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Analytic Proof that 1+1+1+1+1+⋯=-1/12 📂Functions

Analytic Proof that 1+1+1+1+1+⋯=-1/12

Theorem

1+1+1+1+1+=nN1n0=ζ(0)=12 \begin{align*} & 1 + 1 + 1 + 1 + 1 + \cdots \\ =& \sum_{n \in \mathbb{N}} {{ 1 } \over { n^{0} }} \\ =& \zeta (0) \\ =& -{{ 1 } \over { 2 }} \end{align*}

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 nN1\sum_{n \in \mathbb{N}} 1 is expressed as the Dirichlet series nN1n0\displaystyle \sum_{n \in \mathbb{N}} {{ 1 } \over { n^{0} }}, and it is calculated as the function value ζ(0)\zeta (0) of its analytic continuation, the Riemann Zeta Function ζ\zeta. 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)=12 \zeta (0) = - {{ 1 } \over { 2 }} .

Proof1

Since the Riemann Zeta Function is analytic from C{1}\mathbb{C} \setminus \left\{ 1 \right\}, it is continuous from s=0s = 0 and therefore ζ(0)=lims0ζ(s) \zeta (0) = \lim_{s \to 0} \zeta (s)

Riemann’s Functional Equation: ζ(s)=2sπs1sin(πs2)Γ(1s)ζ(1s) \zeta (s) = 2^{s} \pi^{s - 1} \sin \left( {{ \pi s } \over { 2 }} \right) \Gamma (1-s) \zeta (1-s)

The function value of the Gamma Function Γ(1z)\Gamma (1-z) at z=0z=0 is Γ(1)=0!=1\Gamma (1) = 0! = 1, so when s0s \to 0, 2s1πs11πΓ(1s)1 2^{s} \to 1 \\ \pi^{s-1} \to {{ 1 } \over { \pi }} \\ \Gamma (1-s) \to 1 Meanwhile, the limit at lims0\displaystyle \lim_{s \to 0} sin(πs2)0ζ(1s) \sin \left( {{ \pi s } \over { 2 }} \right) \to 0 \\ \zeta (1-s) \to \infty will be canceled.

Taylor Expansion of the Sine Function: sinx=n=0x2n+1(2n+1)!(1)n \sin x=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ 2n+1 } }{ (2n+1)! }{ { (-1) }^{ n } } }

Laurent Series of the Riemann Zeta Function: ζ(s)=1s1+n=0γn(1s)nn!,s1 \zeta (s) = {{ 1 } \over { s-1 }} + \sum_{n=0}^{\infty} \gamma_{n} {{ (1-s)^{n} } \over { n! }} \qquad , s 1

Here γn\gamma_{n} is defined as the nnth Stieltjes constant as follows.

γn:=limmk=1m((logk)nk(logm)nn+1) \gamma_{n} := \lim_{m \to \infty} \sum_{k=1}^{m} \left( {{ \left( \log k \right)^{n} } \over { k }} - {{ \left( \log m \right)^{n} } \over { n+1 }} \right)

Following the Taylor expansion of sin(πs2)\displaystyle \sin \left( {{ \pi s } \over { 2 }} \right) and the Laurent series of ζ(1s)\zeta (1-s),

lims0ζ(s)=lims02sπs1sin(πs2)Γ(1s)ζ(1s)=lims011πsin(πs2)1ζ(1s)=1πlims0sin(πs2)ζ(1s)=1πlims0[πs2π3s348+][1s+n=0γnsnn!]=1πlims0[π2+O(s)]=12 \begin{align*} \lim_{s \to 0} \zeta (s) =& \lim_{s \to 0} 2^{s} \pi^{s - 1} \sin \left( {{ \pi s } \over { 2 }} \right) \Gamma (1-s) \zeta (1-s) \\ =& \lim_{s \to 0} 1 \cdot {{ 1 } \over { \pi }} \sin \left( {{ \pi s } \over { 2 }} \right) \cdot 1 \cdot \zeta (1-s) \\ =& {{ 1 } \over { \pi }} \lim_{s \to 0} \sin \left( {{ \pi s } \over { 2 }} \right) \zeta (1-s) \\ =& {{ 1 } \over { \pi }} \lim_{s \to 0} \left[ {{ \pi s } \over { 2 }} - {{ \pi^{3} s^{3} } \over { 48 }} + \cdots \right] \left[ -{{ 1 } \over { s }} + \sum_{n=0}^{\infty} \gamma_{n} {{ s^{n} } \over { n! }} \right] \\ =& {{ 1 } \over { \pi }} \lim_{s \to 0} \left[ - {{ \pi } \over { 2 }} + O (s) \right] \\ =& - {{ 1 } \over { 2 }} \end{align*}