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Just another simple inequality - 21/05/2013

Posted: Tue May 21, 2013 8:14 am

\text{Let's define } p_n \text{ as the n-th prime number}\\
\text{For example } p_1=2 ; p_2 =3;p_3 = 5;...\\
\text{Can you prove that }p_n < 2^{2^{n}} ?\\
\text{}

Re: Just another simple inequality - 21/05/2013

Posted: Tue Jun 11, 2013 3:49 pm
OK. I will attempt to prove this through mathematical induction.
Sketch of Proof:
BASIS STEP: This inequality holds true for n=1.
Inductive Step: Assuming that this inequality holds true for n, it must also be true for n+1. (Note: I borrowed Bertrand's conjecture halfway down the line. It was completely proven in 1850, but I'm not smart enough to prove it independently. Feel free to give credit to someone else for this).

Re: Just another simple inequality - 21/05/2013

Posted: Tue Jun 11, 2013 4:41 pm
Thank you for sharing your solution and approach! I just took a look and I do agree about your statements! I really like them and translating the problem to one of Bertrand-Chebyshev theorems is beautiful consequence. You definitely took the credit of the one who solved it first (it’s not necessary to prove the theorems you applies, stating them with HREF’s is enough).

Problem closed, the next one will be published tomorrow!

Re: Just another simple inequality - 21/05/2013

Posted: Sat Jun 15, 2013 9:56 pm
Yes, we used a "brute force" theorem and proved a much stricter equation pn < 2^n, but I don't think this was the point of the original problem.