Help with convergence/divergence

[Song]

Sigma (-1)^n / (ln n)^n

First I start using the ratio test (because I want to test if it's absolutely conver. or conditionally conver.), then I get the limit = 1 so it's inconclusive by the ratio test.
But from the alternating series test, it's convergent. So it's conditionally convergent.

Is my process right?

 

[Song]

limit absolute value of nln(n)/ (n+1)(ln(n+1))
--->n/(n+1) so the limit is 1.

 

[courtrigrad]

yes you are correct. it is conditionally convergent, because it satisfies the alternating series test but fails the ratio test.

 

[StatusX]

As you've written it in the first post, $(\ln(n))^n\neq n \ln(n)$, it's $\ln(n^n)= n \ln(n)$. Which did you mean?

 

[Song]

sigma n=1 to infinite 1/(n*ln(n))
sigma n=1 to infinite (-1)^n /(n*ln(n))

The second one is convergent, how about the first one? It's convergent as well?

 

[Song]

To StatusX,
Find the limit for absolute value of nln(n)/ (n+1)(ln(n+1))
I cross out ln(n) on the top with the ln(n+1) at the bottom, so it left with n/(n+1). Do you mean this one?
For the first one, ln(n)^n is at bottom. So it's same as nln(n).

 

[courtrigrad]

$$\sum_{n=1}^{\infty} \frac{1}{n\ln n}$$.

Use the integral test.$$\lim_{x\rightarrow \infty}\frac{n\ln n}{(n+1)(\ln(n+1))} = 1$$

 

[Song]

I see. It's divergent for the second one but is convergent for the first one because of Alternating series test.

Thanks a lot.

 

[Song]

Here's another question.

Sigma n=1 to infinity (-1)^(n-1)/n^p

For what values of p is serie convergent?

I have p>0, can p be 0 in this case?

 

[StatusX]

I'm saying there's a difference between $(\ln(n))^n$ and $\ln(n^n)$, and the one you've written in you're first post is not equal to $n \ln(n)$.

 

[courtrigrad]

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{p}}$$. p can't be 0, because then $$a_{n}$$ wouldn't be decreasing, and it would fail the alternating series test. Thus $$p>0$$.

 

[Song]

To StatusX,
I see what you mean now.
Let me re-do my problem.
Thanks by the way.

 

[courtrigrad]

For $$\sum_{n=1}^{\infty}\frac{(-1)^{n} }{(\ln n)^{n}}$$, you could use the root test. So it is absolutely convergent.

 

[Song]

Thank you.
Here's a question I don't understand.

Suppose that sigma n=0 to the infinity Csubn*x^n converges when x=-4 and diverges when x=6. What can be said about the convergence or divergence of the following series?

a) Sigma n=0 to infinity Csubn
.....

Can someone give me a hint on this one?

 

[courtrigrad]

https://www.physicsforums.com/showthread.php?t=134683

 

[Song]

Thanks. But I'm not sure if I get the concept.

 

[Song]

a)con.
b)div.
c)con.
d)div.

 

[courtrigrad]

yeah that's right

 

[Song]

I sort of get the concept...but not exactly sure. Thank you so much though.

 

[Song]

But for (d) from https://www.physicsforums.com/showthread.php?t=134683
should it be conver. by the alternating series test?

 

[courtrigrad]

I don't think so, because $$a_{n}$$ is not decreasing.

 

[Song]

Yeah. That's right.
How do you do the integral 2 to infinity lnx/x^2?
Integration by parts?

 

[courtrigrad]

yeah int by parts