10252004, 01:09 PM  #1 (permalink) 
Location: Waterloo, Ontario

Queer properties of infinite sums...
Actually, this is a precursor to a post that I really want to make and, actually, I've wanted to make this post for some time now but haven't gotten around to it and now it looks like the best days of this forum are gone. I don't recognize all the interesting characters who could appreciate what I'd like to explain. Oh well, maybe they're still lurking. Here goes...
...the precursor, that is. Are we all aware of what an infinite sum is? The harmonic series is a series of the form: 1 + 1/2 +1/3 + ... + 1/n. Surprisingly, even though the terms limit to zero, this series doesn't converge. It will sum indefinitely high. However, if you alternate the terms, like so: 1  1/2 + 1/3  ... + (1)^n/n (order of operations apply), then this does converge! In fact, it can be shown that any series, where the term sign alternates and the absolute values of the terms monotonically decrease, will converge to a finite value. Can everyone see why this is true? Can a stronger statement be made? The claim I'd like to make will probably leave most physicists incredulous. I've tried to explain it to some before but have only been met with skepticism. Hopefully, many math enthusiasts will undestand me. For now, I just want to know what the board is currently like... 
10252004, 01:37 PM  #2 (permalink) 
Junkie
Location: In the land of ice and snow.

We are studying infinite sums right now in my calc class. They seem like a nice change of pace from the standard integration problems we had been doing. I can't yet see why your alternating and decreasing monotonic sequence converges when summed, but give me a week more of class and i will probably be able to.
Let's see: 1  1/2 = 1/2 = .5 1  1/2 + 1/3 = 5/6 = 0.83333 1  1/2 + 1/3  1/4 = 7/12 = 0.58333 1  1/2 + 1/3  1/4 + 1/5 = 47/60 = 0.78333 1  1/2 + 1/3  1/4 + 1/5  1/6 = 37/60 = 0.61666 I don't have a general explanation as to why it converges, but based on its initial behavior it seems to be approaching a specific number. 
10252004, 02:43 PM  #3 (permalink) 
Tilted
Location: Sydney, Australia

A stanger fact is that there are infinite sums such that if you change the order of summation then the value they converge to changes. I'm also doing infinite sums right now in my calc class and should probably understand how to set one up but don't.
I'll hunt around for an example later. 
10252004, 04:48 PM  #5 (permalink) 
Addict

An example of molloby's idea appears on page 77 of Goldberg's Methods of Real Analysis (second edition):
We have that (1) \sum_{n=1}^\infty (1)^{n+1}/n = L = \log 2. (where L is the sum of the series). Now, divide L by 2, giving L/2 = 1/2  1/4 + 1/6  1/8 + ..., and thus (2) L/2 = 0 + 1/2  0  1/4 + 0 + 1/6  0  1/8 + ... . Adding (2) and (1) gives (3) 3/2 L = 1 + 1/3  1/2 + 1/5 + 1/7  1/4 + ... . You will notice that (3) is just a rearrangement of (1), yet it sums to 3/2 L = 3/2 \log 2. The book goes on to give the following theorem: Let L be a conditionally convergent series of real numbers. Then for any real number x, there is a rearrangement of L which converges to x. However, if the series is absolutely convergent, then all rearrangements sum to the same value. 
10252004, 06:09 PM  #7 (permalink)  
Location: Waterloo, Ontario

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So, if you are a conditionally convergent series, then the sum of your terms of like sign must diverge (in other words, the sum of all the negative terms must diverge negatively, while all positive terms diverge positively). Otherwise, you must be absolutely convergent. I mean, if only your negative terms diverge, how can you be convergent? Vice versa for your positive terms, they must diverge as well. If both signs of terms converge then how are you not absolutly convergent? So, now we have established that the sum of terms of like signs must diverge. Now, pick a real number x, any x! (even x factorial!) If x > 0, then you can sum just enough positive terms to so that that sum will be greater than x. Start adding just enough negative terms so that the sum is less than x. You know you need, at least, one negative term to do this. Now, do the same thing you were doing with the positive terms, before. That is, add just enough terms to exceed x. Again, you know you need at least one term to do this. This is very important. Again, sum the negative terms the same way you were, before. Repeat this algorithm so that the sum will waffle between being greater than and less than the target number, x. Because we know that the absolute value of the terms in the series limits to 0 (or would people like me to prove that?), we know that this order of terms will sum arbitrarily close to the target number. Furthermore, we know that everytime you switch signs you will use, at least, one term of that sign and that you must, repeatedly and endlessly switch signs. Thus, we know that we will use all the terms from both signs and that will make this a true reordering of our series. Without loss of generality, you can see that this works regardless of what value x is. Thus, a conditionally convergent series can be reordered to sum to any value. QED I hope you enjoyed it! 

11072004, 03:11 AM  #9 (permalink) 
Crazy
Location: San Diego, CA

KnifeMissile  I'm a little bit confused. You'll have to fogive me, as my infinite sum theory is a bit rusty.
First of all, what do you mean by "just over x." If it's a distinct number from x, then there are an infinite number of real numbers between it and x, so really any number greater than x would do. If it's not a distinct number from x, then it is x, and so you're just adding and subtracting 0 over and over, in which case the only number you can form is x by rearranging. Also, I do recall that there are infinite sums that do not converge to a number, but also do not diverge to infinity. Think about this one: 2  2 + 2  2 + 2  2 + ... At any one point it's either 2 or 0, but I certaintly wouldn't say it converges to 1. Could you try and clarify your proof a bit? Maybe write out some equations and examples? I'm also a bit curious about WHAT you're trying to prove. I had to read to the end just to find any kind of conclusion... if you clearly stated what you were trying to prove at the beginning, it would make reading it much easier.
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11072004, 08:25 AM  #11 (permalink)  
Rookie
Location: Oxford, UK

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It's an interesting concept though  but I guess intuitions can be appeased with the idea that you're essentially asking for the difference between two infinite (divergent) series. Their values being undefined it seems reasonable that depending on exactly how you ask the question you can get any answer you like.
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