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

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Anonymous

1-1+1-1+1-... = 1/2 2016-02-06 13:49:46 Post No. 7838110

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1-1+1-1+1-... = 1/2 2016-02-06 13:49:46 Post No. 7838110

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So what you do, is assuming this converges.

Like so: x = 1-1+1-1...

1 - x = x

x = 1/2.

This works in a similar way to how imaginary numbers allow us to find true answers.

so whats the problem?

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>>7838110

the problem is that it doesn't converge

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> assume OP is not a faggot

> OP being a faggot, there must be a god somewhere to allow two different realities at the same time

> checkmate atheists

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>>7838165

this tbqh familia

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It doesn't converge. In order for a geometric series to converge, the absolute value of common ratio must be less than 1. In this case, the common ratio is -1 which does not meet the conditions of a convergent geometric series

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It doesn't converge. Don't post here unless you know simple math.

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>>7838110

Using that, prove that the product of all prime numbers is 4 pi^2.

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>>7838248

Define "pi" as one fourth of the product of all prime numbers

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>>7838680

One of those argument is actually correct. |x| <1 (STRICTLY LESS) for a geometric series. So that's why you cant use 1/(1-r)

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>>7838680

Except it doesn't actually converge. The 1/2 value only represents the value it should converge to but you never actually get there no matter how long you compute the sum.

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Ok so the first problem is your phrasing. Instead of saying "assuming this converges", you could instead consider [math]x = \sum_{n \in N} (-1)^n[/math] as a formal sum

You can then try to define the operations the way you did here implicitely, and then try to make sense of what algebraic structure you are dealing with. I guess some inconstancies will show up sooner or later.

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>>7838248

I want to believe sooooo badly.

But how will you even...

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>>7838855

Except with each step in the series you mention (1+1/2+1/4+1/8...), you actually do get closer to 2. OP's series does not do this. It just hops back and forth between 0 and 1, forever in limbo.

>you sound like a religious person.

Nice strawman.

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>>7838110

For this kind of stuff, I recommend reading Terry Tao's article on it

https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

The problem with this is getting to work is finding a rigorous method of doing so. There is one, which is where we define the Riemann Zeta function for a domain of all complex numbers with real part greater than 1, then use the process of analytic continuation to extend the domain. In doing so, we get some weird results, but it is a rigorous process.

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>>7838855

The problem with your approach is that I can make that series have a value of anything I want. First, sum the first 2119372 positive ones. Then, substract -1, then add 1, then substract -1, etc, you get the drill. The series will not converge to 1/2 as it's always oscillating within 2119372 and 2119371.

So the problem is not that it doesn't exist, but that it is not consistent. Imaginary numbers do not exist, but they have a rigorous, consistent construction, and within the field of algebra they make all the sense of the world (we can build new fields adding to Q the roots of a polynomial irreducible in Q. The fact that those roots are complex are not really a difference).

If you want to be rigorous, see >>7838888. However, take into account that, for a non-converging infinite series, you will be able to find a series of manipulations that make it "converge" to whatever the hell you want.

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>>7838855

> no converging series ever really get to the end. like 1+1/2+1/4+1/8... never really equal 2.

Whether it "gets to the end" doesn't matter. What matters is that the series has a limit (which is 2).

In case you're unfamiliar with that term, I'll spell it out for you:

Given a sequence of elements x[n], and some value L, if for all k>0 there exists a natural number N such that for all n>N, |x[n]-L|<k, then L is a limit of the sequence.

(If a sequence has a limit, the limit is unique. Not all sequences have a limit, but no sequence can have more than one limit. This fact is in no way relevant here, btw).

In the case of the above sequence (i.e. the sequence of sums, x[n] = sum[0<=i<=n](2^-i), so x[0]=1, x[1]=1+1/2, x[2]=1+3/4, etc), it's clear that 2 is a limit of the sequence. Specifically, for any k>0, let N=-ceil(log[2](k)); clearly, |x[n]-2|<k for all n>N.

That's what matters. Not hand-waving bullshit about "getting to the end".

Conversely, the series 1-1+1-1+... clearly does not have a limit. Specifically, 1/2 is *not* a limit of the sequence, because for any value of k less than 1/2, there is no N such that |x[n]-1/2|<k for all n>N. For any other would-be limit, the situation is even worse (i.e. the smallest value of k for which the limit criterion holds is even larger).

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>>7838888

He's right, OP. Quads don't lie.

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>>7839148

This anon has taken analysis. Listen to this anon.

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-1/12=1+2+3+4+5+...=1+(1+1)+(1+1+1)+(1+1+1+1)+(1+1+1+1+1)+...=1+1+1+1+… right?

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