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 !