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.



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


