Series converges or diverges

• Nov 13th 2009, 02:57 PM
superman69
Series 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 PM
Plato
Quote:

Originally Posted by superman69
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)

Oh, come on. Think about it.
$\cos(n\pi)$ is just a devious way to say $(-1)^n$.
• Nov 13th 2009, 03:10 PM
Prove It
Quote:

Originally Posted by superman69
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)

The test for absolute convergence would be better.

Note that $\cos{(n\pi)} = (-1)^n$.

So $|\cos{(n\pi)}| = |(-1)^n| = 1$.

Test the convergence of

$\sum_{n = 1}^\infty \left|\frac{\cos{(n\pi)}}{n^{\frac{4}{5}}}\right|$.

If this converges, so does the original series.
• Nov 13th 2009, 03:22 PM
Plato
Quote:

Originally Posted by Prove It
The test for absolute convergence would be better.

Actually that test, the absolute convergence, has nothing to do with this question.
Apply the alternating series test.
• Nov 13th 2009, 04:20 PM
tonio
Quote:

Originally Posted by Prove It
The test for absolute convergence would be better.

Note that $\cos{(n\pi)} = (-1)^n$.

So $|\cos{(n\pi)}| = |(-1)^n| = 1$.

Test the convergence of

$\sum_{n = 1}^\infty \left|\frac{\cos{(n\pi)}}{n^{\frac{4}{5}}}\right|$.

If this converges, so does the original series.

Indeed, but the problem is that this series does not converge absolutely so this test won't help at all. Leibnitz will do the work, though.

Tonio
• Nov 13th 2009, 11:12 PM
superman69
Quote:

Originally Posted by tonio
Indeed, but the problem is that this series does not converge absolutely so this test won't help at all. Leibnitz will do the work, though.

Tonio

So this does not converge at all?
• Nov 14th 2009, 01:20 AM
Krizalid
meaning doesn't converge at all as "conditional convergence," yes.
• Nov 14th 2009, 01:54 AM
HallsofIvy
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 $\frac{1}{n^{\frac{4}{4}}}$ goes to 0 as n goes to infinity. Therefore, that series does converge.
• Nov 14th 2009, 02:41 AM
Prove It
Quote:

Originally Posted by HallsofIvy
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 $\frac{1}{n^{\frac{4}{4}}}$ goes to 0 as n goes to infinity. Therefore, that series does converge.

I believe you mean $\frac{1}{n^{\frac{4}{5}}}$.

And I KNEW my logic was right about checking for absolute convergence.
• Nov 14th 2009, 02:46 AM
tonio
Quote:

Originally Posted by superman69
So this does not converge at all?

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 $\sum\limits_{n=1}^\infty a_n$ is said to be absolutely convergent or to converge absolutely if $\sum\limits_{n=1}^\infty|a_n|$ converges , and it's said to converge conditionally if $\sum\limits_{n=1}^\infty a_n$ converges but $\sum\limits_{n=1}^\infty |a_n|$ 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 $\sum\limits_{n=1}^\infty \left|\frac{cosn\pi}{n^{4\slash 5}}\right|$ 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 AM
tonio
Quote:

Originally Posted by Prove It
I believe you mean $\frac{1}{n^{\frac{4}{5}}}$.

And I KNEW my logic was right about checking for absolute convergence.

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 AM
mr 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.