# Help with convergence/divergence

1. Oct 11, 2006

### 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?

2. Oct 11, 2006

### Song

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

3. Oct 11, 2006

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

4. Oct 11, 2006

### 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?

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

### 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?

6. Oct 11, 2006

### 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).

7. Oct 11, 2006

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

Use the integral test.

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

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

### 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.

9. Oct 11, 2006

### 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?

10. Oct 11, 2006

### 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 $n \ln(n)$.

11. Oct 11, 2006

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

12. Oct 11, 2006

### Song

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

13. Oct 11, 2006

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

14. Oct 11, 2006

### 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?

15. Oct 11, 2006

16. Oct 11, 2006

### Song

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

17. Oct 11, 2006

### Song

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

18. Oct 11, 2006

yeah thats right

19. Oct 11, 2006

### Song

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

20. Oct 12, 2006