What Value of x Causes the Series Ʃ(1/n^x) to Diverge?

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SUMMARY

The series Ʃ(1/n^x) diverges for values of x less than or equal to 1 and converges for values greater than 1. Specifically, Ʃ(1/n) diverges while Ʃ(1/n^2) converges, establishing that the critical threshold for divergence is at p = 1. This conclusion is supported by the integral test, which confirms that the series converges if p > 1. Understanding this relationship is essential for analyzing the convergence of series in mathematical analysis.

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Ʃ(\frac{1}{n}) diverges but Ʃ(\frac{1}{n^2}) converges but both corresponding sequences converge (I think). So at what rate does a sequence have to converge for the corresponding series to converge.

Or asked differently what is the largest value x can be for Ʃ(\frac{1}{n^x}) to diverge
 
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gottfried said:
Ʃ(\frac{1}{n}) diverges but Ʃ(\frac{1}{n^2}) converges but both corresponding sequences converge (I think). So at what rate does a sequence have to converge for the corresponding series to converge.

Or asked differently what is the largest value x can be for Ʃ(\frac{1}{n^x}) to diverge

The dividing line is at p = 1. ##\sum \frac 1 {x^p}## converges if ##p>1##, as you can verify by the integral test.
 
Thanks.
 

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