Prove me wrong
Prove me wrong
It's not porn, it's gore.
But the series diverges
Prove you right first
It's a sum of positive numbers, so the result, if it exists, must be positive.
Thus it can't be -1/12
Logic.
Next.
It diverges...
And it isnt even a cauchy series so it simply can not converge.
There i proved you wrong
This was never the famous result.
F[0]=1
F[n]=n
F[oo]=oo!=-1/12
I'm stupid that was wrong, kek
F[0]=1
F[n]=F[n-1]+n
F[n]=(n+n2)/2+1
F[oo]=(oo+oo2)+1€oo
F[oo]=(oo+oo2)/2+1€oo
what is my bot doing
Suppose [math] s_N=\sum_{n=1}^Nn\to-\frac{1}{12} [/math]. Then [math] s_N,s_{N-1}\to-\frac{1}{12} [/math], and [math] s_{N}-s_{N-1}=N\to 0\,; [/math] a contradiction.
Burden of Proof is on you.
well fooooooooooooooooo to you too
Fuck off erryone knows this shit is a divergent series
Come on
Come the fucking on
t. brainlet who doesn't know what analytic continuation is
Diverges by integral test.
I don't know man, that picture sure does get me hard.
Not OP but here's some fun:
A = 1 -1 +1 -1 ...
1 - A = 1 -(1 - 1 + 1...) = A
1 = 2A
A = 1/2
----
B = 1 - 2 + 3 - 4...
B + B = 1 + (1 - 2) + (-2 + 3) + (3 - 4)...
B + B = 1 - 1 + 1 - 1... = A
2B = 1/2
B = 1/4
----
S = 1 + 2 + 3 + 4 + 5...
S - B = (1-1) + (2 - (-2)) + (3-3) + (4 - (-4)) ...
S - B = 0 + 4 + 0 + 8 + 0 + 12....
S - B = 4(1+2+3+4...)
S - B = 4S
-B = 3S
-1/12 = S
Then stop abusing the notation of equality. The sum of all natural numbers is not "equal" to -1/12 in the usual sense. They are in a different equivalence relation.
but then you can't add up divergent series in the classical sense this picture and your representations suggest. This doesnt mean its wrong since this is the representation zeta(-1) but its not like this makes any sense in regard to sums.
it's a false statement
zeta(-1) = -1/12
but thats cause of analytic continuation and has nothing to do with that sum
>but thats cause of analytic continuation and has nothing to do with that sum
Bingo
It's important to understand that just because two functions yield the same results on a certain range then they're automatically the same function.
For example, the geometric series on x yields the same result as 1/(1-x) if |x| < 1, but in no way does that imply that the geometric series function and the 1/(1-x) function are the same function.
a=0
b=negative infinite
s=positive infinite
a=0|1
b=+-inf
s=inf
also as recursive function with recurrence equation solution:
a[n]=Ca+((-1)^n - 1)/2
b[n]=Cb+((-1)^n (2 n + 1) - 1)/4
s[n]=Cs+n (n + 1)/2