Estimate for the $2n$-th consecutive prime number

Question

Using result from calculus:(1) $ \log(x\log x)>\log\vartheta \cdot \log(2x\log 2x)$ for $2\leq \vartheta<e$ ($e$-Euler number) and $ x>x(\vartheta )>2$, I am able to prove $ p_{2n}<\frac{2n}{\log(\vartheta )}\cdot \log(p_{n})$ for $n>n(\vartheta)$, where $p_{n}$ is the $n$-th prime number. The proof does not work for $\vartheta=e$, since inequality (1) is obviously not valid in this case. But data suggests also stronger result:
$ p_{2n}<2n\cdot \log(p_{n})$ for $n>52$. How can it be proved?

Answer 1

Let's use some standard bounds on the $n$th prime, as found in this paper by Pierre Dusart.

We have $$ p_{2n} \leq 2n[\ln(2n)+\ln(\ln(2n))-0.9484] = 2n\left[\ln(n)+\ln\left(\frac{2}{e^{0.9484}}\ln(2n)\right)\right] $$ using an inequality stated on page 414 of the cited paper, valid for $n\geq 19509$. On the other hand, we have $$2n\ln(p_n)\geq 2n\ln[n(\ln(n)+\ln(\ln(n))-1)]=2n[\ln(n)+\ln(\ln(n\ln(n)/e))] $$ by the main result of that paper, for $n\geq 2$.

Thus, your desired inequality will be true once $n\geq 19509$ and when $$\frac{2}{e^{0.9484}}\ln(2n)<\ln\left(\frac{n\ln(n)}{e}\right). $$ That last inequality becomes true when $n\geq 13$. Then a simple check will show that the inequality you want is valid in the region $52\leq n\leq 19509$.