Determine whether the following series converges or diverges.http://hosted.webwork.rochester.edu/...2ae67f7a81.png

I'm thinking its converges because of the comparison test?(Thinking)

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- Nov 13th 2009, 02:57 PMsuperman69Series converges or diverges
Determine whether the following series converges or diverges.

http://hosted.webwork.rochester.edu/...2ae67f7a81.png

I'm thinking its converges because of the comparison test?(Thinking)

- Nov 13th 2009, 03:10 PMPlato
- Nov 13th 2009, 03:10 PMProve It
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. - Nov 13th 2009, 03:22 PMPlato
- Nov 13th 2009, 04:20 PMtonio
- Nov 13th 2009, 11:12 PMsuperman69
- Nov 14th 2009, 01:20 AMKrizalid
meaning doesn't converge at all as "conditional convergence," yes.

- Nov 14th 2009, 01:54 AMHallsofIvy
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. - Nov 14th 2009, 02:41 AMProve It
- Nov 14th 2009, 02:46 AMtonio

Hehe (Rofl)...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...(Cool)

Tonio - Nov 14th 2009, 02:49 AMtonio
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 - Nov 14th 2009, 03:45 AMmr fantastic
Since the OP has been answered and since I don't want to see this thread become a debate on whether or not absolute convergence is relevant (it's

**not**relevant), thread closed.