 My Math Forum Bored in math class proof featuring prime numbers
 User Name Remember Me? Password

 Number Theory Number Theory Math Forum

 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. $\displaystyle Hx = \frac{\sum_{n=1}^\infty \frac{1}{n} - \sum_{}^{x-1} H}{Px}$ $\displaystyle \sum_{n=1}^\infty \frac{1}{n} = 1 + \sum_{}^\infty H$ 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. $\displaystyle H(x)P(x) - H(x+1)P(x+1) = \sum_{n=1}^\infty \frac{1}{n} - \sum_{}^{x-1} H - \sum_{n=1}^\infty \frac{1}{n} + \sum_{}^{x} H$ $\displaystyle H(x)P(x) - H(x+1)P(x+1) = H(x)$ $\displaystyle \frac{P(x) - 1}{P(x+1)} = \frac{H(x+1)}{H(x)}$ 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 $\displaystyle 1 = \frac{1}{2} \sum_{x=1}^\infty \prod_{n=1}^{x-1} \frac{P(n)-1}{P(n+1)}$ I latter refined this $\displaystyle 1=\sum_{x=1}^\infty \frac{1}{P(x)} \prod_{n=1}^{x-1} \frac{P(n)-1}{P(n)} = \frac{1}{2} + \frac{1}{3}\frac{1}{2} + \frac{1}{5}\frac{1}{2}\frac{2}{3} + \frac{1}{7}\frac{1}{2}\frac{2}{3}\frac{4}{5} ...$ 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 $\displaystyle 1=\sum_{x=N}^\infty \frac{1}{P(x)} \prod_{n=N}^{x-1} \frac{P(n)-1}{P(n)}$ I haven't seen this around so just though I would post my thinking if I made any mistakes please point them out. Tags bored, class, featuring, math, numbers, prime, proof Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Benu Academic Guidance 4 March 31st, 2013 09:58 PM rummi Calculus 1 October 21st, 2012 12:48 PM Dudealadude Abstract Algebra 1 January 28th, 2012 02:03 AM aric Algebra 5 November 24th, 2009 07:21 AM Gelembjuk Computer Science 5 November 26th, 2008 08:45 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      