# Why is RH optimum?

### Re: Why is RH optimum?

Some Food for Thought:

'Prime-counting function', by
https://en.m.wikipedia.org/wiki/Prime-counting_function;

'Computing $$\pi(x)$$ Analytically' by
Prof. David J. Platt.

Please see attached file for details.
Attachments 1203.5712.pdf
Guest

### Re: Why is RH optimum?

Guest wrote:Given the values, $$z_{n}$$ and $$p_{n}$$, how do we correctly derive equation two,

2. $$f(n, p_n, z_n) =$$ $$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}}$$, \Leftrightarrow

from equation one,

1. $$\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}$$ = 0?

We hope that our latest question is a big question which leads us eventually to deeper insights about Riemann's great work on prime number theory: his lofty quest to prove the Prime Number Theorem and his very lofty quest to compute most accurately the number of primes less than or equal to any appropriate real value, x.

The derivation question is not difficult to answer as we shall soon see.

$$\zeta(z_{n} ) = \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}}$$
=
Guest

### Re: Why is RH optimum?

We are having some technical difficulties cutting and pasting... We shall try again tomorrow.
Guest

### Re: Why is RH optimum?

Given the values, $$z_{n}$$ and $$p_{n}$$, how do we correctly derive equation two,

2. $$f(n, p_n, z_n) =$$ $$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}}$$,

from equation one,

1. $$\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}$$ = 0?

We hope that our latest question is a big question which leads us eventually to deeper insights about Riemann's great work on prime number theory: his lofty quest to prove the Prime Number Theorem and his very lofty quest to compute most accurately the number of primes less than or equal to any appropriate positive real value, x.

The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and $$p_n$$), Simple Nontrivial Zeros (z and $$z_n$$) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p $$≤ k^{1/2}$$.

HS (v1): $$\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n}$$;

RZF (v1): $$\ \ \zeta(z_{n} ) = \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}}$$.

HS (v2): $$\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n}$$;

RZF (v2): $$\ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ \ = \ \ \sum_{n=1}^{\infty }\frac{1}{p_n^{z_n}} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{(j*p_n)^{z_n}}$$.

We need to check our derivations. And we have more work...
Guest

### Re: Why is RH optimum?

Guest wrote:Given the values, $$z_{n}$$ and $$p_{n}$$, how do we correctly derive equation two,

2. $$f(n, p_n, z_n) =$$ $$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}}$$,

from equation one,

1. $$\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}$$ = 0?

We hope that our latest question is a big question which leads us eventually to deeper insights about Riemann's great work on prime number theory: his lofty quest to prove the Prime Number Theorem and his very lofty quest to compute most accurately the number of primes less than or equal to any appropriate positive real value, x.

The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and $$p_n$$), Simple Nontrivial Zeros (z and $$z_n$$) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p $$≤ k^{1/2}$$.

HS (v1): $$\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty$$;

RZF (v1): $$\ \ \zeta(z_{n} ) = \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0$$.

HS (v2): $$\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty$$;

RZF (v2): $$\ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ \ = \ \ \sum_{n=1}^{\infty }\frac{1}{p_n^{z_n}} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{(j*p_n)^{z_n}} = 0$$.

We need to check our derivations. And we have more work...
Guest

### Re: Why is RH optimum?

Guest wrote:
Guest wrote:Given the values, $$z_{n}$$ and $$p_{n}$$, how do we correctly derive equation two,

2. $$f(n, p_n, z_n) =$$ $$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}}$$,

from equation one,

1. $$\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}$$ = 0?

We hope that our latest question is a big question which leads us eventually to deeper insights about Riemann's great work on prime number theory: his lofty quest to prove the Prime Number Theorem and his very lofty quest to compute most accurately the number of primes less than or equal to any appropriate positive real value, x.

The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and $$p_n$$), Simple Nontrivial Zeros (z and $$z_n$$) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p $$≤ k^{1/2}$$.

HS (v1): $$\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty$$;

RZF (v1): $$\ \ \zeta(z_{n} ) = \ \ \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0$$.

HS (v2): $$\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \ \ 1 + \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty$$;

RZF (v2): $$\ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ = \ \ 1 + \ \ \sum_{n=1}^{\infty }\frac{1}{p_n^{z_n}} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{(j*p_n)^{z_n}} = 0$$.

We need to check our derivations. And we have more work...
Guest

### Re: Why is RH optimum?

Given the values, $$z_{n}$$ and $$p_{n}$$, how do we correctly derive equation two,

2. $$f(n, p_n, z_n) =$$ $$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}}$$,

from equation one,

1. $$\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}$$ = 0?

