Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $ S= \sum 1/n log^1n log^2n log^3n ..log^{TL(n)}n $.

Is it convergent when $n$ runs on integers say above 2 ?

$log^i n$ denotes the i'th iterate of $log$ (in base 2 ) of $n$, $log^2n$ means $loglogn$ .

$T(n)$ is the tower of $n$ (stack of $n$ 2's) that is $T(1)=2$ , $T(n+1)=2^{T(n)}$.

$TL(n)$ is the towerian log:
$ TL(n) = Sup ( k : T(k) <= n < T(k+1) ) $.

MOTIVATION : Generalizing the following that are called Bertrand series (I think):

$\sum 1/n$ is the harmonic serie , $\sum 1/nlogn$ , $\sum 1/nlognlog^2n $ and $\sum 1/nlognlog^2nlog^3n $ are all known to be divergent.

Here the product of iterated logs is pushed as far as possible and its size depends on the parameter $n$.

share|improve this question
add comment

1 Answer

up vote 8 down vote accepted

The sum diverges. This is Putnam Problem A4, 2008.

share|improve this answer
    
Thank you for this fast and apparently accurate answer. I had made up this problem ten years ago didn't ask the right people. –  Jérôme JEAN-CHARLES Jan 5 '10 at 2:44
    
Do you agree at a glance that the result (divergence) remains the same if we use floor function to round everything inside integer. Meaning replacing any the Log function by floor(log). Floor(x) being the highest integer less than x or equal to it. –  Jérôme JEAN-CHARLES Sep 25 '10 at 15:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.