Help with convergence/divergence

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

Thanks a lot.
 
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
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
\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.
 
  • #12
To StatusX,
I see what you mean now.
Let me re-do my problem.
Thanks by the way.
 
  • #13
For \sum_{n=1}^{\infty}\frac{(-1)^{n} }{(\ln n)^{n}}, you could use the root test. So it is absolutely convergent.
 
  • #14
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
Thanks. But I'm not sure if I get the concept.
 
  • #17
a)con.
b)div.
c)con.
d)div.
 
  • #18
yeah that's right
 
  • #19
I sort of get the concept...but not exactly sure. Thank you so much though.
 
  • #21
I don't think so, because a_{n} is not decreasing.
 
  • #22
Yeah. That's right.
How do you do the integral 2 to infinity lnx/x^2?
Integration by parts?
 
  • #23
yeah int by parts
 

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