Determine the sum of the following series. \[ \sum_{n=1}^{\infty}\left(\sin \left(\frac{-4}{n}\right)-\sin \left(\frac{-4}{n+1}\right)\right) \]

Let \( S = \sum_{n=1}^{\infty}\left(\sin \left(\frac{-4}{n}\right)-\sin \left(\frac{-4}{n+1}\right)\right) \). We can interpret this sum as a telescoping series. To see this, we can rewrite \( S \) by expanding the terms: \[ S = \left( \sin \left(\frac{-4}{1}\right) - \sin \left(\frac{-4}{2}\right) \right) + \left( \sin \left(\frac{-4}{2}\right) - \sin \left(\frac{-4}{3}\right) \right) + \left( \sin \left(\frac{-4}{3}\right) - \sin \left(\frac{-4}{4}\right) \right) + \ldots \] Notice that in this series, most terms cancel each other: \[ S = \sin \left(\frac{-4}{1}\right) - \lim_{n \to \infty} \sin \left(\frac{-4}{n+1}\right) \] Now, let's calculate \( \sin \left(\frac{-4}{1}\right) \): \[ \sin \left(\frac{-4}{1}\right) = \sin(-4) \] Next, we consider the term \( \lim_{n \to \infty} \sin \left(\frac{-4}{n+1}\right) \). As \( n \) approaches infinity, \( \frac{-4}{n+1} \) approaches \( 0 \), and thus: \[ \lim_{n \to \infty} \sin \left(\frac{-4}{n+1}\right) = \sin(0) = 0 \] Putting this all together, we have: \[ S = \sin(-4) - 0 = \sin(-4) \] Using the odd property of sine, which states \( \sin(-x) = -\sin(x) \), we can further simplify: \[ \sin(-4) = -\sin(4) \] Therefore, the sum of the series is: \[ \boxed{-\sin(4)} \]