Maths as we know it could be wrong

this only works for |x| < 1
im guessing either the pic youı posted is wrong, or x^n becomes so small that it doesn'T matter

>>8977716
It's authors privilege.
He could have let s = 1+x+...x^n-1

>>8977720
OP here. You're right, I didn't specify this was for convergent series. Thought it was obvious enough to not have to mention here, but yeah, if x^n is so small that it doesn't matter, then it is only a good approximation. And even then, if that's the case then by that logic you get inconsistent results if you try the sx^2 case, and so on..

>>8977716
>Why is this proof actually a thing? Clearly, multiplying s by x will create an x^(n+1) term at the end.
Yes, you are right.
That "Proof" is bullshit.
It has to go like this:

Sn=1+x+...+x^n
xSn=x+x^2+...+x^n+x^(n+1)
S-xSn=1-x^(n+1)
Sn(1-x)=1-x^(n+1)
for x =/= 1
Sn=[1-x^(n+1)]/[1-x]

1+x+x^2+... is by definition the limit as n goes to infinity of sn.
If |x|<1 then x^(n+1) goes to 0.
Thus lim sn = 1/(1-x)

it's poorly written, but i guess the n is meaningless as the series is infinite.

>>8977732
Thank you anon, well written.
Kinda wished this wasn't the case though. i like the idea of creating a shitstorm over a well-established field - makes things more interesting

>>8977716
Wrong.
When your x > 1, 1 + x + x^2 + ... on and on to infinity IS infinity.

Therefore you can't just subtract one, add parenthesis, or even do something as benign as multiply by x/x, because infinity DOES not obey the same rules.
As such, his (2) is illegal.

>>8977716
This is a very fucking shitty proof, what's the source?

>>8977716
This proof is literally wrong, he multiplies x to s but forgets to make the last term x^(n+1) instead of just x^n

>>8977716
I hope you didn't make this image because it's wrong and retarded in multiple ways

>>8977737
>A random brainlet making an incorrect proof
>creating a shitstorm over a well-established field
You're going to have to try harder than that

>>8977716

People 6000 years from now will find it cute how wrong we are in our understanding of the universe with our archaic and primitive understanding of math.

>>8977716
This is a horrible explanation of this proof.

[math]s[/math] represents a partial sum, and (5) should say [math]s=s_n=\frac{1-x^{n+1}}{1-x}[/math]. The "infinite" sum is just the limit of [math]\{s_n\}[/math] which any calc student can check is [math]\frac{1}{1-x}[/math].