1 + 2 + 3 + 4 + … = − 1
12
What is going on ?
Let's begin with f(z) defined as
f(z) ≡ 1 −z + z2 −z3 + …
This function is convergent for |z| < 1. On the other hand, another function, g(z), defined as
g(z) ≡ 1
z+1
 ,
converges in the entire plane except for z = −1. However, in |z| < 1,
f(z) ≡ g(z),
hence, g(z) is analytic continuation of f(z). Therefore, if z = −2 is substituted into f(z), it follows
1 + 2 + 4 + 8 + 16 + … = −1.
Now, define the Riemann zeta function, ζ(z), as
ζ(z) ≡ 1
1z
 + 1
2z
 + 1
3z
 + 1
4z
 + … = ∞
∑
n=1 
 1
nz
 .
The function, ζ(z), can be shown to be convergent for ℜ(z) > 1. (Example: ζ(2) = [(π2)/6]) Let's denote analytic continuation of ζ(z) as ―ζ(z). It can be shown that for ℜ(z) > 1,
-
ζ
 
 (z) ≡ ζ(z).
while ζ(z) cannot be evaluated for z = −1, ―ζ(−1) can be evaluated using the following functional relationship,
-
ζ
 
 (z) = 2z πz−1 Γ(1 − z) sin ( πz/2)  
-
ζ
 
 (1−z),
as
-
ζ
 
 (−1) = − 1
12
 .
Therefore, one can claim that
ζ(−1) = 1 + 2 + 3 + 4 + 5 + … =  − 1
12
 .


File translated from TEX by TTH, version 4.03.
On 18 Jan 2021, 22:04.