# least common multiple $\lim\sqrt[n]{[1,2,\dotsc,n]}=e$

The least common multiple of $1,2,\dotsc,n$ is $[1,2,\dotsc,n]$, then

$$\lim_{n\to\infty}\sqrt[n]{[1,2,\dotsc,n]}=e$$

we can show this by prime number theorem, but I don't know how to start

I had learnt that it seems we can find the proposition in G.H. Hardy's number theory book, but I could not find it.

I am really grateful for any help

• Is this homework? This looks like exercise 3.2 in Hildebrands notes, page 103. – punctured dusk Jun 14 '14 at 19:27
• They are not the same – Rene Schipperus Jun 14 '14 at 19:42
• – lhf Jun 14 '14 at 19:44

## 2 Answers

Let's look how the least common multiple evolves.

If $n > 1$ is a prime power, $n = p^k$ ($k \geqslant 1$), then no number $< n$ is divisible by $p^k$, but $p^{k-1} < n$, so $[1,2,\dotsc,n-1] = p^{k-1}\cdot m$, where $p\nmid m$. Then $[1,2,\dotsc,n] = p^k\cdot m$, since on the one hand, we see that $p^k\cdot m$ is a common multiple of $1,2,\dotsc,n$, and on the other hand, every common multiple of $1,2,\dotsc,n$ must be a multiple of $p^k$ as well as of $m$.

If $n > 1$ is not a prime power, it is divisible by at least two distinct primes, say $p$ is one of them. Let $k$ be the exponent of $p$ in the factorisation of $n$, and $m = n/p^k$. Then $1 < p^k < n$ and $1 < m < n$, so $p^k\mid [1,2,\dotsc,n-1]$ and $m\mid [1,2,\dotsc,n-1]$, and since the two are coprime, also $n = p^k\cdot m \mid [1,2,\dotsc,n-1]$, which means that then $[1,2,\dotsc,n] = [1,2,\dotsc,n-1]$.

Taking logarithms, we see that for $n > 1$

\begin{align} \Lambda (n) &= \log [1,2,\dotsc,n] - \log [1,2,\dotsc,n-1]\\ &= \begin{cases} \log p &, n = p^k\\ \;\: 0 &, \text{otherwise}.\end{cases} \end{align}

$\Lambda$ is the von Mangoldt function, and we see that

$$\log [1,2,\dotsc,n] = \sum_{k\leqslant n} \Lambda(k) = \psi(n),$$

where $\psi$ is known as the second Chebyshev function.

With these observations, it is clear that

$$\lim_{n\to\infty} \sqrt[n]{[1,2,\dotsc,n]} = e\tag{1}$$

is equivalent to

$$\lim_{n\to\infty} \frac{\psi(n)}{n} = 1.\tag{2}$$

It is well-known and easy to see that $(2)$ is equivalent to the Prime Number Theorem (without error bounds)

$$\lim_{x\to\infty} \frac{\pi(x)\log x}{x} = 1.\tag{3}$$

To see the equivalence, we also introduce the first Chebyshev function,

$$\vartheta(x) = \sum_{p\leqslant x} \log p,$$

where the sum extends over the primes not exceeding $x$. We have

$$\vartheta(x) \leqslant \psi(x) = \sum_{n\leqslant x}\Lambda(n) = \sum_{p\leqslant x}\left\lfloor \frac{\log x}{\log p}\right\rfloor\log p \leqslant \sum_{p\leqslant x} \log x = \pi(x)\log x,$$

which shows - the existence of the limits assumed -

$$\lim_{x\to\infty} \frac{\vartheta(x)}{x} \leqslant \lim_{x\to\infty} \frac{\psi(x)}{x} \leqslant \lim_{x\to\infty} \frac{\pi(x)\log x}{x}.$$

For $n \geqslant 3$, we can split the sum at $y = \frac{x}{(\log x)^2}$ and obtain

$$\pi(x) \leqslant \pi(y) + \sum_{y < p \leqslant x} 1 \leqslant \pi(y) + \frac{1}{\log y}\sum_{y < p < x}\log p \leqslant y + \frac{\vartheta(x)}{\log y},$$

whence

$$\frac{\pi(x)\log x}{x} \leqslant \frac{y\log x}{x} + \frac{\log x}{\log y}\frac{\vartheta(x)}{x} = \frac{1}{\log x} + \frac{1}{1 - 2\frac{\log \log x}{\log x}}\frac{\vartheta(x)}{x}.$$

Since $\frac{1}{\log x}\to 0$ and $\frac{\log\log x}{\log x} \to 0$ for $x\to \infty$, it follows that (once again assuming the existence of the limits)

$$\lim_{x\to\infty} \frac{\pi(x)\log x}{x} \leqslant \lim_{x\to\infty} \frac{\vartheta(x)}{x},$$

and the proof of the equivalence of $(1)$ and $(3)$ is complete.

• Very nice exposition! +1 – Markus Scheuer Feb 10 '16 at 18:53
• It was a pleasure to read this answer. It is truly a textbook quality work. – Prism Mar 18 '16 at 0:03

(Not an answer, just a suggestion.)

It would seem that the power of a prime $$p$$ in $$[1,2,\dots,n]$$ is $$\left\lfloor \frac{\log n}{\log p}\right\rfloor.$$ Start of proof: $$\prod_{p\leq n} p^{\frac{1}{n}\left\lfloor \frac{\log n}{\log p}\right\rfloor} = e^{\frac{1}{n}\sum_{p\leq n} \log p \left\lfloor \frac{\log n}{\log p}\right\rfloor}$$

So we need to show that: $$S_n=\frac{1}{n}\sum_{p\leq n} \log p \left\lfloor \frac{\log n}{\log p}\right\rfloor\to 1\text{ as }n\to\infty$$

The crudest estimate of the terms is:

$$\log n - \log p < \log p \left\lfloor \frac{\log n}{\log p}\right\rfloor \leq \log n$$

But that doesn't seem to be good enough. It does show that the limsup is less than $$e$$.

We can certainly see that $$\limsup S_n\leq 1$$ from the prime number theorem, since:

$$S_n\leq \frac{1}{n}\sum_{p\leq n}\log n = \frac{\pi(n) n}{\log n}\to 1$$

• Psst! Chebyshev says $\psi$ could be interesting. Von Mangoldt agrees. – Daniel Fischer Jun 14 '14 at 19:57
• I remember doing this exercise and struggling with the liminf and limsup too. Can you imagine the frustration? ;-) – punctured dusk Jun 14 '14 at 19:58
• As Daniel suggested, the limit follows from $\theta(n)\leqslant\Sigma\cdots\leqslant\pi(n)\log n$, where $\Sigma\log p=\theta(x)\sim\pi(x)\log x\sim x$. (No $\psi$ or $\Lambda$, however.) – punctured dusk Jun 14 '14 at 20:13
• @barto $\log [1,2,\dotsc,n] = \psi(n)$. – Daniel Fischer Jun 14 '14 at 20:20