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, [itex](\ln(n))^n\neq n \ln(n)[/itex], it's [itex]\ln(n^n)= n \ln(n)[/itex]. 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

[tex] \sum_{n=1}^{\infty} \frac{1}{n\ln n} [/tex].

Use the integral test.


[tex] \lim_{x\rightarrow \infty}\frac{n\ln n}{(n+1)(\ln(n+1))} = 1 [/tex]

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 [itex](\ln(n))^n[/itex] and [itex]\ln(n^n)[/itex], and the one you've written in you're first post is not equal to [itex]n \ln(n) [/itex].

courtrigrad

[tex] \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{p}} [/tex]. p cant be 0, because then [tex] a_{n} [/tex] wouldnt be decreasing, and it would fail the alternating series test. Thus [tex] p>0 [/tex].

Song

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

courtrigrad

For [tex] \sum_{n=1}^{\infty}\frac{(-1)^{n} }{(\ln n)^{n}} [/tex], 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

http://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.