Can we deduce that $\lim_{n\to\infty}(\log(p_{n+1})/\log(p_{n}))=1$

Question

Let $p_{k}$ denotes the sequence of prime numbers. We know that $$\lim_{n\to\infty}(p_{n+1}/p_{n})=1$$

Can we deduce that $$\lim_{n\to\infty}\log(p_{n+1})/\log(p_{n})=1$$ where $\log$ is the natural logarithm.

Answer 1

The second limit does indeed tend to $1$, because $$\frac{\log p_{n+1}}{\log p_n}=\log_{p_n}p_{n+1}<\log_{p_n}2p_n=1+\log_{p_n}2$$ Here we have used Bertrand's postulate. $\log_{p_n}2$ tends to $0$ and the result follows.

Answer 2

Yes, it does follow: $$\frac{\log p_{n+1}}{\log p_n}=1+\frac{\log\frac{p_{n+1}}{p_n}}{\log p_n}\stackrel{n\to\infty}\longrightarrow \left[1+\frac{\log 1}{\infty}\right]=1$$