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Divergent series

by happyg1
Tags: divergent, series
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happyg1
#1
Nov10-05, 11:01 AM
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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,
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quasar987
#2
Nov10-05, 01:22 PM
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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]).
happyg1
#3
Nov10-05, 03:52 PM
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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

happyg1
#4
Nov10-05, 04:05 PM
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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?
NateTG
#5
Nov10-05, 04:29 PM
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Quote Quote by happyg1
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?
Yes.
[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]
quasar987
#6
Nov10-05, 07:01 PM
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You should do it my way. It's much more instructive then mechanically using a test anyway.


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