Thread: 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?

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 converges if and only if does?

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

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?

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.