Convergence of Infinite Series: A Comparison Test Approach

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SUMMARY

The discussion centers on the convergence of two infinite series: \(\sum_{n=1}^{\infty} \cos(\frac{\pi}{2n})\) and \(\sum_{n=1}^{\infty} \sin(\frac{\pi}{2n})\). The first series diverges, while the convergence of the second series remains uncertain. A comparison test is suggested as a method to analyze the second series, utilizing the inequality \(\frac{1}{2}x \leq \sin(x)\) for small values of \(x\). This approach provides a pathway to determine the behavior of the sine series as \(n\) approaches infinity.

PREREQUISITES
  • Understanding of infinite series and convergence tests
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Knowledge of the comparison test for series convergence
  • Basic calculus concepts, including limits and inequalities
NEXT STEPS
  • Study the comparison test for series convergence in detail
  • Research the properties of trigonometric functions as \(n\) approaches infinity
  • Explore the implications of the inequality \(\frac{1}{2}x \leq \sin(x)\) in series analysis
  • Examine other convergence tests, such as the ratio test and root test
USEFUL FOR

Students studying calculus, mathematicians interested in series convergence, and educators looking for examples of comparison tests in infinite series.

Barbados_Slim
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I was doing my math homework when I started thinking about the sums of two infinite series.
I determined that the sum of the first series [itex]\sum_{n=1}^{\infty} cos(\frac{\pi}{2n})[/itex] diverges. I could not figure out whether or not the series [itex]\sum_{n=1}^{\infty} sin(\frac{\pi}{2n})[/itex] converges or diverges. I think it diverges but I'm unsure because as n approaches infinity each term in the series approaches zero.
Any help would be much appreciated. Thank you in advance
 
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Try to do a comparison test. Try to use the inequality:

[tex]\frac{1}{2}x\leq \sin(x)[/tex]

which holds for small x.
 

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