Determine whether the following series converges or diverges.
I'm thinking its converges because of the comparison test?
The test for absolute convergence would be better.
Note that $\displaystyle \cos{(n\pi)} = (-1)^n$.
So $\displaystyle |\cos{(n\pi)}| = |(-1)^n| = 1$.
Test the convergence of
$\displaystyle \sum_{n = 1}^\infty \left|\frac{\cos{(n\pi)}}{n^{\frac{4}{5}}}\right|$.
If this converges, so does the original series.
One does not normally use "does not converge at all" to mean "conditional convergence"!
I'm not sure why the argument about "absolute convergence" got started, superman69, but the very first post told you what you wanted to know. That is an alternating sequence and the absolute value of the terms $\displaystyle \frac{1}{n^{\frac{4}{4}}}$ goes to 0 as n goes to infinity. Therefore, that series does converge.
Hehe ...perhaps a little confusion because of the words, but to "converge absolutely" is a mathematical term which comes from "absolute convergence": a series $\displaystyle \sum\limits_{n=1}^\infty a_n$ is said to be absolutely convergent or to converge absolutely if $\displaystyle \sum\limits_{n=1}^\infty|a_n|$ converges , and it's said to converge conditionally if $\displaystyle \sum\limits_{n=1}^\infty a_n$ converges but $\displaystyle \sum\limits_{n=1}^\infty |a_n| $ doesn't.
Now, there's a lemma that states that if a series is absolutely convergent then it is (conditionally) convergent, but with the proposed example of $\displaystyle \sum\limits_{n=1}^\infty \left|\frac{cosn\pi}{n^{4\slash 5}}\right|$ we can't use this lemma since this series does not converge absolutely.
That's what I meant...
Tonio
Your logic may have been outstanding but here checking for absolute convergence doesn't really help since the series is NOT absolutely convergent (meaning the series of the absolute value of the general temr is a divergent positive series).
This has nothing to do, as far as I can see, with right or wrong logic.
Tonio