 Posers and Puzzles

1. 07 Jun '07 19:30
Let S equal the infinite series 1-1+1-1+1-1+1.... S can equal:

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

S=1-(1-1)-(1-1)-(1-1)-(1-1)...
S=1-0-0-0-0...
S=1

However, there's one more:

S=1-(1-1+1-1+1-1+1-1...)
S=1-S
2S=1
S=0.5

But how can S=0.5 if the series is only adding and subtracting 1s?
2. 07 Jun '07 19:58
because you added S=1 and S=0 and divided by 2. thats 0.5 then.
3. 07 Jun '07 20:08
Originally posted by twilight2007
Let S equal the infinite series 1-1+1-1+1-1+1.... S can equal:

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

S=1-(1-1)-(1-1)-(1-1)-(1-1)...
S=1-0-0-0-0...
S=1

However, there's one more:

S=1-(1-1+1-1+1-1+1-1...)
S=1-S
2S=1
S=0.5

But how can S=0.5 if the series is only adding and subtracting 1s?
As I recall, any divergent series can be rearranged to give any limit you choose, as you just demonstrated.
4. 07 Jun '07 20:32
Originally posted by twilight2007
Let S equal the infinite series 1-1+1-1+1-1+1.... S can equal:
If you first add inifinit terms of 1's giving infinity, and then subtracts infinit number of 0's, you still got infinity.

As said before, you can get whatever result.

If you see this as a series of sums S(n) = 1 + 0 + 1 + ... with n terms, and try to find limit of S(n) as n aproaches infinity you find no limit at all. This kind of series has no limit.
5. 07 Jun '07 22:25
Originally posted by FabianFnas
If you first add inifinit terms of 1's giving infinity, and then subtracts infinit number of 0's, you still got infinity.

As said before, you can get whatever result.

If you see this as a series of sums S(n) = 1 + 0 + 1 + ... with n terms, and try to find limit of S(n) as n aproaches infinity you find no limit at all. This kind of series has no limit.
But you're not subtracting 0's - you're subtracting 1's.
6. 08 Jun '07 06:01
Right, my bad. I repost, now probably right:

If you see this as a series of sums S(n) = 1 - 1 + 1 - 1 + ... with n terms, and try to find limit of S(n) as n aproaches infinity you find no limit at all. This kind of series has no limit.
7. 08 Jun '07 11:10
It only makes sense to talk about a sum to infinite if the series converges, and this one doesn't.
8. 09 Jun '07 05:141 edit
If we look at the 1-1+1-1+1... again. Every two is a minus term, every two is a plus term, right?
The question at hand is, what is the sum if we have infinite number of terms?

On way to find out the answer is to count the plus terms and count the minus terms and see which is the most.
If we have more plus terms, then the sum is positive.
If we have more minus terms, then the sum is negative.
If we have equal amounts of plus and minus terms, then we have zero as an answer.
This line of reasoning is valid when we have an finite number of terms. But is it valid in this case?

So, the key to the solution of this problem is:
Which is the most in the series? The number of plus terms or the number of minus terms? Or are the number the same?
9. 09 Jun '07 11:30
You'd have the same number of minus terms as plus terms right?

What if the sum was 1+1-1+1+1-1+1+1-1...

If that was an infinite serise, then surely the number of +1s would be the same as the number of -1s (infinite) and therefore would be equal to 0.
10. 09 Jun '07 11:39
Originally posted by Jake Ellison
You'd have the same number of minus terms as plus terms right?

What if the sum was 1+1-1+1+1-1+1+1-1...

If that was an infinite serise, then surely the number of +1s would be the same as the number of -1s (infinite) and therefore would be equal to 0.
You got my point correctly.
Something like a paradox, isn't it?
11. 12 Jun '07 16:421 edit
Originally posted by FabianFnas
You got my point correctly.
Something like a paradox, isn't it?
To an infinite sum to converge, that is to have a sum, its general term must tend to zero. This a necessary but not sufficent condition. A series can be written like this S u_n
For instance S(1/2)^n where S denotes a sum from zero to infinity. This is a convergent series whose value is 2 and the general term is (1/2)^n
in the 1-1+1-1... series the general term is (-1)^n and the limit oh this general term isn't zero so the series can't be convergent. This is was said before but the mess up continued so this is my help to clean it up.

lim u_n=0 being a necessary but not sufficient condition means that if a series converges than its general term necessarily tends to zero as n goes to infinity, but not being suficient means that there are series whose general term goes to zero but never the less diverges. S(1/n) being one simple example.
12. 12 Jun '07 17:13
To an infinite sum to converge, that is to have a sum, its general term must tend to zero. This a necessary but not sufficent condition. A series can be written like this S u_n
For instance S(1/2)^n where S denotes a sum from zero to infinity. This is a convergent series whose value is 2 and the general term is (1/2)^n
in the 1-1+1-1... series ...[text shortened]... whose general term goes to zero but never the less diverges. S(1/n) being one simple example.
Of course you're right. I've never doubted that. My math professor says it, the book says it, I say it myself.

But just for the fun of it...

Suppose you have a series S with terms of this looks:
S(n) = -1 if n is prime and +1 otherwise.
What is the sum of S when n goes from 1 to infinity?

One way to see it is to count the primes and compare this number with the number of the other numbers.
(The sum of S(n) when n goes from 1 to 20 is -1-1-1+1-1+1-1+1+1+1-1+1-1+1+1+1-1+1-1+1=+2. The higher n the higher sum.)
Surely there are more number that is not primes than the number of primes, right?
Wrong! There are exactly the same number primes as there are non-primes.

So we are actually led to believe that the sum of S = 0 !!!

But this is just for the fun of it. Somewhere this reasoning is wrong. One can't do like that.
The limit of S when n approaches infinity is not defined at all, of the same reasons you wrote.
13. 13 Jun '07 13:45
Originally posted by FabianFnas
Of course you're right. I've never doubted that. My math professor says it, the book says it, I say it myself.

But just for the fun of it...

Suppose you have a series S with terms of this looks:
S(n) = -1 if n is prime and +1 otherwise.
What is the sum of S when n goes from 1 to infinity?

One way to see it is to count the primes and compare thi ...[text shortened]... e limit of S when n approaches infinity is not defined at all, of the same reasons you wrote.
And that's right! 😏
14. 13 Jun '07 23:48
Originally posted by Jake Ellison
You'd have the same number of minus terms as plus terms right?

What if the sum was 1+1-1+1+1-1+1+1-1...

If that was an infinite serise, then surely the number of +1s would be the same as the number of -1s (infinite) and therefore would be equal to 0.
I just wanted to comment on this. The statement made at the end is not true. If it were, then I could argue the same for this infinite series:

1-1+1-1+1-1...

By the same logic, if this series were infinite, there would be an equal number of +'s and -'s, and again this would equal zero. But the original series is just this series with a 1 in front of it:

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

So does this mean that this series equals 1 and 0? Obviously not. The idea is that when dealing with infinity, you cannot classify it as an even or an odd number, so you can't say that there would be an equal amount of +'s as -'s. We just don't know, which is why the series diverges.

Anyway, crazyblue answered the problem in the second post, so the rest of the thread was not really relevant to the question.
15. 14 Jun '07 22:19
What's the fuss? The answer = 0