Step-by-step Solution:

The given series is: \[ \frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \cdots \] Rewriting the sum: \[ \sum_{n=1}^{\infty} \frac{2n + 1 - 1}{(2n! + 1)} \] \[ \sum_{n=1}^{\infty} \frac{2n + 1}{2n! + 1} \cdot \frac{1}{2n! + 1} \] The series can be simplified as: \[ \sum_{n=1}^{\infty} \frac{1}{2n! + 1} - \frac{1}{2n! + 1} \] This simplifies to the following: \[ 1 + 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} - \frac{1}{6!} + \cdots = e^{-1} \] Thus, the value of the series is: \[ {e^{-1}} \]