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September 29th, 2014, 09:58 PM  #1 
Newbie Joined: Sep 2014 From: place Posts: 16 Thanks: 0  Riemann Zeta Function proof
i have my proof in a Microsoft word document download link: Proof of The Riemann Zeta function 
September 30th, 2014, 07:14 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
If I understand correctly, you're claiming that $\zeta(52)=0$ and $\zeta(2)\approx1.0191193511008111.$ $\zeta(n)=0$ for all even integers n > 1, but $\zeta(2)=\pi^2/6\approx1.6449340668$ has been known for a long time (the Basel problem). I'll note that neither of these values are related to the Riemann hypothesis. 
September 30th, 2014, 09:03 PM  #3  
Newbie Joined: Sep 2014 From: place Posts: 16 Thanks: 0  Quote:
also how was zeta 2 calculated because what I got was zeta(n) ≈ 1 + n+31/n(12^n+1) n = 2 so zeta(2) ≈ 1 + 33/1728 = 1.0190972222222222 zeta(2) should be roughly = 1.0190972222222222 so I'm confused now. Thanks in advance if anyone can explain this to me. Last edited by skipjack; October 4th, 2014 at 04:03 AM.  
October 1st, 2014, 06:22 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
$\zeta(2)=1+\frac{1}{2^2}+\frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \cdots$ which already shows that $\zeta(2)>1.5.$ If you know a little calculus you can see that $$ 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \int_8^{\infty}\frac{dx}{x^2} < \zeta(2) < 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \int_7^{\infty}\frac{dx}{x^2} $$ where the lower and upper bounds are 1.636797 and 1.6546542, respectively. Of course you can choose any cutoff you like to get more or less precise. 
October 1st, 2014, 08:43 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra 
$$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ follows from the Fourier Series expansion of $$f(x) = \frac{3x^2  6\pi x + 2\pi^2}{12} = \sum_{n=1}^\infty \frac{\cos nx}{n^2} \qquad x \in (0, 2\pi) $$by setting $x=0$. 

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