# (-1)^∞ =? -1/2?

1. Nov 1, 2013

2. Nov 1, 2013

### Office_Shredder

Staff Emeritus
Can you tell us why you think this is a meaningful expression? Where did you get it from?

3. Nov 1, 2013

### Staff: Mentor

To be clear you're saying the lim n-> infiinity (-1)^n = -1/2 where n is {1,2,3...} right?

as n progresses from 1 to N it will produce an oscillating sequence of numbers -1, 1, -1, 1, -1 ...

so it looks if you add an even number you get 0 and if you add an odd number you get -1 so you've averaged them together to get -1/2.

This isn't how series summation is done for this kind of series and I don't think it has an answer because it doesn't converge mathematically.

4. Nov 1, 2013

### arildno

There exist, OP, FUNCTIONS that may have a divergent series as its argument*, and that function assigns a unique number to that divergent sum that in some forms of technical cases is called a "sum". Borel summation and Abel summation are examples of this.

But, and this is important:
Although such functions can be constructed (and be very useful), and has a number of properties that motivates the use of the "summation" term to designate them, they should not by any standards be CONFUSED with a regular sum.

They are not, they are functions that can have some subclass of divergent series as part of their argument domain.
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*Or, more precisely, having as its argument a sequence of numbers which, if they had been summed in a standard manner, would represent a divergent series

Last edited: Nov 1, 2013
5. Nov 1, 2013

### pwsnafu

Question to the OP: are you asking about a summation or a sequence?

6. Nov 1, 2013

### japplepie

well, for its supposed "sum" to agree with 1/2, for 1-1+1-.... and finite (-1)^n's cancel out.

terms with infinite n must be:
-not an integer
-could be written as (or is actually) 1 term

7. Nov 1, 2013

### pwsnafu

Well if you are talking about the sum, then it's called Grandi's series, and has Cesaro summation of 1/2.

8. Nov 1, 2013

### japplepie

That's right, but I'm trying to figure out why it isn't commutative.

1-1+1-... = 1/2
-1+1-1+..= -1/2

9. Nov 2, 2013

### Staff: Mentor

expanding my earlier post

so for the first one: it looks if you add an even number you get 0 and if you add an odd number you get 1 so you've averaged them together to get 1/2.

and for the second one: it looks if you add an even number you get 0 and if you add an odd number you get -1 so you've averaged them together to get -1/2.

10. Nov 2, 2013

### japplepie

Yes, I see that the half is from getting an average.

Adding (both positive and negative numbers) is commutative, but why isn't this commutative?

I would be perfectly fine if the "sum" was 0, since +0=-0 but it's not; it's 1/2.

That's what every method of summing divergent series spit out.

11. Nov 2, 2013

### pwsnafu

Just because an operation is commutative with a finite number of terms, it does not follow that it is commutative with an infinite number of terms.

Edit: Expanding on this. By the Riemann rearrangement theorem any conditionally convergent series can be rearranged to form any value. Hence commutativity is not necessarily true for convergent series, let alone divergent ones.

Last edited: Nov 2, 2013
12. Nov 2, 2013

### japplepie

That is exactly what I'm looking for!
thanks