gottfried said:Ʃ([itex]\frac{1}{n}[/itex]) diverges but Ʃ([itex]\frac{1}{n^2}[/itex]) 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([itex]\frac{1}{n^x}[/itex]) to diverge
The convergence rate of a series refers to how quickly the sequence of partial sums approaches a certain value, known as the limit, as more terms are added to the series. In simpler terms, it describes how fast the series is approaching a particular value.
The convergence rate of a series can be determined by analyzing the behavior of the terms in the series and their relationship to each other. This can involve techniques such as the ratio test, the root test, or the comparison test.
In a convergent series, the sequence of partial sums approaches a specific value as more terms are added, while in a divergent series, the sequence either tends towards infinity or does not have a defined limit. Essentially, a convergent series has a finite sum while a divergent series does not.
The convergence rate of a series can greatly impact its sum. A series with a fast convergence rate will have a smaller margin of error and therefore a more accurate sum, while a series with a slow convergence rate may have a larger margin of error and a less precise sum.
No, a series cannot have a convergence rate of 0. This would imply that the series is either constantly increasing or constantly decreasing, which would make it divergent. A series must have a non-zero convergence rate in order to converge to a specific value.