Help with convergence/divergence


by Song
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Song
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#1
Oct11-06, 02:04 PM
P: 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?
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Song
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#2
Oct11-06, 02:13 PM
P: 47
limit absolute value of nln(n)/ (n+1)(ln(n+1))
--->n/(n+1) so the limit is 1.
courtrigrad
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#3
Oct11-06, 02:37 PM
P: 1,239
yes you are correct. it is conditionally convergent, because it satisfies the alternating series test but fails the ratio test.

StatusX
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#4
Oct11-06, 03:17 PM
HW Helper
P: 2,566

Help with convergence/divergence


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
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#5
Oct11-06, 03:19 PM
P: 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?
Song
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#6
Oct11-06, 03:23 PM
P: 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).
courtrigrad
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#7
Oct11-06, 03:29 PM
P: 1,239
[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
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#8
Oct11-06, 03:33 PM
P: 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.
Song
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#9
Oct11-06, 03:42 PM
P: 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?
StatusX
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#10
Oct11-06, 03:43 PM
HW Helper
P: 2,566
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
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#11
Oct11-06, 03:46 PM
P: 1,239
[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
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#12
Oct11-06, 03:46 PM
P: 47
To StatusX,
I see what you mean now.
Let me re-do my problem.
Thanks by the way.
courtrigrad
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#13
Oct11-06, 03:53 PM
P: 1,239
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
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#14
Oct11-06, 06:45 PM
P: 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?
courtrigrad
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#15
Oct11-06, 06:46 PM
P: 1,239
http://www.physicsforums.com/showthread.php?t=134683
Song
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#16
Oct11-06, 07:03 PM
P: 47
Thanks. But I'm not sure if I get the concept.
Song
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#17
Oct11-06, 07:05 PM
P: 47
a)con.
b)div.
c)con.
d)div.
courtrigrad
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#18
Oct11-06, 07:08 PM
P: 1,239
yeah thats right


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