Convergence Test for Alternating Series and When to Use It

[ichilouch]

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(4^k/8^k) would result to divergence
since lim of (4^k/8^k) 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.
[

 

[arildno]

That is perfectly OK for the alternating 1/k-series!

"since lim of (4^k/8^k) is infinity"
Is it?

 

[ichilouch]

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?