TITLE: How prove that $P_{2n}<(n+1)^2$
QUESTION [8 upvotes]: 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

REPLY [4 votes]: 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.