Taylor Series. Just the fact that we can take practically any function and write it as a polynomial is mind-blowing. Not to mention you can use them to prove e^(ix)+1=0, which ties for second for me along with 1+2+3+4...=-1/12

1+2+3+4...=-1/12

Remind me again how this can be even remotely possible?  There seems to be no way to get that to work

My question is, how can 1+2+3+4... ever end up being less than 1??  You are starting with 1 and only adding positive numbers, how can you get negative?  It seems mathematically impossible O.o

it seems that Eulers formula only deals with trig and complex numbers Can i see some sort of mathematical proof or explanation?

It's here: http://en.wikipedia.org/wiki/Euler's_formula#Proofs It makes no sense, but apparently it makes sense O.o

That is because even math is theoretical
But think of this... when working with mod systems, negatives are really positive. So, say you are working in mod 8... you would count {0,1,2,3,...,6,7,0,1...}...

Now also consider the numbers that are not infinity, but by performing a mathematical operation, you still get that number (so n+a=n, n*a=n, et cetera).

Now say you work in mod r... -1 really equals r-1. Taking all of this into account, working with mod (r) where r is equal to one of the before mentioned "infinite" numbers, the very largest number is in fact -1. No matter how high you count, it has been proven you will never reach these numbers, but in a mod system, -1 means you are just 1 away from that number!

Anywho, that has nothing to do with the proof that 1+2+3+4... is less than 1, but it is still interesting, yes?

∭^∞_-∞(√(x²+y²+z²)e^{-(x²+y²+z²))})dxdydz =2ㅠ

The reason why is very fun to figure out, but should be rather obvious to some people