Jtdylktuy

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

$$lim_{ntoinfty}sqrt[n]{[1,2,dotsc,n]}=e$$

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

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 $pnmid m$. Then $[1,2,dotsc,n] = p^kcdot m$, since on the one hand, we see that $p^kcdot m$ 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 $p^kmid [1,2,dotsc,n-1]$ and $mmid [1,2,dotsc,n-1]$, and since the two are coprime, also $n = p^kcdot m mid [1,2,dotsc,n-1]$, 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

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

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

With these observations, it is clear that

$$lim_{ntoinfty} sqrt[n]{[1,2,dotsc,n]} = etag{1}$$

is equivalent to

$$lim_{ntoinfty} frac{psi(n)}{n} = 1.tag{2}$$

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

$$lim_{xtoinfty} frac{pi(x)log x}{x} = 1.tag{3}$$

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

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

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

$$vartheta(x) leqslant psi(x) = sum_{nleqslant x}Lambda(n) = sum_{pleqslant x}leftlfloor frac{log x}{log p}rightrfloorlog p leqslant sum_{pleqslant x} log x = pi(x)log x,$$

which shows - the existence of the limits assumed -

$$lim_{xtoinfty} frac{vartheta(x)}{x} leqslant lim_{xtoinfty} frac{psi(x)}{x} leqslant lim_{xtoinfty} frac{pi(x)log x}{x}.$$

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

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

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

$$lim_{xtoinfty} frac{pi(x)log x}{x} leqslant lim_{xtoinfty} frac{vartheta(x)}{x},$$

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

(Not an answer, just a suggestion.)

It would seem that the power of a prime $p$ in $[1,2,dots,n]$ is $$leftlfloor frac{log n}{log p}rightrfloor.$$
Start of proof:
$$prod_{pleq n} p^{frac{1}{n}leftlfloor frac{log n}{log p}rightrfloor} = e^{frac{1}{n}sum_{pleq n} log p leftlfloor frac{log n}{log p}rightrfloor}$$

So we need to show that:
$$S_n=frac{1}{n}sum_{pleq n} log p leftlfloor frac{log n}{log p}rightrfloorto 1text{ as }ntoinfty$$

The crudest estimate of the terms is:

$$log n - log p < log p leftlfloor frac{log n}{log p}rightrfloor leq log n$$

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

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

$$S_nleq frac{1}{n}sum_{pleq n}log n = frac{pi(n) n}{log n}to 1$$