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October 23rd, 2015, 08:13 AM  #1 
Newbie Joined: Dec 2014 From: A plane of mathematical matter Posts: 6 Thanks: 0  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. $\displaystyle \sum_{n=1}^\infty \frac{1}{n} $ I had decided to separate everything into parts so that you could represent it like this. $\displaystyle Hx = \frac{\sum_{n=1}^\infty \frac{1}{n}  \sum_{}^{x1} H}{Px} $ $\displaystyle \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. $\displaystyle H(x)P(x)  H(x+1)P(x+1) = \sum_{n=1}^\infty \frac{1}{n}  \sum_{}^{x1} H  \sum_{n=1}^\infty \frac{1}{n} + \sum_{}^{x} H $ $\displaystyle H(x)P(x)  H(x+1)P(x+1) = H(x) $ $\displaystyle \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 $\displaystyle 1 = \frac{1}{2} \sum_{x=1}^\infty \prod_{n=1}^{x1} \frac{P(n)1}{P(n+1)} $ I latter refined this $\displaystyle 1=\sum_{x=1}^\infty \frac{1}{P(x)} \prod_{n=1}^{x1} \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 $\displaystyle 1=\sum_{x=N}^\infty \frac{1}{P(x)} \prod_{n=N}^{x1} \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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bored, class, featuring, math, numbers, prime, proof 
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