We hope that our latest question is a big question which leads us eventually to deeper insights about Riemann's great work on prime number theory: his lofty quest to prove the Prime Number Theorem and his very lofty quest to compute most accurately the number of primes less than or equal to any appropriate positive real value, x.

The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and $$p_n$$), Simple Nontrivial Zeros (z and $$z_n$$) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p $$≤ k^{1/2}$$.

HS (v1): $$\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty$$;

RZF (v1): $$\ \ \zeta(z_{n} ) = \ \ \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0$$.

HS (v2): $$\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \ \ 1 + \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty$$;

RZF (v2): $$\ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ = \ \ 1 + \ \ \sum_{n=1}^{\infty }\frac{1}{p_n^{z_n}} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{(j*p_n)^{z_n}} = 0$$.

We need to check our derivations. And we have more work...

Confirmed Observations:

The Riemann Hypothesis is extraordinarily exceptional since all nontrivial and simple zeros of RZF are accounted for. And there are no exceptions or 'stray' (off the critical line, Re(z) = 1/2) nontrivial zeros possible.

Furthermore, we have the following equations too:

$$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ = \infty$$ and $$\ \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty$$;

$$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ = \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0$$;

$$\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty$$ and $$\ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty$$;

$$\sum_{n=1}^{\infty }\frac{1}{p_n^{z_n}} \ \ + \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{(j*p_n)^{z_n}} = -1$$.

Dave.
Guest

### Re: Why is RH optimum?

Notes:

From the equations,

$$\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty$$ (Euler's Proof) and $$\ \ \ \ \sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty$$,

we compute,

$$\lim_{n \to \infty} \frac{\sum_{n=1}^{\infty }\frac{1}{p_n}}{\sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n}} \rightarrow 0$$.

https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes
Guest

### Re: Why is RH optimum?

Oops!

The equation,

$$\sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty$$, is wrong!

The above equation is adding too many (infinitely many) identical composite numbers... We shall fix the problem and post a revision soon.
Guest

### Re: Why is RH optimum?

Guest wrote:Oops!

The equation,

$$\sum_{j=2}^{\infty } \sum_{n=1}^{\infty }\frac{1}{j*p_n} = \infty$$, is wrong!

The above equation is adding too many (infinitely many) identical composite numbers... We shall fix the problem and post a revision soon.

The correct equation is

$$\sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m^z}} = \infty$$.
Guest

### Re: Why is RH optimum?

An Update:

Given the values, $$z_{n}$$ and $$p_{n}$$, how do we correctly derive equation two,

2. $$f(n, p_n, z_n) =$$ $$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}}$$,

from equation one,

1. $$\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}$$ = 0?

We hope that our latest question is a big question which leads us eventually to deeper insights about Riemann's great work on prime number theory: his lofty quest to prove the Prime Number Theorem and his very lofty quest to compute most accurately the number of primes less than or equal to any appropriate positive real value, x.

The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and $$p_n$$), Simple Nontrivial Zeros (z and $$z_n$$) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p $$≤ k^{1/2}$$.

HS (v1): $$\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty$$;

RZF (v1): $$\ \ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0$$.

HS (v2): $$\sum_{k=1}^{\infty }\frac{1}{k} \ = \ \ 1 \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} \ \ + \ \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ \ = \infty$$;

RZF (v2): $$\ \ \zeta(z) = \ \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ = \ \ 1 \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m^z}} \ \ + \ \ \ \sum_{n=1}^{\infty }\frac{1}{{p_n}^{z}} \ \ \ = 0$$.

We need to check our derivations. And we have more work...

Confirmed Observations:

The Riemann Hypothesis is extraordinarily exceptional since all nontrivial and simple zeros of RZF are accounted for. And there are no exceptions or 'stray' (off the critical line, Re(z) = 1/2) nontrivial zeros possible.

Furthermore, we have the following equations too:

$$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ = \infty$$ and $$\ \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty$$;

$$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ = \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0$$;

$$\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty$$ and $$\ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} = \infty$$;

$$\sum_{n=1}^{\infty }\frac{1}{p_n^{z}} \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m^z}} \ \ = -1$$.

Notes:

From the equations,

$$\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty$$ (Euler's Theorem) and $$\ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} = \infty$$,

we compute as $$n \rightarrow \infty$$,

$$\ \ \ \frac{\sum_{n=1}^{\infty }\frac{1}{p_n}}{ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}}} \ \ \rightarrow 0$$.

https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes
Guest

### Re: Why is RH optimum?

Guest wrote:An Update:

Given the values, $$z_{n}$$ and $$p_{n}$$, how do we correctly derive equation two,

2. $$f(n, p_n, z_n) =$$ $$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}}$$,

from equation one,

1. $$\zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}$$ = 0?

