1+2+3+4+5+6+... = -1/12
 

My brain hurts.

Infinite series: When the sum of all positive integers is a small negative fraction.

I actually heard this disspelled by someone way smarter than I am the other day. His reply was basically this:

They are saying if you can show that the sum of all positive integers has a solution, we can prove it is -1/12. Well, if you can show my wife is a man, I can prove she is Alan Alda.

In other words, it's wrong.

The whole thing goes awry to me when they use the Grandi's series staircase bit. The convert a series of positive INTEGERS into a series of positive REAL numbers. They take 1-1+1-1 and turn it into a plot that converts them to real numbers, and that's NOT the original premise.

Or to take MJ4H's example, you can prove his wife is Alan Alda by importing a picture of her, overwriting it with Alan Alda's picture, and then showing it.

Or my favorite example:

3+4=5

if 3, 4, and 5 are vectors, rather than the positive integers portrayed in the original problem.

Sometimes I look at stuff like this and think the people that come up with it have to be huddled away in a room full of wires and screens and scraps of paper with illegible scribbles all over them, drinking black coffee religiously and sitting in a fetal position while nodding their head and babbling like a person who just suffered a concussion.

Or if you're using extremely small values of 3 and 4.

They go from 1+2+3+4+5+... to 1-2+3-4+5-6+... in the article.

These are not the same two series, and I am not sure how they go from one to the other, or how one proves the other.

Not saying they don't, but on the face of it, this doesn't make sense, and the article, while pleasantly written, does little to explain how it jumps from one series to the other and comes to the conclusion that the sum total for both is -1/12.

No, they are the same series. The ... represents continuing numbers to infinity. So the last number that is shown doesn't matter, it just needs to represent that the next number in the series is one higher.

FTA:

The video explains it.

Ah ha... I can't watch the video here at work.

After thinking it out some, I can see how the two differently phrased series would still essentially "go to infinity" (diverge) and more or less result in the same thing, although I haven't dived it into enough to be ready to agree with you, cartman, that they are the same series.

I see this as more of a "slight of hand" trick. If the series is infinite, then you can't really come up with an addition summation without picking an endpoint. If you pick an endpoint to get a sum, we're not really talking about infinity anymore.

http://en.wikipedia.org/wiki/Convergent_series

edit: Sorry, read that as "if a series" in infinite. If you are specifically referring to this, then, withdrawn.

I should have been more specific... The series in question is divergent, not convergent. The only thing you can do is come up with a partial sum using alternative methods as portrayed in the video. One of those alternative methods is a form of averaging that happens to tge eird summation in its name. That doesn't make it a sum that that little equals signs tries to make us think.

Agree with you.

Stop listening to anyone the moment they say they will prove something by introducing fuzzy operators that don't mean anything at all. Articles like this are written by stupid people trying to act smart.

Even approximations are fine if you keep within the constraints of those approximations, but saying the divergent series is your magic presto I can bend math to my whim operator and then substitute it out for real operators in the end is abuse. We have enough idiots running around that do not know how a proof works and can't figure out you can solve complex problems through transforming them into understood problems.

+1

Does 3+4=5 or does 3+4=7? It is actually both.

Let 3+4=z.
We want to solve for z.
Let x=3.
x+4=z
x+(x+1)=z
2x+1=z
(2x+1)-6=(z)-6
2x-5=z-6
(2x-5)^2=(z-6)^2
(2x-5)(2x-5)=(z-6)(z-6)
4x^2-20x+25=z^2-12z+36
4x^2-20x-11=z^2-12z
x(4x-20)-11=z^2-12z
Since x=3,
3(4(3)-20)-11=z^2-12z
3(12-20)-11=z^2-12z
3(-8)-11=z^2-12z
-24-11=z^2-12z
-35=z^2-12z
z^2-12z+35=0
(z-7)(z-5)=0

So, z=7 or z=5

:eek: :devil:

Shouldn't that actually go....

x(4x) - x(20) - 11 = z^2 - 12z

That works too... :p

x(4x) - x(20) - 11 = z^2 - 12z
Since x=3,
(3)(4(3)) - 3(20) - 11 = z^2 - 12z
3(12) - 60 - 11 = z^2 - 12z
36 - 60 - 11 = z^2 - 12z
36 - 71 = z^2 - 12z
-35 = z^2 - 12z
z^2 - 12z + 35 = 0
(z-7)(z-5) = 0

:devil:

I am onto your evil witchcraft. :lol:

A follow up was posted with some clarifications and retractions.

Follow-up: The Infinite Series and the Mind-Blowing Result.