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abhi@maths
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1. Find whether the following series converges or diverges or is oscillatory
[tex]\sum1/(n^3*(sin^2 n))[/tex]
[tex]\sum1/(n^3*(sin^2 n))[/tex]
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The convergence of a series refers to the behavior of the sum of its terms as the number of terms increases. A series is said to converge if the sum of its terms approaches a finite value as the number of terms approaches infinity.
To determine if a series converges or diverges, one can use various tests such as the comparison test, ratio test, root test, or the integral test. These tests involve analyzing the behavior of the terms in the series as the number of terms increases.
The series 1/(n^3*(sin^2 n)) converges, as it can be proven using the comparison test. By comparing it to the series 1/n^3, which is a known convergent series, we can conclude that 1/(n^3*(sin^2 n)) also converges.
The convergence of this series has practical implications in various fields of science and engineering, such as in the analysis of electrical circuits and the study of oscillating systems. Understanding the convergence of this series can also aid in solving mathematical problems and equations that involve similar series.
Yes, the convergence of this series has real-world applications in fields such as signal processing, control theory, and physics. In these fields, the series 1/(n^3*(sin^2 n)) is used to model and analyze the behavior of systems that exhibit oscillatory behavior.