Convergence/Divergence of Given Series Using Alternating Series test.

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 9K views
carlodelmundo
Messages
133
Reaction score
0

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?
 
Physics news on Phys.org
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!