# 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).