\( \sum _ { n = 2 ^ { n ^ { 2 } } - 1 } ^ { \infty } \frac { 2 } { n ^ { 2 } } \)
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Bonus Knowledge
First, let's delve into the series you're looking at! The expression \( \sum_{n = 2^{n^{2}} - 1}^{\infty} \frac{2}{n^{2}} \) begins at a rather unique starting point, as \( n \) takes values on the order of \( 2^{n^{2}} \). This suggests that for most \( n \), the terms rapidly increase towards infinity. The term \( \frac{2}{n^{2}} \) drops off quickly as \( n \) increases, making our series concentrate contributions at relatively smaller values of \( n \). Now, on to the mathematics! You can realize that while the series technically runs infinitely, due to the nature of \( n \), the effective number of terms actually sums over very few large \( n \). Since the terms decrease like \( \frac{1}{n^{2}} \), and for values like \( n \) around \( 2 \) at our lower bounds, the series converges reminiscent of the classic \( p \)-series tests where \( p > 1 \) ensures convergence! Always keep in mind that shifting indices can be a clever approach to unpack the wonders of such sums!