1. **State the problem:** Approximate an interval for the sum of the convergent alternating series $$\sum_{n=1}^\infty (-1)^n \frac{2}{n^2}$$ using the Alternating Series Error Bound after the first 6 terms. 2. **Recall the Alternating Series Error Bound:** For an alternating series $$\sum (-1)^n a_n$$ with decreasing positive terms $$a_n$$ converging to $$S$$, the error after $$N$$ terms satisfies $$|S - S_N| \leq a_{N+1}$$. 3. **Calculate the 7th term:** $$a_7 = \frac{2}{7^2} = \frac{2}{49} \approx 0.0408$$ 4. **Sum of first 6 terms:** Calculate $$S_6 = \sum_{n=1}^6 (-1)^n \frac{2}{n^2}$$: $$S_6 = -\frac{2}{1^2} + \frac{2}{2^2} - \frac{2}{3^2} + \frac{2}{4^2} - \frac{2}{5^2} + \frac{2}{6^2}$$ $$= -2 + \frac{2}{4} - \frac{2}{9} + \frac{2}{16} - \frac{2}{25} + \frac{2}{36}$$ $$= -2 + 0.5 - 0.2222 + 0.125 - 0.08 + 0.0556 = -1.6216$$ 5. **Apply the error bound:** $$S$$ lies within $$S_6 \pm a_7$$: $$-1.6216 - 0.0408 \leq S \leq -1.6216 + 0.0408$$ $$-1.6624 \leq S \leq -1.5808$$ 6. **Compare with given bounds:** The given bounds $$-1.6625 \leq S \leq -1.5809$$ closely match our calculation, confirming the approximation. **Final answer:** $$-1.6625 \leq S \leq -1.5809$$