# Homework Help: Absolute Convergent, Conditionally Convergent?

1. May 13, 2012

### knowLittle

1. The problem statement, all variables and given/known data
$\sum _{n=2}\dfrac {\left( -1\right) ^{n}} {\left( \ln n\right) ^{n}}$

3. The attempt at a solution
I have applied the Alternating Series test and it shows that it is convergent. However, I need to show that it's either absolute conv. or conditionally conv.

Next, I tried the root test:
$\lim _{n\rightarrow \infty }\sqrt {\left| \left( \dfrac {1} {\ln n}\right) ^{n}\right| }$, **Correction; this is the root of n**
Now, I'm tempted to use direct comparison with 1/n harmonic series and show divergence. However, I don't know, if I am allowed to do this since the root test tells me that then I have to find the limit of my An and classify it accordingly.
According to the root test:
if the limit >1 or infinity it diverges
if the limit <1 it converges absolutely.
if the limit =1 it's inconclusive.

2. May 13, 2012

### LCKurtz

What do you get when you simplify$$\lim_{n\rightarrow\infty}\root\scriptstyle ^n\of {\frac 1 {(\ln n)^n}}$$Then what happens as $n\rightarrow \infty$? Don't answer that by talking about some series test. This is just the limit of a sequence. What do you get?

3. May 13, 2012

### knowLittle

$$\lim_{n\rightarrow\infty}\root\scriptstyle ^n\of {\frac 1 {(\ln n)^n}}$$ =0.

Then, by the root test my series converges absolutely.

4. May 13, 2012

### LCKurtz

Do you have an aversion to showing your work? How did you get zero?

5. May 13, 2012

### knowLittle

$\lim _{n\rightarrow \infty }\left( \dfrac {1} {\ln n}\right) ^{\dfrac {n} {n}}=\lim _{n\rightarrow \infty }\dfrac {1} {\ln n}=0$

6. May 13, 2012

### LCKurtz

Yes, thank you. So you see the root test was perfect for this problem. Quick and easy and instantly gives convergence. Nothing else needed.

7. May 13, 2012

Thank you.