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October 23rd, 2015, 08:13 AM   #1
Joined: Dec 2014
From: A plane of mathematical matter

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Bored in math class proof featuring prime numbers

So while I was scribbling in my college calculus class I was just goofing around playing with the harmonic series.
\sum_{n=1}^\infty \frac{1}{n}
I had decided to separate everything into parts so that you could represent it like this.
Hx = \frac{\sum_{n=1}^\infty \frac{1}{n} - \sum_{}^{x-1} H}{Px}
\sum_{n=1}^\infty \frac{1}{n} = 1 + \sum_{}^\infty H
The reasoning for this is because everything is composed of prime factors. The first H would have $\displaystyle \frac{1}{2}$ factor of each. Then in following H it would have $\displaystyle \frac{1}{3}$ as the common factor, but you need to remove the existing H values so you don't get repeats $\displaystyle \frac{1}{6}$ would be part of the first H.

I then looked at trying to rewrite the harmonic series in these terms. So I wanted to see the relationship between H and its successor.

H(x)P(x) - H(x+1)P(x+1) = \sum_{n=1}^\infty \frac{1}{n} - \sum_{}^{x-1} H - \sum_{n=1}^\infty \frac{1}{n} + \sum_{}^{x} H

H(x)P(x) - H(x+1)P(x+1) = H(x)

\frac{P(x) - 1}{P(x+1)} = \frac{H(x+1)}{H(x)}

We know that $\displaystyle H(1)=\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n}$

So plugging everything back into the modified harmonic series.

$\displaystyle \sum_{n=1}^\infty \frac{1}{n} = 1 + (\frac{1}{2}) \sum_{n=1}^\infty \frac{1}{n} + (\frac{1}{2})(\frac{1}{3}) \sum_{n=1}^\infty \frac{1}{n} ... + ((\frac{1}{2})(\frac{1}{3})...(\frac{P(x)-1}{P(x+1)})) \sum_{n=1}^\infty \frac{1}{n}$

Then I realized I can divide out the harmonic series.

$\displaystyle 1 = \frac{1}{\sum_{n=1}^\infty \frac{1}{n}} + (\frac{1}{2}) + (\frac{1}{2})(\frac{1}{3}) ... + ((\frac{1}{2})(\frac{1}{3})...(\frac{P(x)-1}{P(x+1)}))$

This is essentially zero
$\displaystyle 0 = \frac{1}{\sum_{n=1}^\infty \frac{1}{n}}$

$\displaystyle 1 = (\frac{1}{2}) + (\frac{1}{2})(\frac{1}{3}) ... + ((\frac{1}{2})(\frac{1}{3})...(\frac{P(x)-1}{P(x+1)}))$

You can express this as
1 = \frac{1}{2} \sum_{x=1}^\infty \prod_{n=1}^{x-1} \frac{P(n)-1}{P(n+1)}

I latter refined this
1=\sum_{x=1}^\infty \frac{1}{P(x)} \prod_{n=1}^{x-1} \frac{P(n)-1}{P(n)} = \frac{1}{2} + \frac{1}{3}\frac{1}{2} + \frac{1}{5}\frac{1}{2}\frac{2}{3} + \frac{1}{7}\frac{1}{2}\frac{2}{3}\frac{4}{5} ...

Then I discovered a revelation in this series. Looking at the above series I can it by 2 and subtract 1 leaving $\displaystyle 1 = \frac{1}{3} + \frac{1}{5}\frac{2}{3} + \frac{1}{7}\frac{2}{3}\frac{4}{5} + \frac{1}{11}\frac{2}{3}\frac{4}{5}\frac{6}{7} ...$

Looking at the series again I can it by 3 and subtract 1 then divide by 2 leaving $\displaystyle 1 = \frac{1}{5} + \frac{1}{7}\frac{4}{5} + \frac{1}{11}\frac{4}{5}\frac{6}{7} + \frac{1}{13}\frac{4}{5}\frac{6}{7}\frac{10}{11} ...$

The first term is one over the prime number and in every following term contains the prime - 1 over the prime. So we will be able to do this forever allowing me to refine my series to start at any N natural number

1=\sum_{x=N}^\infty \frac{1}{P(x)} \prod_{n=N}^{x-1} \frac{P(n)-1}{P(n)}

I haven't seen this around so just though I would post my thinking if I made any mistakes please point them out.
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