We hope that our latest question is a big question which leads us eventually to deeper insights about Riemann's great work on prime number theory: his lofty quest to prove the Prime Number Theorem and his very lofty quest to compute most accurately the number of primes less than or equal to any appropriate positive real value, x.

The derivation question is not difficult to answer as we shall soon see.

Keywords/Key Ideas: Fundamental Theorem of Arithmetic (FTA), Basic Fact of Primality, Harmonic Series (HS), Riemann Zeta Function (RZF), Simple Primes (p and $$p_n$$), Simple Nontrivial Zeros (z and $$z_n$$) of RZF, Critical Line, and Riemann Hypothesis (RH).

We shall be very mindful of the following basic fact of primality while pondering the following derivations of HS and RZF and the truth of RH.

Basic Fact of Primality:
For all k > 1, where k is any positive integer, there exists a prime number, p, such that p|k
where p $$≤ k^{1/2}$$.

HS (v1): $$\ \ \sum_{k=1}^{\infty }\frac{1}{k} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty$$;

RZF (v1): $$\ \ \ \zeta(z_n) = \sum_{k=1}^{\infty }\frac{1}{k^{z_{n}}} \ \ = \sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ + \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0$$.

HS (v2): $$\sum_{k=1}^{\infty }\frac{1}{k} \ = \ \ 1 \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} \ \ + \ \ \ \sum_{n=1}^{\infty }\frac{1}{p_n} \ \ \ = \infty$$;

RZF (v2): $$\ \ \zeta(z) = \ \sum_{k=1}^{\infty }\frac{1}{k^{z}} \ = \ \ 1 \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m^z}} \ \ + \ \ \ \sum_{n=1}^{\infty }\frac{1}{{p_n}^{z}} \ \ \ = 0$$.

We need to check our derivations. And we have more work...

Confirmed Observations:

The Riemann Hypothesis is extraordinarily exceptional since all nontrivial and simple zeros of RZF are accounted for. And there are no exceptions or 'stray' (off the critical line, Re(z) = 1/2) nontrivial zeros possible.

Furthermore, we have the following equations too:

$$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m} \ \ = \infty$$ and $$\ \ \ \sum_{l=1}^{\infty }\frac{1}{l*p_n} = \infty$$;

$$\sum_{∀m\ge 1\ s.t. \ gcd(p_n, m)=1}^{\infty} \frac{1}{m^{z_n}} \ \ = \ \ \sum_{l=1}^{\infty }\frac{1}{(l*p_n)^{z_n}} = 0$$;

$$\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty$$ and $$\ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} = \infty$$;

$$\sum_{n=1}^{\infty }\frac{1}{p_n^{z}} \ \ + \ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m^z}} \ \ = -1$$.

Notes:

From the equations,

$$\sum_{n=1}^{\infty }\frac{1}{p_n} \ \ = \infty$$ (Euler's Theorem) and $$\ \ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}} = \infty$$,

we compute as $$n \rightarrow \infty$$,

$$\ \ \ \frac{\sum_{n=1}^{\infty }\frac{1}{p_n}}{ \sum_{\forall m>1 \ s.t. \ m\notin \mathbb{P} }^{\infty }{\frac{1}{m}}} \ \ \rightarrow 0$$.

https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes
Guest

### Re: Why is RH optimum?

I apologize for the sloppy (mistaken) mathematics posted here. As we know, mathematical research is hard work. So please continue to help me find my flaws whenever possible. Thank you!

The proof of Riemann Hypothesis is the cumulative work of many outstanding or great mathematicians, Euler, Gauss, Riemann, Hardy, Littlewood, Siegel, Erdős, Montgomery (Go Blue!), Lagarias (Go Blue!), Odlyzko, and others. And I am trying my best to make a significant contribution too.

Thank you for your patience! And Go Blue!

David Cole.
Guest

### Re: Why is RH optimum?

We conclude that the Riemann Hypothesis is true! Do you agree?

David Cole,

https://www.researchgate.net/profile/David_Cole29.
Guest

### Re: Why is RH optimum?

Guest wrote:We conclude that the Riemann Hypothesis (RH) is true! Do you agree?

David Cole,

https://www.researchgate.net/profile/David_Cole29.

If we are still in doubt about the validity of RH, let's do some computations.

Let z = a + i * b, and let $$\bar{z}$$ = a - i * b. Note: $$\bar{z}$$ is the complex conjugate of z.

Therefore, z + $$\bar{z}$$ = 1 which implies 2a = 1 which implies a = Re(z) = Re($$\bar{z})$$ = 1/2 which implies RH, and we are done.

