violate any mathematical rule like associative?

edit:

I mean accepting the series =0 as the only valid result and all the other ones are incorrect.

 

[FactChecker]

Since in the first case you are telling us there are extra +1 that do not have their corresponding -1.

No. They are the exact same numbers in the exact same order. The associative law has just been applied differently in the two cases. By the associative property, that is not allowed to give different answers.

 

[ImperialThinker]

The series is not convergent though it is bounded. So the series sum cannot be determined.
However, in case of divergent series we don't usually determine the entire sum, the ramanujan sum of the series is 1/2

 

[PeroK]

Every time you represent T you forget to show you added an extra +1 to S, let me show that extra +1 and all your confusions will dissapear:

T= ...-1+1-1+1+1-1+1-1+1
S=...-1+1-1+1-1+1-1+1-1

One last try, the I give up!

Suppose you had never posted this thread and I defined T = -1 + 1 -1 +1 ... (without any reference to S). Why can't I conclude that T = 0?

Then you come along with S = 0. Well, your series is just T with a 1 at the front. I would then say S = 1 + T. So, I would say T = 0 and S = 1. And, you would say S = 0 and T = -1. But, we would only be saying this because of the series we had calculated first.

(In fact, if you look at your original post, you DID specify T. The original series you posted started with a -1, not with a 1. So, which is it? Which one is "really" the series that adds to 0 and which is the sequence that adds to 0 with an extra term at the front?).

Also, in any case, why isn't:

S = 1 - 1 + 1 - 1 ...
T = -1 + 1 - 1 + 1 ...
S+T = 0 + 0 ... = 0

Finally, what about

S' = a - a + a - a ... = 0

You're saying that S' = 0, but only when a is positive? But surely you can group a - a = 0 for any a, regardless of whether a is +/-?

Surely S' = 0 for any a?

 

[zoki85]

Who says that every infinite series must have sum?

 

[Mark44]

In math, a summation has a valid total only if changing the order or clustering of terms does not change the number that the partial sums eventually converge to. The associative and commutative properties of addition are not violated by the math definition of an infinite sum.

What I think is accepting the sum is = 0 does not generate any inconsistency nor absurd result and I would like you to demonstrate where I am wrong since I am not even rearranging terms.

But you are grouping the terms in a certain way. In an absolutely convergent series, reordering or grouping don't change the sum. The series in question here is not absolutely convergent, so reordering and grouping aren't valid.

I have a question for you, assuming the reesult is 0, Does it violate any mathematical rule like associative?

You can't just assume that the series in question is convergent.

The question has been asked and answered, so I am closing this thread.