I encountered the following exercise in the beautiful book Neverending Fractions by Borwein, van der Poorten, Shallit, and Zudilin. They refer to the book The Distribution of Prime Numbers by A.E. Ingham.(Chapter I, Theorem 3). But I was not able to link Theorem 3 to the Exercise in this book! Perhaps I do not understand what they mean. Anyway, let me state their question.
Prove that the following two statements are equivalent :
- (The Prime Number Theorem) For real $x$, let ${\pi}(x)$ be the number of primes less than or equal to $x$. Then, $${{\lim}_{n\rightarrow{\infty}}}{\frac{\pi{(x)}}{x/{\log}x}}=1.$$
- Let ${d_n} = {\rm{lcm}}(1,2,\cdots ,n).$ Then, $${{\lim}_{n\rightarrow{\infty}}}{\frac{{\log}({d_n})}{n}}=1.$$ Not sure how to even approach this!
For each prime $p\le n$, $d_n$ contains the highest power $\alpha_p$of $p$ that is $\le n$. Thus
$$\log d_n=\sum_{p\le n}\log p^{\alpha_p}\le\sum_{p\le n}\log n=\pi(n)\log n\;.$$ This establishes that the limit in $2$. is at most the limit in $1$.
To prove that it’s also at least the limit in $1$., you could try to show that the proportion of primes $\le n$ that are greater than, say, $n\exp\left(-\sqrt{\log n}\right)$, goes to $1$, as does the ratio of their logarithm to $\log n$:
$$\frac{\log\left(n\exp\left(-\sqrt{\log n}\right)\right)}{\log n}=\frac{\log n-\sqrt{\log n}}{\log n}\to_{n\to\infty}1\;.$$