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abhi@maths
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1. Find whether the following series converges or diverges or is oscillatory
\sum1/(n^3*(sin^2 n))
\sum1/(n^3*(sin^2 n))
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Convergence of \sum1/(n^3*(sin^2 n)) Series
- Thread starter abhi@maths
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SUMMARY
The series \(\sum \frac{1}{n^3 \sin^2 n}\) converges due to the comparison with the convergent p-series \(\sum \frac{1}{n^3}\). Since \(\sin^2 n\) is always positive, it ensures that the terms of the series remain bounded above by \(\frac{1}{n^3}\). Therefore, the convergence of \(\sum \frac{1}{n^3}\) directly implies the convergence of \(\sum \frac{1}{n^3 \sin^2 n}\). This conclusion is established through the direct comparison test in series convergence analysis.
PREREQUISITES- Understanding of series convergence tests, particularly the comparison test.
- Familiarity with p-series and their convergence criteria.
- Basic knowledge of trigonometric functions, specifically the behavior of \(\sin n\).
- Mathematical notation and manipulation of infinite series.
- Study the comparison test for series convergence in detail.
- Learn about p-series and their convergence properties, focusing on p-values.
- Explore the behavior of trigonometric functions in series, particularly \(\sin n\) and \(\cos n\).
- Investigate other series involving trigonometric functions and their convergence or divergence.
Mathematicians, students studying real analysis, and anyone interested in series convergence, particularly in the context of trigonometric functions.
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