# Thread: Series converges or diverges

1. ## Series converges or diverges

Determine whether the following series converges or diverges.

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

2. Originally Posted by superman69
Determine whether the following series converges or diverges.

I'm thinking its converges because of the comparison test?
Oh, come on. Think about it.
$\cos(n\pi)$ is just a devious way to say $(-1)^n$.

3. Originally Posted by superman69
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 $\cos{(n\pi)} = (-1)^n$.

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.

4. 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.

5. 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

6. 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?

7. meaning doesn't converge at all as "conditional convergence," yes.

8. 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.

9. 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.

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

Hehe ...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...

Tonio

11. 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

12. 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.