# Finding $\limsup\limits_{n\to\infty}\space\Big (\frac {π(2n+1)}{π(2n)}\Big)^n$

Let $\pi(x)$ denote the prime counting function.Then what is $\lim\limits_{n\to\infty}sup\space\Big (\frac {π(2n+1)}{π(2n)}\Big)^n$ ?

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For every odd integer $n$, $\pi(n+1)=\pi(n)$. This rules out the limit you propose. –  Did Nov 8 '12 at 11:18
Even after the correction, for most even numbers $2n$, $\pi(2n) = \pi(2n+1)$. Would $\limsup$ be applicable, perhaps? –  Arthur Nov 8 '12 at 14:15
@Arthur: Is it okay now ? –  Souvik Dey Nov 9 '12 at 5:23
But didn't Charles post a computation of the limsup several hours ago? –  Gerry Myerson Nov 9 '12 at 5:39
@GerryMyerson: Oh , right , I did not notice. –  Souvik Dey Nov 9 '12 at 6:20

$\pi(x)$ is unbounded above, and $\pi(2n+1)=\pi(2n)$ except on a subsequence of density 0. So $$\liminf_{n\to\infty}n\log\frac{\pi(2n+1)}{\pi(2n)}=\liminf_{n\to\infty}n\log1=0.$$
On the other side it suffices to look at the subsequence of primes: $$\limsup_{n\to\infty}n\log\frac{\pi(2n+1)}{\pi(2n)}=\limsup_{n\to\infty}\frac{p_{n+1}-1}{2}\log\frac{n+1}{n}=\limsup_{n\to\infty}\frac{p_{n+1}-1}{2}\left(\frac{1}{n}+O\left(\frac{1}{n^2}\right)\right)=+\infty$$