1. **Problem:** Determine whether the series $$\sum_{n=1}^\infty \frac{2 - \sin^2(n)}{e^n}$$ converges or diverges. 2. **Test and formula:** We use the Comparison Test. Since $\sin^2(n)$ is bounded between 0 and 1, $2 - \sin^2(n)$ is bounded between 1 and 2. 3. **Intermediate work:** $$1 \leq 2 - \sin^2(n) \leq 2$$ Therefore, $$\frac{1}{e^n} \leq \frac{2 - \sin^2(n)}{e^n} \leq \frac{2}{e^n}$$ 4. **Explanation:** The series $$\sum_{n=1}^\infty \frac{1}{e^n}$$ is a geometric series with ratio $\frac{1}{e} < 1$, so it converges. By the Comparison Test, since $$\sum \frac{1}{e^n}$$ converges and $$\frac{2 - \sin^2(n)}{e^n}$$ is bounded above by a constant multiple of $$\frac{1}{e^n}$$, the original series converges. **Final answer:** The series $$\sum_{n=1}^\infty \frac{2 - \sin^2(n)}{e^n}$$ converges by the Comparison Test.