Prove the Harmonic Series diverges to infinity.

[Guest]

FYI: Harmonic series (mathematics);

A SHORT(ER) PROOF OF THE DIVERGENCE OF THE HARMONIC SERIES.

Enjoy!

[Guest]

FYI:

[tex]\sum_{k=1}^{ \infty } \frac{1}{k} \ge 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ... = 1 + 1 + 1 + 1 + ... = 1 + 2 + 3 + 4 + 5 + ...
= 2 + 3 + 5 + 7 + 11 + 13 + ... = 2 + 4 + 6 + 8 + ... = 3 + 6 + 9 + 12 + 15 + ... = 5 + 10 + 15 + 20 + 25 + ... = p + 2p + 3p + 4p + 5p + ... = \infty[/tex] where p is any positive prime number.

Right?

[Guest]

FYI:

[tex]\sum_{k=1}^{ \infty } \frac{1}{k} \ge 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ... = 1 + 1 + 1 + 1 + ... = 1 + 2 + 3 + 4 + 5 + ...
= 2 + 3 + 5 + 7 + 11 + 13 + ... = 2 + 4 + 6 + 8 + ... = 3 + 6 + 9 + 12 + 15 + ... = 5 + 10 + 15 + 20 + 25 + ... = p + 2p + 3p + 4p + 5p + ... = \infty[/tex] where p is any positive prime number.

Right?

What is acceptable?

[tex]\sum_{k=1}^{ \infty } \frac{1}{k} -[/tex] error [tex]= p + 2p + 3p + 4p + 5p + ... = \infty[/tex] where p is any positive prime number.

[Guest]

FYI:

[tex]\sum_{k=1}^{ \infty } \frac{1}{k} \ge 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ... = 1 + 1 + 1 + 1 + ... = 1 + 2 + 3 + 4 + 5 + ...
= 2 + 3 + 5 + 7 + 11 + 13 + ... = 2 + 4 + 6 + 8 + ... = 3 + 6 + 9 + 12 + 15 + ... = 5 + 10 + 15 + 20 + 25 + ... = p + 2p + 3p + 4p + 5p + ... = \infty[/tex] where p is any positive prime number.

Right?

What is acceptable?

[tex]\sum_{k=1}^{ \infty } \frac{1}{k} -[/tex] error [tex]= p + 2p + 3p + 4p + 5p + ... = \infty[/tex] where p is any positive prime number.

What is best?

[tex]\sum_{k=1}^{ \infty } \frac{1}{k} = \sum_{i=1}^{ \infty } p_{i } +[/tex] error [tex]= \infty[/tex] where [tex]p_{i }[/tex] is ith positive prime number.

Why?

"We, humans, learn, but Nature knows."

Two Important Reasons:

1. It is logical and consistent that we apply the basic prime number test to each term of the above series excluding the error term since the error term is not a positive integer (right?). The basic prime number test computes [tex]p_{i } ^{ \frac{1}{2} }[/tex] for all [tex]i \ge 1[/tex], and it is used to determine the primality of each appropriate term.


2.

In a nutshell, [tex]\zeta(z = 1) = \sum_{k=1}^{\infty }\frac{1}{k} = \infty[/tex] if and only if [tex]\zeta(z = \frac{1}{2} ± bi ) = \sum_{k=1}^{\infty }\frac{1}{k^{ \frac{1}{2} ± bi}} = 0.[/tex]

[Guest]

Oops!

I think I created a mess ([tex]\sum_{k=1}^{ \infty } \frac{1}{k} = \sum_{i=1}^{ \infty } p_{i } +[/tex] error [tex]= \infty[/tex]...) with the last post since there is no apparent correlation between primes and zeta zeros... Furthermore, primes (p) and non-primes (2p, 3p, 4p, ...) coexist...

Sorry! I'll think about the mess I created and how to fix it. Goodbye!

Dave.

[Guest]

The Fix:

[tex]\sum_{k=1}^{ \infty } \frac{1}{k} = p + 2p + 3p + 4p + 5p + ...[/tex] + error [tex]= \infty[/tex] where p is any positive prime number....

[Guest]

FYI: Why does the harmonic series diverge but the p-harmonic series converge?

[Guest]

The determinants n x n increasing along the diagonal form an alternating natural series? Super beautiful: https://t.me/math_code/676514 !

[Guest]

The Fix:

[tex]\sum_{k=1}^{ \infty } \frac{1}{k} = p + 2p + 3p + 4p + 5p + ...[/tex] + error = [tex]\sum_{k=1}^{ \infty }kp[/tex] + error = [tex]\infty[/tex] where p is any positive prime number....

