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?

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?

Using result from calculus:(1) $ log(xlogx)>log\vartheta \cdot log(2xlog2x)$ 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?

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?