# Prove the Harmonic Series diverges to infinity.

### Prove the Harmonic Series diverges to infinity.

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Harmonic Series diverge.jpg (3.51 KiB) Viewed 95 times
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### Re: Prove the Harmonic Series diverges to infinity.

FYI:

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

Right?
Guest

### Re: Prove the Harmonic Series diverges to infinity.

Guest wrote:FYI:

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

Right?

What is acceptable?

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

### Re: Prove the Harmonic Series diverges to infinity.

Guest wrote:
Guest wrote:FYI:

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

Right?

What is acceptable?

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

What is best?

$$\sum_{k=1}^{ \infty } \frac{1}{k} = \sum_{i=1}^{ \infty } p_{i } +$$ error $$= \infty$$ where $$p_{i }$$ 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 $$p_{i } ^{ \frac{1}{2} }$$ for all $$i \ge 1$$, and it is used to determine the primality of each appropriate term.

2.
Guest wrote:In a nutshell, $$\zeta(z = 1) = \sum_{k=1}^{\infty }\frac{1}{k} = \infty$$ if and only if $$\zeta(z = \frac{1}{2} ± bi ) = \sum_{k=1}^{\infty }\frac{1}{k^{ \frac{1}{2} ± bi}} = 0.$$
Guest

### Re: Prove the Harmonic Series diverges to infinity.

Guest wrote:Oops!

I think I created a mess ($$\sum_{k=1}^{ \infty } \frac{1}{k} = \sum_{i=1}^{ \infty } p_{i } +$$ error $$= \infty$$...) 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

### Re: Prove the Harmonic Series diverges to infinity.

The Fix:

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

Guest

### Re: Prove the Harmonic Series diverges to infinity.

The determinants n x n increasing along the diagonal form an alternating natural series? Super beautiful: https://t.me/math_code/676514 !
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Det's ={0, -1, 2, -3, 4, -5, 6, -7, 8, -9, 10, ..., n (-1)^n, ...}
Det's ={0, -1, 2, -3, 4, -5, 6, -7, 8, -9, 10}.jpg (47.32 KiB) Viewed 2 times
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