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

Ask by Hodges Conner. in Chile
Jan 19,2025

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Answer

La suma parece tener una notación confusa. Si quieres sumar \( \frac{2}{n^{2}} \) desde \( n = 2 \) hasta el infinito, la expresión correcta sería: \[ \sum_{n=2}^{\infty} \frac{2}{n^{2}} = \frac{\pi^{2}}{3} - 2 \] Si necesitas ayuda con otra suma, por favor proporciona más detalles.

Solution

Parece que hay una confusión en la notación de la suma que has proporcionado. La expresión que has escrito es: \[ \sum_{n = 2^{n^{2}} - 1}^{\infty} \frac{2}{n^{2}} \] En esta notación, el índice de la suma \( n \) aparece tanto en los límites inferiores como superiores de la suma, lo cual genera una dependencia circular y no es una forma estándar de expresar una serie. Normalmente, el índice de la suma debe ser una variable que incrementa de manera independiente dentro de los límites establecidos. Si la intención era sumar desde un valor específico de \( n \) hasta el infinito, por ejemplo, desde \( n = 2 \) hasta \( \infty \), la suma se expresaría así: \[ \sum_{n = 2}^{\infty} \frac{2}{n^{2}} \] Esta serie es una serie p con \( p = 2 \), la cual es convergente. De hecho, sabemos que: \[ \sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6} \] Por lo tanto, \[ \sum_{n=2}^{\infty} \frac{2}{n^{2}} = 2 \left( \sum_{n=1}^{\infty} \frac{1}{n^{2}} - 1 \right) = 2 \left( \frac{\pi^{2}}{6} - 1 \right) = \frac{\pi^{2}}{3} - 2 \] Si tu intención era otra, por favor proporciona más detalles o corrige la notación para poder asistirte mejor.

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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!

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