Bernoulli numbers are the coefficients of the formula of a sum of powers of natural numbers.
For example, a formula of a sum of natural numbers can be expressed by two Bernoulli numbers.
$$
1 + 2 + 3 + \cdots + n=\frac{1}{2}B_0 n^2 + B_1 n^1
$$
Zeta function can be expressed by these Bernoulli numbers.
$$
\zeta(1s) =  \frac{B_s}{s}
$$
I explain the mechanism that the sum of natural numbers of the zeta function converges to 1/12 on this page.
•Why doesn't the sum of natural numbers of the zeta function diverge to infinity? [2014/2/8]
Mathematician Pietro Mengoli of Italy issued a problem of the following series in 1644.
$$ \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots $$Mathematician Jacob Bernoulli of Switzerland tried to obtain the value, but he couldn't. This problem has been referred to as the Basel problem since his hometown was the Basel. Euler who was 28 years old solved this difficult problem in 1735, and he became famous suddenly. The value is shown below.
$$ \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6} $$Because the value contains the ratio of a circle's circumference to its diameter, this is a surprising result. He got a more surprising result in 1749.
This expression is very mysterious. I wonder why it doesn't diverge to infinity. Mathematician Ramanujan of India calculated it in the following way in 1913.
At first, we will calculate the following values.
We will shift and add it.

Then, we will calculate the following values.
We will shift and add it.

Finally, we will calculate the following values.
We multiply the above formula by four.
We subtract 4S from S as follows.

If the sum of natural numbers converges the above result is correct. But the sum doesn't converge the result is not correct. How can we have the sum converged? I think the sum of natural numbers is damped and oscillating very slowly as the following graph.
We define the following damped oscillation function.
$$ H_n(x)=\sum_{k=1}^n k \exp(kx)\cos(kx) $$The function equals to the sum of natural numbers for the finite n at the limit of x = 0.
$$ \lim_{x \to 0 + }\sum_{k=1}^n k \exp(kx)\cos(kx) = 1 + 2 + 3 + \cdots + n $$However, the function equals to 1/12 for the infinite n.
$$ \lim_{x \to 0+} \sum_{k=1}^\infty k \exp(kx)\cos(kx)=\frac{1}{12} $$We are able to confirm the surprising result by the online calculator WolframAlpha. If you click the following URL, you will calculate it.
I would like to explain the mechanism in the following article.
I would like to explain the mechanism in the following paper for the person who is familiar with mathematics.
We can calculate the Casimir effect without divergence by using the result of the paper.
NS found the following formula and published it in 2015/3/27.
$$ \zeta(12t)=\lim_{x \to 0 + } \sum_{k=1}^\infty \frac{1}{k^{12t}} \exp(k^tx)\cos(k^tx) $$ For example, ζ(3)=1/120 is shown below.Riemann defines the following zeta function in 1859
$$ \zeta(s)=\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots $$Then, Riemann defined the following analytic continuation of the above zeta function.
$$ \zeta(s)=\Gamma(1s)\oint_{C} \frac{dz}{2\pi iz} z^s \frac{e^z}{1e^z} $$We call the above function global zeta function. Generally, normal zeta function is different from the global zeta function. So,we express the normal zeta function by using a letter Z in order to distinguish two functions.
$$ Z(s)=\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots $$He had the following reflection formula for the global zeta function.
$$ \zeta(1s) = \frac{2}{(2\pi)^s}\;\;\Gamma(s)\;\;\cos \Bigl(\frac{\pi}{2}s \Bigr) \; \zeta(s) $$We can derive the −1/12 of the mysterious formula from the reflection formula. We can see the graph of the global zeta function in the following page.
The picture at the beginning of this page is the above picture rotated 90 degrees to the right and enlarged. It seems like a mountain with the background of the blue sky.
Riemann conjectured the following thing.
This is called "Riemann Hypothesis." Some of nontrivial zeros ρ_{m} are shown below.
Have you calculated the sum of the powers of the natural numbers? For example, the sum of n natural numbers of powers 0, 1, 2, ..., 6 are shown below.
$$ 1^0 + 2^0 + 3^0 + \cdots + n^0 = \frac{1}{1} n^1 $$ 
$$ 1^1 + 2^1 + 3^1 + \cdots + n^1 = \frac{1}{2} n^2 + \frac{1}{2} n^1 $$ 
$$ 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{1}{3} n^3 + \frac{1}{2} n^2 + \frac{1}{6} n^1 $$ 
$$ 1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{1}{4} n^4 + \frac{1}{2} n^3 + \frac{1}{4} n^2 $$ 
$$ 1^4 + 2^4 + 3^4 + \cdots + n^4 = \frac{1}{5} n^5 + \frac{1}{2} n^4 + \frac{1}{3} n^3\frac{1}{30} n^1 $$ 
$$ 1^5 + 2^5 + 3^5 + \cdots + n^5 = \frac{1}{6} n^6 + \frac{1}{2} n^5 + \frac{5}{12} n^4\frac{1}{12} n^2 $$ 
$$ 1^6 + 2^6 + 3^6 + \cdots + n^6 = \frac{1}{7} n^7 + \frac{1}{2} n^6 + \frac{1}{2} n^5\frac{1}{6} n^3 + \frac{1}{42} n^1 $$ 
Jacob Bernoulli published the Bernoulli numbers in 1713. They are numbers such as the following.
n  B_{n} 