In a nutshell, [tex]\zeta(z = 1) = \sum_{k=1}^{\infty }\frac{1}{k} = \infty[/tex] if and only if [tex]\zeta(z = \frac{1}{2} ± bi ) = \sum_{k=1}^{\infty }\frac{1}{k^{ \frac{1}{2} ± bi}} = 0.[/tex]

There are infinitely many prime numbers, and there are also infinitely many simple nontrivial zeta zeros (z) with Re(z) = [tex]\frac{1}{2}[/tex].

Go figure! Go Blue!

[Guest]

Given [tex]\zeta(z = \frac{1}{2} ± bi ) = \sum_{k=1}^{\infty }\frac{1}{(kp)^{ \frac{1}{2} ± bi}} = 0[/tex] and p = 79, compute b?

[Guest]

Useful Link: Nontrivial zeros of Riemann Zeta Function

[Guest]

Given
[tex]\zeta(z = \frac{1}{2} ± bi, p) = \sum_{k=1}^{\infty }\frac{1}{(kp)^{ \frac{1}{2} ± bi}} = 0[/tex]

and p = 79, compute b.

[Guest]

Key Idea: p is a prime!

More simply, we have in a nutshell (the source of all prime numbers and the detector of prime numbers),

[tex]\sum_{k=1}^{ \infty } \frac{1}{k} -[/tex] error [tex]= \sum_{k=1}^{ \infty }kp = \infty[/tex] if and only if [tex]\zeta(z = \frac{1}{2} ± bi, p) = \sum_{k=1}^{\infty }\frac{1}{(kp)^{ \frac{1}{2} ± bi}} = 0[/tex] where p is any positive prime number.

Moreover, there are infinitely many prime numbers, and there are also infinitely many simple nontrivial zeta zeros (z) with Re(z) = [tex]\frac{1}{2}[/tex].

Go figure! Go Blue!

[Guest]

Key Idea: p is a prime number!

More simply, we have in a nutshell (the source of all prime numbers and the detector of prime numbers),

[tex]\zeta(z = 1) -[/tex] error [tex]= \sum_{k=1}^{ \infty } \frac{1}{k} -[/tex] error [tex]= \sum_{k=1}^{ \infty }kp = \infty[/tex] if and only if [tex]\zeta(z = \frac{1}{2} ± bi, p) = \sum_{k=1}^{\infty }\frac{1}{(kp)^{ \frac{1}{2} ± bi}} = 0[/tex] where p is any positive prime number.

Moreover, there are infinitely many prime numbers, and there are also infinitely many simple nontrivial zeta zeros (z) with Re(z) = [tex]\frac{1}{2}[/tex].

Go figure! Go Blue!

[Guest]

UPDATE:

FYI: "At the death of Riemann, a note was found among his papers, saying "These properties of ζ(z) (the function in question) are deduced from an expression of it which, however, I did not succeed in simplifying enough to publish it." We still have not the slightest idea of what the expression could be. As to the properties he simply enunciated, some thirty years elapsed before I was able to prove all of them but one [the Riemann Hypothesis itself]."

— Jacques Hadamard, The Mathematician's Mind, VIII. Paradoxical Cases of Intuition.

Hmm. That expression referred to by Riemann (on deducing the properties of ζ(z)) must be the source of all prime numbers, the Harmonic Series. Right?

We recall the important divergent Harmonic Series:

ζ(z=1) = [tex]\sum_{k=1}^{ \infty } \frac{1}{k} = \infty[/tex].

...
A Recap:

Key Idea: p is a prime number!

More simply, we have in a nutshell (the source of all prime numbers and the detector of prime numbers),

[tex]\zeta(z = 1) -[/tex] error [tex]= \sum_{k=1}^{ \infty } \frac{1}{k} -[/tex] error [tex]= \sum_{k=1}^{ \infty }kp = \infty[/tex] with (0 < error < 1)

if

[tex]\zeta(z = \frac{1}{2} ± bi, p) = \sum_{k=1}^{N}\frac{1}{(kp)^{ \frac{1}{2} ± bi}} + \gamma(\frac{1}{2} \mp bi) + \sum_{k=1}^{M}\frac{1}{(kp)^{ \frac{1}{2} \mp bi}} + R( \frac{1}{2} ± bi) = 0[/tex]

where p is any positive prime number.

Go figure! Go Blue!

***** The great Riemann Hypothesis (RH) is true! ***** ✌️✌️

Relevant Reference Link:

Riemann Siegel formula

Remark: The second equation is tentative since I do not completely understand all its details and since I fooled myself before. That could happen again. And I am very likely to make minor or important mistakes, too.

We must admit that our second equation is extraordinary, beautiful, and very promising. It must be correct!


Dave.