>1+1+1... > 1+2+3... Mathcucks defend this

>>36565364
Well you see, the plus sign is what's known is a binary operator, taking two inputs and outputting a single value, in this case their sum. These inputs can themselves be the result of another operation, allowing them to be 'chained' indefinitely in the fashion you see there.

>>36565364
>1+1+1... > 1+2+3...

who told you that that's legit? i might as well say staceys love my beta cock

>inb4 -1/12 shitters show up

>>36565364
They're both the same
Infinity

>>36565555
Came here for this post.
Beautiful digits. -1/12-fags btfo.

>>36565605
>not knowing that infinity has multiple layers within itself of sub-infinities

Don't make me laugh, brainlet.

>>36565716
True but those are the same Infinities
No hush yourself brainlet

>>36565364
You can't use these operators on infinite sequences, only to their limits, which are different from the sequences themselves

And 1+2+3... diverges at infinity faster than 1+1+1.... so if anything that would be "greater" as in faster divergence

Yet some idiot mathfags seem to not grasp this simple concept and think limits mean equality

>>36565364
Both are infinity. I don't know where you got the greater than part.

>>36565364
1+2+3+...=(1+1+1+...)+(0+1+2+...)

I hope you are now content with my content asshole buzzkill robot.

>>36566328
are those equal tho?
the left side is basically the right side with an infinite+1 number of ones added so it should be bigger than the right side

When did sonic become a cop?

>>36566245
There are different sized infinities, there are half as many odd numbers as there are odd and even numbers, both are infinite but one is half the size of the other, I think its more about divergence, like big O notation, but what the fuck do i know

>>36566847
>There are different sized infinities
Yes.
>there are half as many odd numbers as there are odd and even numbers, both are infinite but one is half the size of the other
No.

>>36566847
let A(n)=1+1+1+...+1=n
let B(n)=1+2+3+...+n

as n->infinity, A(n) is always equal to the last term of B(n), but B(n) also has the subsequence 1+2+...+(n-1). B(n) is always greater than A(n), so therefore lim n->infinity A(n) < lim n->infinity B(n)