# Divergent series

by happyg1
Tags: divergent, series
 Sci Advisor HW Helper PF Gold P: 4,771 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?
$$\lim_{n \rightarrow \infty} 2 \left( \frac{n}{n+1}\right)^x=2[/itex] But [tex] \frac{(\log n)^x}{(\log n+1)^x} \neq 2 \left( \frac{n}{n+1}\right)^x$$