# Help with convergence/divergence

1. ### Song

47
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?

2. ### Song

47
limit absolute value of nln(n)/ (n+1)(ln(n+1))
--->n/(n+1) so the limit is 1.

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

4. ### StatusX

2,567
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?

Last edited: Oct 11, 2006
5. ### Song

47
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?

6. ### Song

47
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).

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

Use the integral test.

$$\lim_{x\rightarrow \infty}\frac{n\ln n}{(n+1)(\ln(n+1))} = 1$$

Last edited: Oct 11, 2006
8. ### Song

47
I see. It's divergent for the second one but is convergent for the first one because of Alternating series test.

Thanks a lot.

9. ### Song

47
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?

10. ### StatusX

2,567
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 $n \ln(n)$.

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

12. ### Song

47
To StatusX,
I see what you mean now.
Let me re-do my problem.
Thanks by the way.

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

14. ### Song

47
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?

16. ### Song

47
Thanks. But I'm not sure if I get the concept.

17. ### Song

47
a)con.
b)div.
c)con.
d)div.

yeah thats right

19. ### Song

47
I sort of get the concept...but not exactly sure. Thank you so much though.

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