Convergence Test Help

  • Thread starter ichilouch
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  • #1
ichilouch
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Homework Statement


Ʃ cos(k*pi)/k from 1 to infinity.
This is a test for convergence.

and when is the proper time to use the alternating series test
like using it on (-1)k(4k/8k) would result to divergence
since lim of (4k/8k) is infinity and not 0 but the function is really
convergent by the geometric series?

Homework Equations





The Attempt at a Solution


Is it right to do this:

for k is odd cos is negative and for k is even cos is positive
then
cos(k*pi)\k < (-1)k/k
and by alternating series test ;
(-1)k*(1/k) since 1/k is decreasing and lim as 1/k approaches infinity is 0 then
(-1)k1/k converges thus cos(k*pi)/k converges by direct comparison test.
[
 

Answers and Replies

  • #2
arildno
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That is perfectly OK for the alternating 1/k-series! :smile:

"since lim of (4^k/8^k) is infinity"
Is it?
 
  • #3
ichilouch
9
0
Thanks for the reply
So rechecking limit of (4/8)^k as k approaches infinity is 0
Then if it is sine or cosine over a variable, the comparison test for it is (-1)^k?
and if it is sine or cosine squared, the comparison is 1 only?
 

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