# What is the sum of this infinite series?

1. Dec 10, 2014

### jonjacson

Sum= ...- 1 + 1 -1 +1-1+1... until infinite

It is just an infinite sum of -1 plus 1.

Can anyone tell me the sum of this infinite series and a demonstration of that result?

THanks!!

2. Dec 10, 2014

### hilbert2

3. Dec 10, 2014

### jonjacson

So the answer is that there is not a definite answer, in case we can associate a number we could pick the 1/2 number. Is that right?

I wonder what I am doing wrong with this reasoning:

1.- I think the answer is 0, since we can arrange the infinite series in pairs, for every -1 there will always be a 1, (-1 + 1) +(1-1) etc. As every one of those pairs sums 0, the total sum of the infinite seres should be 0. What is wrong? Why is it better 1/2 than 0?

4. Dec 10, 2014

### hilbert2

The series can not be summed in the usual sense. Its partial sums don't approach any particular value, they oscillate between 0 and 1. However, the mean of first n partial sums approaches 1/2 when n is given larger and larger values, so we can assign the series a Cesaro sum of 1/2.

5. Dec 10, 2014

### jonjacson

My partial sums are equal to zero always, I just need to add two numbers to the series to get more and more 0, and I always get 0 as a result. What am I doing wrong?

6. Dec 10, 2014

### hilbert2

Before you can calculate sums of infinite series, you need to define accurately what a "sum" means in the case there are infinitely many terms. I suggest you first make yourself familiar with the epsilon-delta definition of a limit, to see how things are defined in mathematics. If you don't have a strict definition, you can assign that sum ANY integer value by just rearranging terms.

7. Dec 10, 2014

### jonjacson

But the way I used to rearrange terms includes tem all.

I have seen somewhere that they rearranged the terms and said the sum was -1 :

-basically they said -1 (+1 -1)+ (+1-1) etc should be equal to -1 + 0 +0 +0 ... so the sum was -1.

-and viceversa it could be the same with +1

But if you "isolate" +1 I think somewhere else there must be a pair for him which is a -1
If you isolate -1, there must be a +1 somewhere that is his couple

But using my reasoning we are not forgeting any number, it looks fair to me, we just get together all the numbers in pairs, since every sum is 0, the total sum must be 0. It looks so clean that I cannot see where is the fallacy, or the mistake with that reasoning.

8. Dec 10, 2014

### Staff: Mentor

That's not how you need to calculate the partial sums. For your series,
S1 = -1
S2 = S1 + 1 = 0
S3 = S2 - 1 = -1
...
and so on. As already mentioned, the partial sums oscillate between -1 and 0, so the series fails to converge in the usual sense. With alternating series, like this one, you can't group terms as you have done.

9. Dec 10, 2014

### jonjacson

Could you justify why I cannot group terms as I did?

Last edited: Dec 10, 2014
10. Dec 10, 2014

### Staff: Mentor

Here's a quote from one of the Calculus texts I have, in Remark 4 on page 779 of "Calculus and Analytic Geometry, 2nd Ed.," by Abraham Schwartz:
Some definitions:
A series $\sum a_n$ is absolutely convergent if $\sum |a_n|$ converges.
A series $\sum a_n$ is conditionally convergent if $\sum a_n$ converges but $\sum |a_n|$ diverges.

Your series, -1 + 1 - 1 + 1 + ... + ... is not absolutely convergent, because its partial sums increase without bound. Also, since the partial sums of your series oscillate (and therefore don't converge), the series isn't conditionally convergent, either. For these reasons, rearranging the terms in your series isn't valid.

11. Dec 10, 2014

### jonjacson

Ok, so this lead us to more advanced courses. I will search for those to find the answer.

12. Dec 11, 2014

### FactChecker

As hilbert2 said
There is no answer that can be considered authoritatively correct. It is fundamental in math to be able to rearrange a summation without changing the answer. But you can cluster any number of +1s or -1s together to get any integer values repeating as often as you want. So there is no well defined limit.

13. Dec 13, 2014

### jonjacson

Hi,

Could you show me how to rearrange the terms to get +2 as an answer?

Thanks

14. Dec 13, 2014

### WWGD

1+1 +(1-1)+(1-1)+.........

15. Dec 13, 2014

### jonjacson

I am sorry but that does not correspond to the series I was talking about and I will explain why, let's take ANY quantity of numbers from this series, let's say 10:

-1, +1, -1, +1, -1, +1, -1, +1, -1, +1 As we see the proportion of -1 is the same as +1, and for every positive one there is a corresponding negative

if we now take 100 numbers the proportion will be the same, you will have 50 positive ones and 50 negative ones.

If we take any quantity of numbers that proportion is always constant.

Your mistake is you forget that in that series there must be -1 -1 somewhere else, that are the couple for your +1+1.

16. Dec 13, 2014

### FactChecker

It's not a question of 2 being the "answer". There is no answer to that series. You can get 2 to show up over and over as terms are added.
You always have - 1 + 1 - 1 + 1 = + 1 + 1 - 1 - 1. So:
- 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 ....
= (- 1 + 1 - 1 + 1) + (- 1 + 1 - 1 + 1) + (- 1 + 1 - 1 + 1) + ...
Make the replacement of - 1 + 1 - 1 + 1 with + 1 + 1 - 1 - 1 an infinite number of times without changing the "answer" to get:
= (+ 1 + 1 - 1 - 1) + (+ 1 + 1 - 1 - 1) + (+ 1 + 1 - 1 - 1) + ...
= + 1 + 1 - 1 - 1 + 1 + 1 - 1 - 1 + 1 + 1 - 1 - 1 + ...
As you add term after term, you get the sequence of sums: 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 0, .......
So 2 shows up an infinite number of times. So does 1 and 0.
There is no answer to the sum of this series.

17. Dec 13, 2014

### WWGD

But this is not so when you have an infinite collection of numbers. There is no clear cut proportion when you have an infinite collection; the best you can aim for is to have a bijection between the 1's and the -1's , which still holds if you pair numbers as I did.

Basically, you can swap , say the 1st and 4th terms in the series, and then even the proportion
will remain constant :

Original: -1+1 -1+1 -........ Swap 1st and 4th to get:

1+1 -1 -1 + ( -1+1 -........ )

And the proportion remains constant.

EDIT: I don't mean to imply that the series actually formally converges to 2, but that we can

rearrange the terms to have an informal sum equal to 2 ( or to 3, 4, etc. ).

Last edited: Dec 13, 2014
18. Dec 13, 2014

### jonjacson

What you did basically is to show the sum of the series S1= -1+1-1+1-1+1-1... is equal to the series S2= -2 + 2 -2+2 -2+2-2+2 and your reasoning looks good to me, but that is consistent with what I said, that the sum of that series is equal to 0.
Since for every -2 you have a 2, and the sum of this series is S2=(-2+2) + (-2+2)... = 0 +0+0 = 0 since if we add an infinite amount of zeros the result is 0. What am I doing wrong?

I don't understand why the proportion of -1 vs +1 does not hold when the set is infinite, Can you prove that?

19. Dec 13, 2014

### WWGD

EDIT: I think my statement was kind of confusing.
But the proportion actually does hold, just not at every single step of the way: you are just rearranging the order of the terms, so there
are as many -1's as there are 1's, only that these 1's, -1's are in a different spot on the sum.
The infinite number of terms allows you to add an infinite string of (1-1)'s to have a sum "equal" to 2.

20. Dec 13, 2014

### jonjacson

Actually I am confused XD

I don't understand anything, Why (1-1)'s is "equal" to 2?

I said is equal to 0, not to 2.