1. 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?

2. Re: Cauchy Condensation

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?

3. Re: Cauchy Condensation

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

4. Re: Cauchy Condensation

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?

5. 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.