# Help with convergence/divergence

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 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?
 P: 47 limit absolute value of nln(n)/ (n+1)(ln(n+1)) --->n/(n+1) so the limit is 1.
 P: 1,240 yes you are correct. it is conditionally convergent, because it satisfies the alternating series test but fails the ratio test.
HW Helper
P: 2,566

## Help with convergence/divergence

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?
 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?
 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).
 P: 1,240 $$\sum_{n=1}^{\infty} \frac{1}{n\ln n}$$. Use the integral test. $$\lim_{x\rightarrow \infty}\frac{n\ln n}{(n+1)(\ln(n+1))} = 1$$
 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.
 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?
 HW Helper P: 2,566 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 $n \ln(n)$.
 P: 1,240 $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{p}}$$. p cant be 0, because then $$a_{n}$$ wouldnt be decreasing, and it would fail the alternating series test. Thus $$p>0$$.
 P: 47 To StatusX, I see what you mean now. Let me re-do my problem. Thanks by the way.
 P: 1,240 For $$\sum_{n=1}^{\infty}\frac{(-1)^{n} }{(\ln n)^{n}}$$, you could use the root test. So it is absolutely convergent.
 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?
 P: 1,240
 P: 47 Thanks. But I'm not sure if I get the concept.
 P: 47 a)con. b)div. c)con. d)div.
 P: 1,240 yeah thats right

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