"... deepest interrelationships in analysis are of  an arithmetical nature."                    -- Hermann Minkowski.

In the simplest sense, the Riemann Hypothesis (RH) confirms for all positive integers,

k > 1, there exists a prime, p, which divides k such that either

p = k

or

p ≤ k(1/2 = a = Re(s = a ± b * i) )

where s is the nontrivial zero of the Riemann zeta function.

Furthermore, only s + s' = 1 over 0 ≤ a ≤ 1 satisfies the  divergent Euler zeta function (the Harmonic Series) for which there exists infinitely prime numbers (see Euler's proof)  where a = 1/2.  

Note:  s = a + b * i and s' = a - b* i are nontrivial zeta zeros of the Riemann zeta function in the interval, 0 ≤ a ≤ 1.  

And Σ (1 / ks  * 1 / ks' =  1/k(s + s') = 1/k1 =  1/k) (from k = 1 to k = ∞) =  ∞ according to the Fundamental Theorem of Arithmetic and according to Euler's Proof.

Hence, s = 1/2 + b * i and s' = 1/2 - b* i are the only  nontrivial zeta zeros of the Riemann zeta function in the interval, 0 ≤ a ≤ 1.

"When the solution is simple, GOD is answering."  -- Albert Einstein.

Keywords:   Euclid's Theorem (ET) on the existence of infinitely many distinct prime numbers, Fundamental Theorem of Arithmetic (FTA),  and a simple proof which demonstrates the Harmonic Series is a source of all prime numbers.

A simple proof which demonstrates the Harmonic Series (H) is a source of all prime numbers:

By definition, H is Σ 1/k  (from k = 1 to k = ∞) = Γ.  

And we know  Γ >> 1  for some k > 1.

We let η = (∏ k) (from k = 1 to k = ∞) = ∏ piαi (from i = 1 to i = ∞)  =  ∞  where αi → ∞  as  k → ∞ and where each pi is a distinct prime number according to ET and FTA.

Hence, (∏ k) (from k = 1 to k = ∞) * Σ 1/k (from k = 1 to k = ∞)

= Γ * (∏ k) (from k = 1 to k = ∞).

But,   (∏ k) (from k = 1 to k = ∞) * Σ 1/k (from k = 1 to k = ∞)

= Σ(∏ k) / k (from k = 1 to k = ∞) = Σ η/k (from k = 1 to k = ∞)

= Γ * (∏ k) (from k = 1 to k = ∞) = Γ * η.

Therefore, Σ η/k (from k = 1 to k = ∞) = ∞2 >> η = ∞.

Thus,  1/η * Σ η/k (from k = 1 to k = ∞) =  ∞2 / ∞

 =  Σ 1/k (from k = 1 to k = ∞) = Γ = ∞ (q.e.d.)

Note:  Σ η/k (from k = 1 to k = ∞) = ∏ pjθj (from j = 1 to j = ∞) = ∞

where θj → ∞ as k → ∞ and where each pj is a distinct prime number according to ET and FTA.  

Since Σ η/k (from k = 1 to k = ∞) = ∞2 >> η = ∞.,  

η divides  Σ η/k (from k = 1 to k = ∞). 

*****

Hmm.  Think H2O,  the sound Polignac Conjecture, and the Prime Parity Law!    :-)

https://www.researchgate.net/post/Does_the_nth_nontrivial_simple_zero_of_the_Riemann_zeta_function_indicate_the_nth_prime_p_n_occurs_as_a_prime_factor_in_all_multiples_of_p_n

http://www.math.lsa.umich.edu/~lagarias/doc/elementaryrh.pdf

https://en.wikipedia.org/wiki/Riemann_zeta_function

Riemann zeta function - Wikipedia, the free encyclopedia 

https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Harmonic series (mathematics) - Wikipedia, the free encyclopedia 

http://www.math.lsa.umich.edu/~lagarias/doc/mt-holyoke-rev.pdf

From the Riemann zeta function, Σ 1 / ks (from k= 1 to k= ∞), equalling zero or    0 + 0 * i, we derived the following equations:

1.   Σ cos (b * log (k) ) / ka (from k = 2 to k = ∞) = -1

and

2.   Σ sin (b * log (k) ) / ka (from k = 2 to k = ∞) = 0

where 0 ≤ a ≤ 1 and

where s = a ± b*i is a complex number, and s is also the nontrivial zero of the Riemann zeta function.

You have two equations, one and two, with two unknowns, a and b.

*****

In regards to equations, one and two, there's uniqueness associated with the variable a which is of degree one, and there's periodicity associated with the variable b indirectly which is also of degree one. So, whatever is the true value of variable a, it is unique and the only one.  And we know what the true value of variable a is. Variable, a, equals one-half (a=1/2).

And because of the periodicity of the trigonometric functions, cosine and sine, there are infinitely many b values. Each b value is unique, and each one is associated with a = 1/2. So, s = 1/2 +/- b *i   solves the Riemann zeta function.

Conclusions: These observations confirm my proof of the Riemann Hypothesis and all empirical evidence which supports the Riemann Hypothesis.

Furthermore, all nontrivial zeros of the Riemann zeta function are simple.

*****

Keywords:  Equation One and Equation Two (see above), Fundamental Theorem of Arithmetic (FTA) and the Existence of  Essential Prime Number(s) (EEPN), less than or equal to k1/2,  which divide(s) all composite positive integers, k.

Hmm.  Now multiply the equations, one and two,  by the expression, 

(∏ k)a  (from k = 2 to k = ∞).  

And let  λk  = (∏ k)a over all k ∈ N \ {1, k}   where N is the set of all natural numbers or positive integers.  We recall: 0 ≤ a ≤ 1 or a ∈ [0, 1] .

