Absolute Convergent, Conditionally Convergent?

[knowLittle]

Homework Statement

$\sum _{n=2}\dfrac {\left( -1\right) ^{n}} {\left( \ln n\right) ^{n}}$

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.

 

[LCKurtz]

Homework Statement

$\sum _{n=2}\dfrac {\left( -1\right) ^{n}} {\left( \ln n\right) ^{n}}$

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.

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?

 

[knowLittle]

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?

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

Then, by the root test my series converges absolutely.

 

[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?

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

Then, by the root test my series converges absolutely.

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

 

[knowLittle]

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

 

[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.

 

[knowLittle]

Thank you.