# Least common multiple of sequence of integers

Let $a_n$ be a strictly increasing sequence of natural numbers such that $$\lim_{n\to\infty}\frac{a_n}{2^n}=0.$$ Now, define $$l_n=\frac{lcm(a_1,a_2,...,a_{n-1})}{a_n}.$$ My question is: From the definition of $a_n$, can we say that $l_n\to\infty$?

While I haven't been able to find any previous work answering this question, a similar question proves that the least common multiple of a random subset of $\{1,2,...,n\}$ is generally asymptotic to $2^{n(1+o(1))}$ (Found here). Also, I have been unable to find a counterexample or prove that $l_n$ does not go to infinity.

Write the first $10$ terms as follows:

$a_1=2^0,a_2=2^1,a_3=2^0\times 3,a_4=2^2,a_5=2\times 3,a_6=2^3,a_7=2^2\times 3,a_8=2^4,a_9=2^3\times 3,a_{10}=2^9$

Now write the next $10$ terms

$a_{11}=2^8\times 3,a_{12}=2^{10},a_{13}=2^{9}\times 3,a_{14}=2^{11},a_{15}=2^{10}\times 3,a_{16}=2^{12},a_{17}=2^{11}\times 3,a_{18}=2^{13},a_{19}=2^{12}\times 3,a_{20}=2^{18}$

Similarly write the next $10$ terms

$a_{21}=2^{17}\times 3,a_{22}=2^{19},a_{23}=2^{18}\times 3,a_{24}=2^{20},a_{25}=2^{19}\times 3,a_{26}=2^{21},a_{27}=2^{20}\times 3,a_{28}=2^{22},a_{29}=2^{21}\times 3,a_{30}=2^{27}$

Now continue the sequence.

General term: For all $n\geq 1$

$a_{10n}=2^{9n},a_{10n+1}=2^{9n-1}\times 3,a_{10n+2}=2^{9n+1},a_{10n+3}=2^{9n}\times 3,a_{10n+4}=2^{9n+2},a_{10n+5}=2^{9n+1}\times 3,a_{10n+6}=2^{9n+3},a_{10n+7}=2^{9n+2}\times 3,a_{10n+8}=2^{9n+4},a_{10n+9}=2^{9n+3}\times 3$

Clearly for the above defined sequence, $$\lim_{n\to\infty}\frac{a_n}{2^n}=0.$$

But $l_n$ does not approach infinity!!

(As $l_{10n}=(2^{9n-5}\times 3)/2^{9n}=3/32$)