1. **State the problem:** We want to analyze the series $$\sum_{n=1}^\infty \left( \frac{n^2}{2^n} + \frac{1}{n^2} \right)$$ and determine its behavior or sum if possible. 2. **Recall the formulas and rules:** - The series is a sum of two separate series: $$\sum_{n=1}^\infty \frac{n^2}{2^n}$$ and $$\sum_{n=1}^\infty \frac{1}{n^2}$$. - We can analyze each series separately and then add their sums if they converge. - The series $$\sum_{n=1}^\infty \frac{1}{n^2}$$ is a p-series with $$p=2$$, which converges to $$\frac{\pi^2}{6}$$. - The series $$\sum_{n=1}^\infty \frac{n^2}{2^n}$$ is a power series with general term involving $$n^2$$ and geometric denominator, which converges and has a known closed form. 3. **Evaluate the first series:** The formula for $$\sum_{n=1}^\infty n^2 x^n$$ when $$|x|<1$$ is: $$\sum_{n=1}^\infty n^2 x^n = \frac{x(1+x)}{(1-x)^3}$$ Substitute $$x=\frac{1}{2}$$: $$\sum_{n=1}^\infty \frac{n^2}{2^n} = \frac{\frac{1}{2} (1 + \frac{1}{2})}{(1 - \frac{1}{2})^3} = \frac{\frac{1}{2} \times \frac{3}{2}}{\left(\frac{1}{2}\right)^3} = \frac{\frac{3}{4}}{\frac{1}{8}} = \frac{3}{4} \times 8 = 6$$ 4. **Evaluate the second series:** $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.644934$$ 5. **Combine the results:** $$\sum_{n=1}^\infty \left( \frac{n^2}{2^n} + \frac{1}{n^2} \right) = 6 + \frac{\pi^2}{6}$$ **Final answer:** $$\boxed{6 + \frac{\pi^2}{6}}$$