We generate the following two equations:

3.    | Σ λk * cos(b * log (k)) (from k = 2 to k = ∞) |

=  (∏ k)a  (from k = 2 to k = ∞) 

=  β * ∏ pi  (from i = 1 to i = ∞);

Therefore,  

|Σ λk /(∏ pi (from i = 1 to i = ∞) ) * cos(b * log (k)) (from k = 2 to k = ∞) |     = β;

Let βk = λk /(∏ pi (from i = 1 to i = ∞) ).

Note: βk is an integral number if and only if  the variable, a, is a rational number in (0,1).

Therefore, 

4.    | Σ βk * cos(b * log (k)) (from k = 2 to k = ∞) |  =  β,

for all distinct prime numbers, pi,  where

p1 = 2,  p2 = 3,  p3 = 5,  p4 = 7,  ..., ∞,  and where the

integral number, β,  if and only if a = 1/2, is such that β >> k for all k ∈ N.

(Wow!!!...  Praise Lord GOD!!!...  Amen!  :) )

And by similar algebraic manipulation, equation two is transformed to the following equation:

5.    Σ βk * sin(b * log (k)) (from k = 2 to k = ∞) = 0.

Note:  If a ≠ 1/2 in the open interval, (0, 1), then β is an irrational number such that β >> k for all k ∈ N.   Can we demonstrate this important result inductively?    

[Research Project To Answer Above Question:  

We suggest researching the finite cases when  β  is finite with k values which have bounded limits (from k = 2 to some k ∈ N\{1, 2}) which forces β to converge to an integral number according to equations, four and five.  

And determine what the expected zeros (1/2 +/- b * i) are.

How do they compare with the empirical zeta zeros of the Riemann zeta function where k goes to infinity?]

We recall equations, four and five.

4.    | Σ βk * cos(b * log (k)) (from k = 2 to k = ∞) | = β;

5.    Σ βk * sin(b * log (k)) (from k = 2 to k = ∞)    =   0,

where  λk = (∏ k)a over all k ∈ N \ {1, k} with a ∈ (0, 1)

and where  βk =  λk /(∏ pi (from i = 1 to i = ∞) ).

In the important equation five,  the LHS (left hand side) reduces to zero for the right values of variables, a and b, so that in the other important equation four,  the LHS and RHS (right hand side) determine the unique and optimal value of  the variable, a, which is a = 1/2 for all (infinitely many and distinct) appropriate values of variable, b,  according to FTA and according to EEPN.

We hope this additional explanation helps us to accept the validity of the Riemann Hypothesis.  

"... deepest interrelationships in analysis are of an arithmetical nature."         -- Hermann Minkowski.

*****

Romans 8:28, http://biblia.com/verseoftheday/image/Ro8.28

My old account was deleted, but I have returned.  And my proof of RH is true!  Thank Lord GOD!  Amen!   :-)

Reference link:  https://www.researchgate.net/post/Why_is_the_Riemann_Hypothesis_true

RH is concerned with the values of zeta(s) when s is in the so-called critical strip. But the series expansion \sum_n 1/n^s is not valid there. The series converges for Re(s)>1, but not in the critical strip.

@Patrick Solé  

Obviously, you do not understand  my proof.  And I cannot help you.   Ask an expert to assist you, or  just drop the matter ...

@David Cole: obviously you do not understand analytic continuation or RH for that matter. zeta(s) is NOT equal to \sum_n 1/n^s  when s is a nontrivial zero. Sorry.

Patrick Solé

Not true!  And we agree to disagree. You do not accept my proof of RH since you do not understand it.  I accept that.  Now, let it go ...

@David Cole: I am not going to read your long  proof if the statement in the question is already WRONG. Unless its a kind of sophisticated pedagogical joke or thought experiment. There is a series expression for zeta in the critical strip and it is NOT \sum_n 1/n^s. It is an elementary fact that \sum_n 1/n^s diverges for Re(s)

@David Cole: if you do not accept rational criticism you are not a scientist, you are just a dreamer.

In a dream, I proved the famous and important Riemann Hypothesis. The dream came true! Thank GOD!  Amen!  :-)

Calculating with divergent series is a mathematical SIN!

The Riemann Hypothesis (RH) is true!  

It's a SIN to think otherwise.

Q1.  Where are all the primes in the critical strip, [0, 1] ?  

Q2.  Where are all the powers of  primes in the critical strip, [0, 1]?  

Q3.  How are prime numbers and the nontrivial zeros of the Riemann zeta function connected?

Hints:  

Please consider  the Fundamental Theorem of Arithmetic, the Generalized Fundamental Theorem of Algebra,  the Harmonic Series, and the very important PRIME NUMBER THEOREM since we are counting primes and predicting primes approximately.

If one can answer the above questions, then one understands why the Riemann Hypothesis is true!

primes in the critical strip? All primes are >2.

Hint for Q1:    1/2 , 1/3, 1/5, 1/7, 1/11, ...., 1/p  for some prime distinct number, p.  

Hint for Q2:    We have  1/n  where  n = ∏ pik_i   over distinct prime numbers, pi, for some integer,  k_i  ≥  1 .  Note:  p1 = 2,  p2 = 3, p3 = 5, ...

Example:  1/315  = 1/(32 * 5 * 7).

Hint for Q3:  Prime Number Theorem and the Harmonic Series ...

End of Discussion.

https://www.criticalthinking.org/ctmodel/logic-model1.htm

https://plus.google.com/115467095957398372179/posts/giNABPxABGG?sfc=true

Please, unfollow me, Best

https://www.researchgate.net/publication/368574983_Proof_of_Riemann_Hypothesis