Let S1 =1-1+1-1+1-1 …
Then 1-S1 = 1-(1-1+1-1+1-1…) = 0+1-1+1-1+1-1+… =S1
Therefore 1-S1 = S1 and S1 = 1/2 note that < S1 > = 1/2 as well
Let S2 =1-2+3-4+5-6+7-8+…
--------------1-2+3-4+5-6+7-8…
Then 2S2 = 1-1+1-1+1-1 = S1 = 1/2
Therefore S2 = 1/4 Note that < S2 > = 1/4 as well
Let S = 1 + 2 + 3 + 4 + 5 + …
Then S-S2 = 0+4+0+8+0+12+0+16+… = 4(1+2+3+4…) = 4S
Therefore S - 1/4 = 4S and S = …-1/12??? More interesting, < S > is undefined! Not the said sum (-1/12 in this case) like the previous two!
From a mathematical viewpoint the complete arguementation is flawed since none of the sums is convergent and therefore they all dont have a limit.
I’m going to stop you right there, since S1’s value is undefined. Depending on how you pair its terms together, you get two different answers:
S1 = (1-1) + (1-1) + (1-1) + … = 0
S1 = 1 + (-1+1) + (-1+1) + (-1+1) + … = 1
If you rewrite S1 as sum(x=0…y) i^(2x) and evaluate lim(y–>infinity) S1, you will find it does not converge.
As a result, none of the rest of this can be determined to be true or false.
[You come into my house with that sort of mathematical hocus-pocus? Not on my watch! OB]
Oh, it will be good. Good for the brain. Either that or I’ll disappoint you.
Thanks, by the way.
Moved from #whine-cellar to #off-topic and edited title for clarity.
Ok, first of all, if anyone interested want to understand what is going on here, the youtube channel Numberphile has a video on it:
So, the problem I and a few other physicists and mathematicians I’ve talked to have with this “proof” is that you are assuming that these infinite sums (or series) can be treated as numbers, i.e. that they converge.
We can safely say that doing a finite amount of arithmetic operations to a finite amount of numbers will result in a number, but the same just can’t be said about infinite amount of operations and numbers.
For example, when you say
Fine, you defined S1 to be this infinite sum. But we have to ask ourselves: “can I treat S1 as a number?” The answer in this case is no, because the series diverges and
has no meaning, because that is an arithmetic operation (subtraction) between a number (one) and a thing thing that is not a number (S1).
There are a few tests we can apply to a series to find out if it diverges (many of which are listed here). If the series don’t pass even a single test, it diverges and therefone it’s not a number and if it passes, we can’t be sure. It turns out that all of the series mentioned in the OP fail all of the tests I’ve heard of.
But… there is some truth in saying that 1+2+3+4… = -1/12, even if loosely, and there’s even applications of this “result” in quantum field theory as the video pointed out. We just have to dig deeper.
There’s this really cool function we call the Riemann Zeta Function (commonly denoted by 
), defined by
and it converges for x > 1. For example, for x = 2, we get
a result that I’m not going to prove here.
The Riemann Zeta function works fine for any complex numbers, as long as their real part is greater than one, i.e. Re(x) > 1, and therefore can only be defined as I did earlier for this domain. If we use x = -1, a number that is not in the domain of the function, we get
also know as S, as defined in the OP. But this series diverge.
So we ask ourselves: “can’t we alter the form of the Riemann Zeta function in a way as to extend its domain to the entire complex plane, but retaining the same values on the orginal domain?” The answer is yes, and we call this process the analytic continuation of the Riemann Zeta function. The new form of the function is defined through a recursive relation:
where ‘sin’ denotes the standard trigonometric sine function and 
is the Euler Gamma function (another really cool function that ties in with definition of factorial, 
, if n is a non-zero natural number). Note that the Riemann Zeta function appears on both sides of the equation. Through this definition, the Riemann Zeta function has the same values on the old domain Re(z)>1, but now it also makes sense to talk about the function on any other complex value outside of the original domain, like z = -1. And, lo and behold, what happens when we plug in -1 on the new formula?
Exactly. What we defined earlier to be 1+2+3+4+… but had no meaning, now has meaning is it is equal to -1/12. So saying 1+2+3+4… = -1/12 is not completely wrong, but it’s not that right either. What is right is to say that the analytic continuation of a function that is defined by a series can converge, even if the series itself doesn’t.
There, I got it out of my system. But @XIYANGYANG, I must ask, what you meant by <S>? Is it some kind of average?
Sidenote: the other two sums S1 and S2 can be interpreted the same way as a geometric series/the derivate of a geometric series with the arguement being outside the radius of convergence
That’s true, and we can extend the domain of the geometric series too. We know that
is only valid for |z| < 1. But setting z = -1 gives
We know this isn’t right because the series diverges.
The following form represents the analytical continuation of the function to entire complex plane:
for any z0. It’s easy to show this form is consistent with the former definition. Setting z = -1, we can choose z0 so as to make both |z0| and |(z-z0)/(1-z0)| smaller than one, z0 = -1/2 for example, and then we can use the old definition. So
And the same can be done to the derivative of the function to get the other result you mentioned.
Haaaa…Ha…
But wait! If you sub in the the numbers for S1 as EITHER 1 or 0,you still get a finite value for < S >! How is that possible??
If you substitute infinity with a finite number, you get a finite number!
HOW IS THIS POSSIBLE?
If slot 1 in my inventory is a dex potion, and I sub it with a wis potion, SLOT 1 BECOMES A WIS POTION OMG WHAT?
nice. I was pretty sure that’s what was really going on, but I didn’t want to bother arguing this particular point because of both who the OP is, and the associated tone.
see? more nonsense right where that came from.
Again, I’m going to stop you right there.
Both of those substitutions would be wrong, because S1 is of indeterminate value. As a result, you can’t simply substitute in whatever value you choose to.
I prefer brilliant.org, but yes.
