Convergence/Divergence of Given Series Using Alternating Series test.

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SUMMARY

The series from n = 1 to infinity of cos(3nπ) diverges based on the Alternating Series Test and the nth term test for divergence. The terms of the series alternate between -1 and 1, leading to a limit of the sequence a(sub n) that does not exist. Consequently, since the limit does not approach zero, the series diverges. The discussion emphasizes the importance of correctly applying convergence tests in series analysis.

PREREQUISITES
  • Understanding of the Alternating Series Test
  • Familiarity with the nth term test for divergence
  • Knowledge of limits in calculus
  • Ability to express series in LaTeX format
NEXT STEPS
  • Study the Alternating Series Test in detail
  • Learn about the nth term test for divergence
  • Explore convergence tests for series, including the Ratio Test
  • Practice writing mathematical expressions in LaTeX
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Students studying calculus, particularly those focusing on series convergence, as well as educators teaching mathematical analysis and series tests.

carlodelmundo
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Homework Statement



Determine the convergence or divergence of the series: series from n = 1 to infinity of cos (3npi). ((please rewrite in LaTex... idk how?)).


The Attempt at a Solution



I rewrote the series to the following:

series from n = 1 to infinity of (-1)^n ... because the terms of the above series go to -1, 1, -1, 1, -1.

The first requirement of the alternative series test, is to take the limit as n approaches infinity of the sequence a(sub n). Is a(sub n) just equal to 1 in this case? Doesn't that mean it diverges by the nth term test for divergence because the limit is 1 and not equal to 0?
 
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carlodelmundo said:

Homework Statement



Determine the convergence or divergence of the series: series from n = 1 to infinity of cos (3npi). ((please rewrite in LaTex... idk how?)).


The Attempt at a Solution



I rewrote the series to the following:

series from n = 1 to infinity of (-1)^n ... because the terms of the above series go to -1, 1, -1, 1, -1.

The first requirement of the alternative series test, is to take the limit as n approaches infinity of the sequence a(sub n). Is a(sub n) just equal to 1 in this case? Doesn't that mean it diverges by the nth term test for divergence because the limit is 1 and not equal to 0?
This is the right test, but you're a little off. Limit of a_n is not 1, nor is it -1; the limit doesn't exist.
 
Thanks Mark44!
 

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