FYI: Harmonic series (mathematics);
A SHORT(ER) PROOF OF THE DIVERGENCE OF THE HARMONIC SERIES.
Enjoy!
Guest wrote: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 wrote:Guest wrote: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 wrote: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 wrote: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.
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