\( \sum _ { n = 2 ^ { n ^ { 2 } - 1 } } ^ { \infty } \frac { 2 } { n ^ { 2 } } \)

To tackle this intriguing series, let’s begin by breaking down the summation. The upper limit of the summation is infinity, while the lower limit is \( n = 2^{n^2 - 1} \). This expression can yield infinitely many terms due to the complexity of the lower limit. However, the general term in the series, \( \frac{2}{n^2} \), suggests that the terms will decrease rapidly for large \(n\), hinting at convergence. Of course, if we ignore the lower limit for a moment, we know that the series \( \sum_{n=1}^\infty \frac{1}{n^2} \) converges (specifically, it converges to \( \frac{\pi^2}{6} \)). Given how the terms \( \frac{2}{n^2} \) taper off, we can confidently affirm that our series would converge, even if it starts from a higher number. Now, if we want to evaluate the series from its starting point rather than an indefinite series, we would need either numerical methods or deeper analysis of how the terms behave as \( n \) increases. To truly engage with its values or form, we could invoke special techniques or functions to analyze it, perhaps even connecting back to the famous Basel problem. In summary, this intriguing summation leads to considerations in convergence and series behavior, making it a delightful puzzle in mathematical exploration.