Homework Help: Determine Absolute Convergence, Conditionally Convergent, or Divergent

1. May 13, 2012

knowLittle

1. The problem statement, all variables and given/known data
$\sum _{n=1}^{\infty }\dfrac {\left( -3\right) ^{n}} {n^{3}}$

According to Wolfram Alpha the series diverges by the Limit Comparison Test, but I remember that the limit comparison only works with series greater than zero. How is this possible?

2. Relevant equations

3. The attempt at a solution
Also, I have no clue how to use the alternating series test on this one, since it's (-3)^n in the numerator not (-1)^n or (-1)^n+1

2. May 13, 2012

sharks

Isn't that the same as:
$$\sum _{n=1}^{\infty }\frac {3^n(-1)^n}{n^3}$$

3. May 13, 2012

knowLittle

Yes, sorry about that. Now, I'm trying to figure out what to do with it.

4. May 13, 2012

sharks

Use the Alternating Series Test.

5. May 13, 2012

knowLittle

Yeah, I'm trying to prove that 3^n > n^3

6. May 13, 2012

sharks

How do you use the Alternating Series Test? There are 2 steps. The first is finding the limit of $a_n$ and testing if it's zero.

7. May 13, 2012

knowLittle

Step1: lim as "n" approaches infinity of A_n =0
Step2: A_n > A_(n+1)

8. May 13, 2012

sharks

Work out the first step and verify if you get zero, or not.

9. May 13, 2012

knowLittle

Yeah, that's what I am trying. Supposedly, I doesn't go to zero. Then from the test of divergence, it diverges. However, I don't know how to test it as n goes to infinity.
$\lim _{n\rightarrow \infty }\dfrac {3^{n}} {n^{3}}$

I'm trying to prove that 3^n is much bigger than n^3; therefore, it diverges.

10. May 13, 2012

sharks

Use L'Hopital's Rule.

11. May 13, 2012

knowLittle

Great! Thank you.

$\lim _{n\rightarrow \infty }\dfrac {3^{n}} {n^{3}}\dfrac {H} {-}\dfrac {3^{n}} {9n}\dfrac {H} {-}\dfrac {n3^{n-1}} {9}$

Diverges.

Why would Wolfram Alpha state that by the limit comparison test, the series diverges?

12. May 13, 2012

sharks

Applying L'Hopital's Rule, your numerator is wrong. What is the derivative of $3^n$? The derivative of your denominator is also wrong.

13. May 13, 2012

knowLittle

(3^n) ln(3)

14. May 13, 2012

sharks

You'll have to apply L'Hopital's Rule three times to get:
$$\lim_{n\to \infty} \frac{3^n (\ln 3)^3}{6}$$

15. May 13, 2012

knowLittle

You are correct.
Why would Wolfram Alpha state that by the limit comparison test, the series diverges?

16. May 13, 2012

Bohrok

I just put it into WolframAlpha and it said "By the limit test, the series diverges" (not limit comparison).