
#1
Oct1106, 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? 



#2
Oct1106, 02:13 PM

P: 47

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



#3
Oct1106, 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.




#4
Oct1106, 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?




#5
Oct1106, 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? 



#6
Oct1106, 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). 



#7
Oct1106, 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] 



#8
Oct1106, 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. 



#9
Oct1106, 03:42 PM

P: 47

Here's another question.
Sigma n=1 to infinity (1)^(n1)/n^p For what values of p is serie convergent? I have p>0, can p be 0 in this case? 



#10
Oct1106, 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].




#11
Oct1106, 03:46 PM

P: 1,239

[tex] \sum_{n=1}^{\infty} \frac{(1)^{n1}}{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].




#12
Oct1106, 03:46 PM

P: 47

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



#13
Oct1106, 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.




#14
Oct1106, 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? 



#15
Oct1106, 06:46 PM

P: 1,239




#16
Oct1106, 07:03 PM

P: 47

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




#17
Oct1106, 07:05 PM

P: 47

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



#18
Oct1106, 07:08 PM

P: 1,239

yeah thats right



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