Note: z + $$\bar{z}$$ = 1 implies the Harmonic Series, the prime source of our previous analysis from past posts.

However, if we are still in doubt about the validity of RH, let's do the harder computations.

From equation one,

1. $$\ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}$$ = 0,

the RH doubters need to solve the following system of two derived equations of equation one:

1A. $$\ \ \ \sum_{k=2}^{\infty }\frac{1}{k^{a}} * cos(b * log(k))$$ = -1;

1B. $$\ \ \ \sum_{k=2}^{\infty }\frac{1}{k^{a}} * sin(b * log(k))$$ = 0;

And the result is again, a = Re(z) = 1/2 for all the different values of b. And there are infinitely many distinct values of b. Thus, RH is true!
Guest

### Re: Why is RH optimum?

\le
Guest wrote:
Guest wrote:We conclude that the Riemann Hypothesis (RH) is true! Do you agree?

David Cole,

https://www.researchgate.net/profile/David_Cole29.

If we are still in doubt about the validity of RH, let's do some computations.

Let z = a + i * b, and let $$\bar{z}$$ = a - i * b. Note: $$\bar{z}$$ is the complex conjugate of z.

Therefore, z + $$\bar{z}$$ = 1 which implies 2a = 1 which implies a = Re(z) = Re($$\bar{z})$$ = 1/2 which implies RH, and we are done.

Note: z + $$\bar{z}$$ = 1 implies the Harmonic Series, the prime source of our previous analysis from past posts.

However, if we are still in doubt about the validity of RH, let's do the harder computations.

From equation one,

1. $$\ \ \zeta(z) = \sum_{k=1}^{\infty }\frac{1}{k^{z}}$$ = 0,

the RH doubters need to solve the following system of two derived equations of equation one:

1A. $$\ \ \ \sum_{k=2}^{\infty }\frac{1}{k^{a}} * cos(b * log(k))$$ = -1;

1B. $$\ \ \ \sum_{k=2}^{\infty }\frac{1}{k^{a}} * sin(b * log(k))$$ = 0.

And the result is again, a = Re(z) = 1/2 for all the different values of b. And there are infinitely many distinct values of b. Thus, RH is true!

An Update:

1A. $$\ \ \ \sum_{k=2}^{\infty }\frac{cos(b * log(k) + 2j\pi)}{k^{a}}= -1$$ ;

1B. $$\ \ \ \sum_{k=2}^{\infty }\frac{sin(b * log(k)+ 2j\pi)}{k^{a}}= 0$$

where $$0 \le a \le 1$$ (Riemann's Theorem) for all integers, $$j\ge 0$$.
Guest

### Re: Why is RH optimum?

"Nature is a grand optimizer, and therefore, one and only one critical line, Re(z) = 1/2 (the Riemann Hypothesis), is enough to define all primes and all nontrivial zeros of the Riemann Zeta Function." Hmm. Do you agree?
Attachments RH Critical Line.png (11.77 KiB) Viewed 80 times
Guest

### Re: Why is RH optimum?

Hmm.

1A. $$\ \ \ \sum_{k=2}^{\infty }\frac{cos(b * log(k) + 2\pi j)}{k^{a}}= -1$$ ;

1B. $$\ \ \ \sum_{k=2}^{\infty }\frac{sin(b * log(k)+ 2\pi j)}{k^{a}}= 0$$

where $$0 \le a \le 1$$ (Riemann's Theorem) for all integers, $$j\ge 0$$.

Are there some or infinitely many values of j which make equations, 1A and 1B, false?
Guest

### Re: Why is RH optimum?

Guest wrote:Hmm.

1A. $$\ \ \ \sum_{k=2}^{\infty }\frac{cos(b * log(k) + 2\pi j)}{k^{a}}= -1$$ ;

1B. $$\ \ \ \sum_{k=2}^{\infty }\frac{sin(b * log(k)+ 2\pi j)}{k^{a}}= 0$$

where $$0 \le a \le 1$$ (Riemann's Theorem) for all integers, $$j\ge 0$$.

Are there some or infinitely many values of j which make equations, 1A and/or 1B, false?
Guest

### Re: Why is RH optimum?

Hmm. We assume $$\ \ a = \frac{1}{2}$$, and we consider the solutions of the following equations:

1A. $$\ \ \ \sum_{k=2}^{\infty }\frac{cos(b * log(k) + 2\pi j)}{k^{a}}= -1$$ ;

1B. $$\ \ \ \sum_{k=2}^{\infty }\frac{sin(b * log(k)+ 2\pi j)}{k^{a}}= 0$$

for all integers, $$j\ge 0$$.

Are there some or infinitely many values of j which make equations, 1A and/or 1B, false?
Guest

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