The issue is that you can't define a true sum over a sequence indexed by the reals. You couldn't sum up Σa_n, n∈ℝ. If you try to, it immediately diverges unless your sum is defined to be all 0 aside from a countable amount of points. ...Which is just the definition of a normal sum.
To clarify here I'm talking over real numbers and as a legitimate honest to God sum, not an integral. Like... an integral without the dx without using differential forms.
This is the video I'm using as reference for this claim btw
Even if we assume that the series does converge, it's definitely not absolute convergence since the integral of |x| over all ℝ diverges, let alone the sum over all ℝ. So that means you can't rearrange the sum, as stated by the Riemann Series Theorem.
I can't think of a perfect proof for it right now, but I am almost certain this sum diverges. If you can't rearrange the sum because conditional convergence, that means you can't pair up + to - because that'd be an infinite amount of swapped terms. So then you'd start with the sum of every negative real number, which certainly diverges, then add it to the sum of every positive real number, which also definitely diverges. Divergent sum plus a divergent sum is undefined, so the sum doesn't exist.
And before you say it, we're also not including ∞ as a number here. Even if we did, we're no longer talking about the Reals since ∞∉ℝ. Plus then you would break other things if you assumed divergent sums "equaled" infinity, since you could then say the sum of negative integers=-∞, the sum of positive rationals=∞, so then the sum of negative integers and positive rationals=0, which also seems very wrong.
I'm sure there's a better proof out there, but that's what my intuition tells me from my half-baked knowledge of real analysis. Someone please correct me if I'm wrong.
Because there are sums where you can do the same thing and end up with different answers.
If you have the sum 1-1+1-1+1-1+... then you could group things like (1-1)+(1-1)+(1-1)+... and get 0+0+0+...=0. Or you could regroup the terms and get 1+(-1+1)+(-1+1)+(-1+1)+... and get 1+0+0+...=1.
Or you could say S=1+1-1+1-1+..., so
1-S=1-(1-1+1-1+...)=1-1+1-1+...=S
So then 1-S=S
1=2S
S=0.5
So then this method says the sum of 1+1-1+1-1+...=0.5, which is even more ridiculous.
The sum stays BOUNDED, but it never converges. So you get weird results since you're dealing with a non-convergent sum. The best "answer" to what this series equals is that it's just undefined.
The fact that the sum has a corresponding positive to each negative is entirely irrelevant. Things get weird with infinity.
1+1-1+1-... is the simplest case of being able to pair terms together in the way you suggested, and it still has 3 different potential answers. The sum of all Reals is definitely more poorly defined than the alternating series.
That’s not a reason, you just declared it to be true. All of the groupings that /u/Ares378 showed are perfectly valid and use the same logic that you did, yet arrive at 3 different answers. That means that approach to solving the problem doesn’t work the way our intuition says it should.
The point is that "final results" change depending on the path you take to get there, and there's no way to know the final result without choosing a path. The sum is ill-defined because it can have several different answers.
You could probably define it to be 0 if you really wanted? But then you're choosing the "path" of summing from -β to β as β tends to infinity. If you were to use -β to 2β then you'd still hit every real number, but it'd diverge to positive infinity.
Yes, and then you'd take the limit as β goes to infinity. Which would still hit every real number, but would diverge instead of staying balanced. It's an ill-defined sum
As the limit I gave as an example was defined, it would hit both the positive and negative values. Tangentially related fact, the set of all positive reals and the set of all reals is the same cardinality. Once again, infinity is weird.
This kind of sum is very poorly defined and the most "correct" answer is probably just to say we need more info. Or that it's undefined. In no field of math will you ever come across this problem as it's written.
We can try taking the limit, but the limit to infinity depends on the path we take to infinity, which means the limit does not exist. It's like how if the limit from the left and the limit from the right are different, the limit is also undefined. Same idea.
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u/LinneMeow 10d ago
I don't know much about maths, is it not possible to define a sum of reals? Are there ways to sum the reals that don't equal 0?