# divergent series

by happyg1
Tags: divergent, series
 P: 308 Hello, I'm working on this problem: Prove that for any real x, the series SUM n=2 to infinity of 1/(log n)^x diverges. So far, I have applied the test that says that if SUM 2^n*a_2n converges then the series converges. I got: 1/log2*SUM 2^n/n^x I know that 1/n^x converges if x>1, but 2^n will explode, so I'd have convergent (sometimes) times divergent = divergent. Can I do that? Thanks, CC
 Sci Advisor HW Helper PF Gold P: 4,765 No, that is not a propery of series. A counter exemple is $$\sum \frac{1}{n}\frac{1}{n^2}$$ In this case, we have divergent times convergeant = convergeant. What you can do to prove the desired result however, is simply show that for all x in R, the limit as n-->infty of the argument of the serie is not 0. Hence it does not converges, hence it diverges. (You'll have to do the 3 separate cases x<0, x=0 and x>0 [the later being the most difficult]).
 P: 308 I tried it using the ratio test. I got $$\2(\frac{n}{n+1})^x[\tex] and the limit of that is 2>1 so it diverges. can I do it that way? CC P: 308 ## divergent series let me try again...I'm still learning latex.... [tex]\mbox{2}(\frac{n}{n+1})^x$$
there. And the limit of THAT is 2.
Is this correct?
 Quote by happyg1 let me try again...I'm still learning latex.... $$\mbox{2}(\frac{n}{n+1})^x$$ there. And the limit of THAT is 2. Is this correct?
$$\lim_{n \rightarrow \infty} 2 \left( \frac{n}{n+1}\right)^x=2[/itex] But [tex] \frac{(\log n)^x}{(\log n+1)^x} \neq 2 \left( \frac{n}{n+1}\right)^x$$