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

Question

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 $ ?

Answer 1

$\pi(x)$ 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: $$ \limsup_{n\to\infty}n\log\frac{\pi(2n+1)}{\pi(2n)}=\limsup_{n\to\infty}\frac{p_{n+1}-1}{2}\log\frac{n+1}{n}=\limsup_{n\to\infty}\frac{p_{n+1}-1}{2}\left(\frac{1}{n}+O\left(\frac{1}{n^2}\right)\right)=+\infty $$