# Cauchy Condensation

• Aug 31st 2011, 02:06 PM
veronicak5678
Cauchy Condensation
Does 1/ nlognloglogn converge or diverge?

I am trying to show this with cauchy Condensation using 2^k in the new series. I have

reduced it to

(1/log2) E 1/ n(logn+loglog2)

where E is the sum from n=3 to infinity. I think it diverges, and am trying to show that this expression is larger than

E 1/ n(logn) which I have seen diverges, but the extra n in the denominator is messing me up. Can someone show me how to bound this, or if I am going about this the wrong way?
• Aug 31st 2011, 02:26 PM
Drexel28
Re: Cauchy Condensation
Quote:

Originally Posted by veronicak5678
Does 1/ nlognloglogn converge or diverge?

I am trying to show this with cauchy Condensation using 2^k in the new series. I have

reduced it to

(1/log2) E 1/ n(logn+loglog2)

where E is the sum from n=3 to infinity. I think it diverges, and am trying to show that this expression is larger than

E 1/ n(logn) which I have seen diverges, but the extra n in the denominator is messing me up. Can someone show me how to bound this, or if I am going about this the wrong way?

But isn't it true (via the limit comparison test for example) that $\displaystyle \displaystyle \sum\frac{1}{n\log(n)}$ converges if and only if $\displaystyle \displaystyle \sum\frac{1}{n(\log(n)+\log(\log(2)))}$ does?
• Aug 31st 2011, 03:09 PM
veronicak5678
Re: Cauchy Condensation
We haven't talked about the limit comparison test. How else can you show that?
• Aug 31st 2011, 03:45 PM
Drexel28
Re: Cauchy Condensation
Quote:

Originally Posted by veronicak5678
We haven't talked about the limit comparison test. How else can you show that?

Can you explain intuitively why it's true?
• Aug 31st 2011, 03:57 PM
veronicak5678
Re: Cauchy Condensation
The first series has terms that are always greater than those of the second, so if the second were to diverge to infinity, the greater series also would.