Does $\frac{\mbox{lcm}(1,2,\dots,n+1)}{\mbox{lcm}(1,2,\dots,n)}\to\infty$?

Question

I feel that the answer to the title quesiton is "yes". However, I tried using different bounds on such least common multiples to prove this with no luck. Any input on this is highly appreciated.

Answer 1

The answer is "no", the limit does not exist, because (now I'm just collecting the comments already made...) consider the series $(2p_n-1)_n$, where $p_1,p_2,\dots$ denote all primes. We have $$\operatorname{lcm}(1,\dots,2p_n) = \operatorname{lcm}(1,\dots,2p_n-1),$$ since $p_n$ is smaller than $2p_n$ (so $p_n$ is already in $1,\dots,2p_n$). So the quotient of the lcm is $1$.

If you take the series $(p_n-1)_n$, then there is a "new prime" in the numerator, so $$\operatorname{lcm}(1,\dots,p_n) = p_n \operatorname{lcm}(1,\dots,p_n-1).$$ Therefore our series of lcm-quotients goes to $\infty$, since there are infinitely many primes.

This means the limit does not exist, the limes inferior is $1$ (since $\operatorname{lcm}(1,\dots,n)$ divides $\operatorname{lcm}(1,\dots,n+1)$), the limes superior is $\infty$.

Answer 2

The limit doesn't exist, as has been pointed out.

However, I'm inclined to interpret the question a bit more loosely. Let $f(n) = lcm(1, 2, \ldots n)$; then it makes sense to ask for some sort of "average" value of $f(n+1)/f(n)$.

Now, $f(n+1)/f(n) = p$ if $n+1$ is a power of a prime $p$, and 1 otherwise. So the non-1 values of $f(n+1)/f(n)$ get larger and larger, but also sparser and sparser, giving hope that there is some kind of average. So we first look at the mean of the first $n$ such quotients, as $n \to \infty$, $$ \lim_{n \to \infty} {1 \over n} \sum_{k=1}^n {f(k+1)/f(k)}. $$ But the sum $\sum_{k=1}^n f(k+1)/f(k)$ is at least the sum of all the primes less than $n$, which grows faster than linearly with $n$, so this limit doesn't exist.

But it seems kind of silly to take the mean of quotients anyway. The more natural limit, I think, is $$ \lim_{n \to \infty} f(n)^{1/n} $$ and this in fact does exist, and has value $e$. Of course this doesn't mean that the answer to your original question is $e$. But it means that $f(n)$ is ``about'' $e^n$; more specifically it follows from the prime number theorem that $\lim_{n \to \infty} (\log f(n))/n = 1$. (I'm quoting this from the encyclopedia of integer sequences and don't remember the proof.) This gives some sense of how quickly $f(n)$ grows.