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the nth prime by $P_{n}$, prove or diprove $$P_{2n}<(n+1)^2$$ $$P_{2}=3<4,P_{4}=7<9,P_{6}=13<16\cdots $$

It is well know that $$P_{n}<2^{2^n}$$

see:http://www.maths.tcd.ie/pub/Maths/Courseware/PrimeNumbers/Primes-I.pdf Thank you everyone

math110
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  • You only need to directly check a few small $n$ because it known that $$P_{2n} < 2n(\ln(2n) + \ln(\ln(2n))), \quad n \ge 3,$$(see the links on http://en.wikipedia.org/wiki/Prime_number_theorem#Approximations_for_the_nth_prime_number). – gammatester Aug 08 '13 at 08:14
  • Thank you ,But This problem is compition problem,I think this problem have other solution – math110 Aug 08 '13 at 08:51

1 Answers1

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We can prove by completely elementary means that for all $n\ge 2$, $$ \frac{n}{6\log n}<\pi(n)<\frac{6n}{\log n}, $$ using estimates $2^n\le \binom{2n}{n}<4^n$ etc., see for example Tom Apostol's book on anayltic number theory. For $n=p_m$, the $m$-th prime, we have $\pi(p_m)=m$, and it follows easily that $p_m<cm\log(m)$ for some positive constant $c$. Then this is smaller than $(m/2+1)^2$ for all $m\ge c_1$. One has to check the small values of $m$, and perhaps compute some suitable constants $c$ and $c_1$, but this should be no problem.

Of course, the question is, what you want to use for a proof. This will depend on your taste, too. In any case, Chebychev type estimates on $\pi(n)$ will yield the desired estimate.

Dietrich Burde
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  • A proof of the upper bound (with a slightly different constant) can be found here: http://math.stackexchange.com/questions/33980/link-summary-proof-of-chebyshev-statement-sum-p-leq-n-frac-log-pp-sim/33995#33995 and the lower bound can be found here: http://math.stackexchange.com/questions/38917/are-there-any-combinatoric-proofs-of-weak-versions-of-bertrands-postulate/38954#38954 – Eric Naslund Sep 22 '13 at 19:04
  • @EricNaslund: thank you for the links, which really give the details and are a pleasure to read (also recommended for the OP). – Dietrich Burde Sep 22 '13 at 20:32