The series of reciprocals of the least common multiples of integers $X_1<X_2<\dots$ converges
$\newcommand{\lcm}{\operatorname{lcm}}$ Note that for $n>1$ the least common multiple $W_n$ of natural numbers $1\leq X_1<X_2<\ldots<X_n$ fulfills \begin{align*} \lcm(X_{n-1},X_n) \leq W_n \end{align*}
The following theorem was originally conjectured by P. Erdös:
Theorem: Let $X_0,\ldots,X_n$ be integers satifying $1\leq X_0<X_1<\ldots<X_n$ then
\begin{align*} \frac{1}{\lcm(X_0,X_1)}+\frac{1}{\lcm(X_1,X_2)}+\cdots+\frac{1}{\lcm(X_{n-1},X_n)}\leq 1-\frac{1}{2^n}\tag{1} \end{align*}
From (1) we obtain by noting that $W_1=\lcm(1,X_1)$
\begin{align*} \frac{1}{W_1}&+\frac{1}{W_2}+\frac{1}{W_3}+\cdots+\frac{1}{W_n}\\ &\leq \frac{1}{W_1}+\frac{1}{\lcm(X_1,X_2)}+\frac{1}{\lcm(X_2,X_3)}+\cdots+\frac{1}{\lcm(X_{n-1},X_n)}\\ &\leq 1-\frac{1}{2^{n}} \end{align*}
and OPs claim follows.
A proof of the theorem was given by D. Borwein in the paper A sum of reciprocals of least common multiples.
Proof (D. Borwein): For $j=1,2,\ldots,n$ let $$S_j=\frac{1}{\lcm(X_0,X_1)}+\cdots+\frac{1}{\lcm(X_{j-1},X_j)}$$
then $\lcm(X_{j-1},X_j)=u_jX_{j-1}=v_jX_{j}$ where $u_j>v_j\geq 1$. Hence \begin{align*} \frac{1}{\lcm(X_{j-1},X_j)}\leq \frac{1}{X_j}\tag{2} \end{align*} It follows from (2) that \begin{align*} \frac{1}{\lcm(X_{j-1},X_j)}\leq(u_j-v_j)\frac{1}{\lcm(X_{j-1},X_j)}=\frac{1}{X_{j-1}}-\frac{1}{X_j}\tag{3} \end{align*} We obtain from (3) \begin{align*} S_j\leq\frac{1}{X_0}-\frac{1}{X_j}\tag{4} \end{align*}
Now we show (1) is valid by considering all possible conditions on $X_0,X_1,\ldots,X_n$
Case 1: $X_n\leq 2^n$. It follows from (4),
\begin{align*} S_n\leq 1 - \frac{1}{X_n}\leq 1-\frac{1}{2^n} \end{align*}
Case 2: $X_j>2^j$ for $1\leq j \leq n$. We obtain by (2) \begin{align*} S_n\leq\frac{1}{X_1}+\frac{1}{X_2}+\cdots+\frac{1}{X_n}<\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^n}=1-\frac{1}{2^n} \end{align*}
Case 3: $X_k\leq2^k$ for some positive integer $k<n$ and $X_j>2^j$ for $k+1\leq j\leq n$. Then, by (2) and (4) \begin{align*} S_n&=S_k+\frac{1}{\lcm(X_{k},X_{k+1})}+\cdots+\frac{1}{\lcm(X_{n-1},X_n)}\\ &<1-\frac{1}{2^k}+\frac{1}{2^{k+1}}+\cdots+\frac{1}{2^n}\\ &=1-\frac{1}{2^n} \end{align*}
We conclude (1) holds in all cases and the theorem follows.
Since each of $X_1,\dots,X_n$ are distinct and divide $W_n$ we have that $$n\leq d(W_n)$$ where $d(n)$ is the number of divisors function. Using the classical result of Severin Wigert that $$\limsup_{n\rightarrow\infty}\frac{\log d(n)}{\log n/\log\log n}=\log2,$$ which is obtained by simply considering integers of the form $\prod_{p\leq x}p$, [See Montgomery and Vaughn chapter 2] we have that $n$ sufficiently large, $$\log n\leq\log d(W_{n})\leq\frac{\log W_{n}}{\log\log W_{n}},$$ and so $$W_{n}\geq n^{\log\log n}$$ when $n$ is sufficiently large. This implies that $$\sum_{n}\frac{1}{W_{n}}$$ converges as desired.