Analytic Proof that 1+1+1+1+1+⋯=-1/12
📂FunctionsAnalytic Proof that 1+1+1+1+1+⋯=-1/12
Theorem
===1+1+1+1+1+⋯n∈N∑n01ζ(0)−21
Explanation
If you only focus on how adding positive numbers results in a negative number, you will never understand this post. The key is that ∑n∈N1 is expressed as the Dirichlet series n∈N∑n01, and it is calculated as the function value ζ(0) of its analytic continuation, the Riemann Zeta Function ζ. If you are going to show an attitude like “Anyway, the equation does not hold, does it?” or “This can show a contradiction, right?” without even trying to understand the proof properly and only taking parts that you can easily handle, it’s worse than not knowing. To be precise, this post introduces a proof solely about the equation
ζ(0)=−21.
Proof
Since the Riemann Zeta Function is analytic from C∖{1}, it is continuous from s=0 and therefore
ζ(0)=s→0limζ(s)
Riemann’s Functional Equation:
ζ(s)=2sπs−1sin(2πs)Γ(1−s)ζ(1−s)
The function value of the Gamma Function Γ(1−z) at z=0 is Γ(1)=0!=1, so when s→0,
2s→1πs−1→π1Γ(1−s)→1
Meanwhile, the limit at s→0lim
sin(2πs)→0ζ(1−s)→∞
will be canceled.
Taylor Expansion of the Sine Function:
sinx=n=0∑∞(2n+1)!x2n+1(−1)n
Laurent Series of the Riemann Zeta Function:
ζ(s)=s−11+n=0∑∞γnn!(1−s)n,s1
Here γn is defined as the nth Stieltjes constant as follows.
γn:=m→∞limk=1∑m(k(logk)n−n+1(logm)n)
Following the Taylor expansion of sin(2πs) and the Laurent series of ζ(1−s),
s→0limζ(s)======s→0lim2sπs−1sin(2πs)Γ(1−s)ζ(1−s)s→0lim1⋅π1sin(2πs)⋅1⋅ζ(1−s)π1s→0limsin(2πs)ζ(1−s)π1s→0lim[2πs−48π3s3+⋯][−s1+n=0∑∞γnn!sn]π1s→0lim[−2π+O(s)]−21
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