# Bertrand's Postulate Proof: n=4000 not true?

I need some advice: I'm dealing with the proof of Bertrand's postulate Proofs from THE BOOK. I understood the proof completly but as I tried to make a drawing of the intersection of the lower und upper boundary (which should be refering to the book at n=4000) I got some strange results.

My result for the intersection is round about n=5000. If someone has an explanation for it, it would be great.

Clearly, from $2^{2n}<2^{20(2n)^{2/3}}$ we may conclude $2n<20(2n)^{2/3}$, then $(2n)^{1/3}<20$ and finally $n<\frac{20^3}2=4000$. I have no idea what you did instead.
I unterstood the following (all the calculus is clear): starting from $4^{\frac{1}{3}n} \leq (2n)^{\sqrt{2n}+1}$ we get the result n=4000, so I think, n=4000 has to satisfy that starting-inequality. Am I wrong? The lower boundary gets an higher value at $n=4000$ then the upper boundary (which is the contradiction). But when I put the inequality into mathematica, I get an other value then $n=4000$ for the intersection of both boundaries. – ulead86 Nov 6 '12 at 17:55
@ulead86: We want an implication of the form $f(n)<g(n)\Rightarrow n<N$. That does not imply that $f(N)<g(N)$ (in fact, necessarily $f(N)\ge g(N)$). – Hagen von Eitzen Nov 7 '12 at 19:50
Hm interesting, that is exactly the wrong implication I made. But can someone explain, why the implication $n<N \rightarrow f(N)<g(N)$ is wrong? – ulead86 Nov 8 '12 at 18:01