# Thread:

1. ## Convergence/divergence

Can someone help me with the following problem...I'm stuck!

Find whether the following series converges/diverges using one of the following tests: Nth Term Test for Divergence, Integral Test, Alternating Series Test, Absolute Convergence, Geometric Series, P-Series, Comparison Test, Limit Comparison Test, Ratio Test, & Root Test

$\displaystyle\Sigma(-1)^n*\frac{3n}{4n-1}$

I tried the A.S.T. but I can't get past how to figure out whether A_(n+1) < A_n. I found out that it doesn't converge absolutely using the absolute convergence & the nth term test. I tried the ratio test but I get a limit of 1..which tells me nothing about the convergence. Can someone explain what would be the best test to use bc I cannot figure it out. Thanks

2. Originally Posted by dbakeg00
Can someone help me with the following problem...I'm stuck!

Find whether the following series converges/diverges using one of the following tests: Nth Term Test for Divergence, Integral Test, Alternating Series Test, Absolute Convergence, Geometric Series, P-Series, Comparison Test, Limit Comparison Test, Ratio Test, & Root Test

$\displaystyle\Sigma(-1)^n*\frac{3n}{4n-1}$

I tried the A.S.T. but I can't get past how to figure out whether A_(n+1) < A_n. I found out that it doesn't converge absolutely using the absolute convergence & the nth term test. I tried the ratio test but I get a limit of 1..which tells me nothing about the convergence. Can someone explain what would be the best test to use bc I cannot figure it out. Thanks
You have already answered your own question. By the basic divergence test since the limit of the $\displaystyle \lim_{n \to \infty}a_n \ne 0$ the series diverges

3. Originally Posted by TheEmptySet
You have already answered your own question. By the basic divergence test since the limit of the $\displaystyle \lim_{n \to \infty}a_n \ne 0$ the series diverges
Well, I took the absolute value as part of testing for absolute convergence, then took the limit..which telles me that is doesn't converge absolutely. Can't it still converge conditionally? How do I take the limit of the original series with that $(-1)^n$ as part of the original series?

4. Originally Posted by dbakeg00
Well, I took the absolute value as part of testing for absolute convergence, then took the limit..which telles me that is doesn't converge absolutely. Can't it still converge conditionally? How do I take the limit of the original series with that $(-1)^n$ as part of the original series?
Yes it can but the basic divergence test states that

if $\displaystyle \lim_{n \to \infty}a_n \ne 0$
Or $\displaystyle \lim_{n \to \infty}a_n$ does not exists then the series diverges!

$\displaystyle \lim_{n \to \infty}\frac{(-1)^n3n}{4n-1}$ this limit does not exist!

so you are done!

5. Originally Posted by dbakeg00
Well, I took the absolute value as part of testing for absolute convergence, then took the limit..which telles me that is doesn't converge absolutely. Can't it still converge conditionally? How do I take the limit of the original series with that $(-1)^n$ as part of the original series?
NO! if $\displaystyle\lim _{n \to \infty } \left| {a_n } \right| \ne 0$ it diverges period.

6. Thanks for the help gentlemen!

7. Originally Posted by Plato
NO! if $\displaystyle\lim _{n \to \infty } \left| {a_n } \right| \ne 0$ it diverges period.
I'm kind of confused about the Nth Term Test for Divergence. You say that it says $\displaystyle\lim _{n \to \infty } \left| {a_n } \right| \ne 0$

But all the books I have on Calculus say this: $\displaystyle\lim _{n \to \infty } {a_n } \ne 0$

Where are you finding the definition of that test with the absolute value bars included?

8. Originally Posted by dbakeg00
I'm kind of confused about the Nth Term Test for Divergence. You say that it says $\displaystyle\lim _{n \to \infty } \left| {a_n } \right| \ne 0$
But all the books I have on Calculus say this: $\displaystyle\lim _{n \to \infty } {a_n } \ne 0$
Where are you finding the definition of that test with the absolute value bars included?
If we know that $\displaystyle\lim _{n \to \infty } \left| {a_n } \right| \ne 0$, does that imply that $\displaystyle\lim _{n \to \infty } {a_n } \ne 0~?$

9. Originally Posted by Plato
If we know that $\displaystyle\lim _{n \to \infty } \left| {a_n } \right| \ne 0$, does that imply that $\displaystyle\lim _{n \to \infty } {a_n } \ne 0~?$
I don't know if that implies what you say or not. What about a case such as $\displaystyle\Sigma(-1)^n*\frac{1}{\sqrt{n}}$ ?

If you take the limit of the absolute value of a_n, then you get a limit = 0, so it would tell you nothing really other than that this series has the potential to converge. If you were to take the limit of a_n (without the absolute value), the limit doesn't exsist, so you could conclude that the original series diverges. Now, talking about the same series, if you use the Alternating Series Test, it tells you that this series converges. You can then test for absolute convergence and find out that it converges conditionally. So I'm confused because if I just took the limit (of a_n above) straight up then I would say that it didn't exist and therefore the series diverges. However, if I did the alternating series test and then tested for absolute convergence, I would find that this series converges conditionally. I'm not sure if I'm misunderstanding something here or what??

10. Originally Posted by dbakeg00
I don't know if that implies what you say or not. What about a case such as $\displaystyle\Sigma(-1)^n*\frac{1}{\sqrt{n}}$ ?
If you take the limit of the absolute value of a_n, then you get a limit = 0, so it would tell you nothing really other than that this series has the potential to converge.
That is exactly correct, it tells nothing.
But that is not what I said. Said nothing about converging to zero.
Said not converging to zero.
The sentence "If P then Q." is equivalent to "If not Q then not P."
Thus "if $(a_n)\to 0$ then $|a_n|\to 0$" is equivalent to "If $|a_n|\not\to 0$ then $(a_n)\not\to 0$."

That is the first test for divergence.
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