0  1 
1  1/2 
2  1/6 
3  0 
4  1/30 
5  0 
6  1/42 
We can express the sum of n natural numbers of powers 0, 1, 2, ..., 6 by Bernoulli numbers as follows.
$$ 1^0 + 2^0 + 3^0 + \cdots + n^0=\frac{1}{1}B_0 n^1 $$ 
$$ 1^1 + 2^1 + 3^1 + \cdots + n^1=\frac{1}{2}B_0 n^2 + B_1 n^1 $$ 
$$ 1^2 + 2^2 + 3^2 + \cdots + n^2=\frac{1}{3}B_0 n^3 + B_1 n^2 + \frac{2!}{1!2!}B_2 n^1 $$ 
$$ 1^3 + 2^3 + 3^3 + \cdots + n^3=\frac{1}{4}B_0 n^4 + B_1 n^3 + \frac{3!}{2!2!}B_2 n^2 $$ 
$$ 1^4 + 2^4 + 3^4 + \cdots + n^4=\frac{1}{5}B_0 n^5 + B_1 n^4 + \frac{4!}{3!2!}B_2 n^3 + \frac{4!}{1!4!}B_4 n^1 $$ 
$$ 1^5 + 2^5 + 3^5 + \cdots + n^5=\frac{1}{6}B_0 n^6 + B_1 n^5 + \frac{5!}{4!2!}B_2 n^4 + \frac{5!}{2!4!}B_4 n^2 $$ 
$$ 1^6 + 2^6 + 3^6 + \cdots + n^6=\frac{1}{7}B_0 n^7 + B_1 n^6 + \frac{6!}{5!2!}B_2 n^5 + \frac{6!}{3!4!}B_4 n^3 + B_6 n^1 $$ 
Bernoulli numbers have two kinds of definition. When the Bernoulli polynomial is B_{n} (x ), they are the following two kinds of definition.
In this page, we adopt the latter definition in order to unite the Bernoulli function that will be explained later.
Generally, we have the following equation for natural number n.
$$ \zeta(n) =  \frac{B_{n+1}}{n+1} $$S. C. Woon of University of Cambridge proposed the following formula for complex s in 1997.
$$ \zeta(1s) =  \frac{B(s)}{s} $$He interpreted Bernoulli numbers B_{n} as Bernoulli function B (s). We call the above formula "formula of Bernoulli function." We can see the graph in the following page.
In this page we express Bernoulli function B (s) as B_{s} based on the traditional notation of Bernoulli numbers.
We can obtain the following equation by substituting "formula of Bernoulli function" to "Riemann's reflection formula."
$$ \zeta(s) = \;\;\frac{(2\pi)^s}{2}\;\;\frac{1}{\Gamma(s+1)} \;\; \frac{1}{\cos(\pi s/2)}\;\;B_s $$The above formula becomes the following formula for an even positive integer s.
$$ \zeta(s) = \;\; \frac{(2\pi)^s}{2}\;\; \frac{1}{s!} \;\;(1)^{\frac{s}{2}+1}\;\; B_s $$The above formula equals to the following formula for even positive integer n.
$$ \zeta(n) = \;\;\frac{(2\pi)^n}{2}\;\; \frac{1}{n!}\;\;(1)^{\frac{n}{2}+1}\;\; B_n $$The above result suggests us that the validity of "Formula of Bernoulli function."
We can derive reflection integral equation by the complex analysis.
$$ \zeta(1s)=\oint_{S^1} \frac{dt}{2\pi i} B(s,t) \zeta(t) $$In the above equation, we used Euler's beta function B (x, y). The method to derive it is explained in the following paper.
We can derive reflection integral equation by the quaternionic analysis.
$$ \zeta(1s)=\oint_{S^1} \frac{dt}{2\pi u} B(s,t) \zeta(t) $$u is the unit quaternion. The method to derive it is explained in the following paper.
We might approach to the Riemann hypothesis by studying the eigenvalues of the reflection integral equation. I would like to remind you that superstring theory began from beta function.
I will explain a new proof in the following paper.
This paper introduced a finite harmonic derivative.
$$ H^{(s)}_n(z) = \sum_{k=1}^n \frac{k^s}{k} \exp(kz) $$A limit of the finite harmonic derivative becomes the harmonic number.
$$ H_n = \sum_{k=1}^n \frac{1}{k} = \lim_{s \to 0} \lim_{z \to 0}H^{(s)}_n(z) $$A limit of the finite harmonic derivative becomes the zeta function.
$$ \zeta(s) = \sum_{k=1}^n \frac{1}{k^s} = \lim_{z \to 0} \lim_{n \to \infty}H^{(1s)}_n(z) $$A limit of the finite harmonic derivative becomes the partition function.
$$ Z(z) = \sum_{k=1}^\infty e^{kz} = \lim_{s \to 1} \lim_{n \to \infty}H^{(s)}_n(z) $$The relation of the functions is shown in the following figure.
Then, zeta function is shown below.
We define the following wave zeta function by quantizing this zeta function.
$$ Z_q(s) = \sum_{k=1}^\infty \frac{1}{k^s} \exp(kq) $$ $$ Z(s) = \sum_{k=1}^\infty \frac{1}{k^s} $$We define the following mean wave zeta function by contour averaging this wave zeta function.
$$ \zeta_q(s)=\oint_{C} \frac{dz}{2\pi i(zq)}Z_z(s) $$Classical limit of the above mean wave zeta function becomes global zeta function.
$$ \lim_{q\to0}\zeta_q(s)=\zeta(s) $$So we can interpret the analytic continuation as an operation to quantize a function and to calculate the contour meanvalue of the function and to calculate a classical limit of the function.
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