
#1
Nov1005, 11:01 AM

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 



#2
Nov1005, 01:22 PM

Sci Advisor
HW Helper
PF Gold
P: 4,768

No, that is not a propery of series. A counter exemple is
[tex]\sum \frac{1}{n}\frac{1}{n^2}[/tex] 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]). 



#3
Nov1005, 03:52 PM

P: 308

I tried it using the ratio test. I got
[tex]\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 



#4
Nov1005, 04:05 PM

P: 308

divergent series
let me try again...I'm still learning latex....
[tex]\mbox{2}(\frac{n}{n+1})^x[/tex] there. And the limit of THAT is 2. Is this correct? 



#5
Nov1005, 04:29 PM

Sci Advisor
HW Helper
P: 2,538

[tex]\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 [/tex] 


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