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While going through F. Beukers proof of irrationality of $\zeta(3)$ I found the inequality $d_{n} < 3^{n}$ for all sufficiently large values of $n$ where $d_{n}$ denotes the LCM of all the numbers $1, 2, 3, \ldots, n$

Now this is easily established (almost obvious) once one assumes the prime number theorem. In fact we can replace $3^{n}$ by $a^{n}$ where $a > e$.

I would like to know if there is any elementary/direct proof of $d_{n} < 3^{n}$ without the recourse to the difficult prime number theorem.

Robert Z
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Paramanand Singh
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  • $\mbox{lcm}(1,2,3,\dots,n)$ is Landau's function $g(n)$. It can be shown that $g(n)<e^{n/e}$. Since $e^{n/e} < 3^n$, we have $g(n) < 3^n$. – lhf Aug 16 '17 at 14:16
  • Note also that $$ d_n = d_{n-1} \cdot \begin{cases} p & \text{if } n = p^k \text{ for some } k\geq 1,\ p\in\mathbb{P} \ 1 & \text{else} \end{cases} $$ – Zubzub Aug 16 '17 at 14:28
  • Related: https://math.stackexchange.com/questions/851328 Interestingly, proving a lower bound seems to be easier than proving an upper bound. Which is also the case in Chebyshev's original proof. – Bart Michels Aug 30 '19 at 12:14

1 Answers1

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An elementary proof of that $d_n=\mbox{lcm}(1,2,3,\dots,n)<3^n$ can be found in this article by D. Hanson: On the product of the primes, Canad. Math. Bull. 15(1972), 33-37. The same paper on researchgate.

Robert Z
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  • The link which you have referred doesn't work :( – C.S. Aug 30 '19 at 11:44
  • @crskhr I edited the link with a new one. You can also try https://www.researchgate.net/publication/267137759_On_the_product_of_the_primes – Robert Z Aug 30 '19 at 12:08
  • Thanks, though not much i could comprehend, i am sure it was useful edit for the users. – C.S. Aug 30 '19 at 15:14