Why is $e$ not a Liouville number?

Using Gauss continued fraction for $\tanh$, it is not difficult to show that the continued fraction of $e$ has the following structure: $$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,\ldots]\tag{1}$$ then by studying the sequence of convergents $\left\{\frac{p_n}{q_n}\right\}_{n\geq 1}$ through $$\left|\frac{p_n}{q_n}-\frac{p_{n+1}}{q_{n+1}}\right|=\frac{1}{q_n q_{n+1}}=\frac{1}{q_n(\alpha_{n+1}q_n+q_{n-1})}\tag{2}$$ and $$ \left|e-\frac{p_n}{q_n}\right| = \left|\sum_{k\geq n}\frac{(-1)^k}{q_k q_{k+1}}\right| \tag{3} $$ we may easily get that there is no rational approximation such that $$ \left|e-\frac{p_n}{q_n}\right|\leq \frac{1}{q_n^4}\tag{4} $$ hence $e$ is not a Liouville number. It is not difficult to use $(1)$ to prove the stronger statement

The irrationality measure of $e$ is $2$.