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?



Estimate for the $2n$-th consecutive prime number
Let's use some standard bounds on the $n$th prime, as found in [this paper](https://www.ams.org/journals/mcom/1999-68-225/S0025-5718-99-01037-6/S0025-5718-99-01037-6.pdf) 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 $52\leq n\leq 19509$.


