1. **Problem:** Determine if the series $$\sum_{n=1}^\infty \frac{1}{n^2}$$ converges or diverges using the Integral Test. 2. **Integral Test conditions:** The function $$f(x) = \frac{1}{x^2}$$ must be positive, continuous, and decreasing for $$x \geq 1$$. Since $$\frac{1}{x^2}$$ is positive, continuous, and decreasing on $$[1, \infty)$$, the Integral Test applies. 3. **Apply the Integral Test:** Evaluate the improper integral $$\int_1^\infty \frac{1}{x^2} \, dx$$. 4. **Compute the integral:** $$\int_1^\infty \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_1^t x^{-2} \, dx = \lim_{t \to \infty} \left[-\frac{1}{x}\right]_1^t = \lim_{t \to \infty} \left(-\frac{1}{t} + 1\right) = 1$$ 5. **Conclusion:** Since the integral converges to 1 (a finite number), by the Integral Test, the series $$\sum_{n=1}^\infty \frac{1}{n^2}$$ converges. Final answer: The series converges.