Series involving repeated logarithm

[pepe98]

For every positive integer k, write
logk(x)=loglog⋯log(x)
where the logarithm has been taken k times. For instance,
log1(x)=log(x),log2(x)=loglog(x),⋯.
Define a function f:N→Rf:N→R by
f(n)=n*log(n)*log2(n)*...*logk[n](n)
where k[n] is the largest integer, s.t. logk[n](n)≥1. Does the series
sum(1/f(n)) converge?
[For log I mean the natural logarithm though it should obviously work for any basis]


I tried the condensation test (since 1/f(n) presents n in the denominator), so I considered the sum of (2^n)1/f(2^n): here the isolated 2^n are cancelled out. Then the denominator(from a certain n) presents the term loglog(2^n)=nlog(2*((log(2))^1/n)), so I have again an isolated n in the denominator: I do the condensation test again, considering (2^n)*(2^(2^n))*f(2^(2^n)). So, from a certain n, in the denominator I have logloglog(2^(2^n)), from which I can take out n. I proceed this way doing condensation test infinite time, and I end up with an infinite sum that I don't know wether it converges or diverges, nor if I can write in a compact way, and I don't know if I am on the right way to solve the problem.

 

[pka]

For every positive integer k, write
logk(x)=loglog⋯log(x) where the logarithm has been taken k times. For instance,
log1(x)=log(x),log2(x)=loglog(x),⋯.
Define a function f:N→Rf:N→R by
f(n)=n*log(n)*log2(n)*...*logk[n](n)
where k[n] is the largest integer, s.t. logk[n](n)≥1. Does the series
sum(1/f(n)) converge?[For log I mean the natural logarithm though it should obviously work for any basis]

The above shows massive confusions on the part of the poster.
In what follows \(n\in\mathbb{Z}^+\) so \(\log ({f^n}) = n\log (f) = \sum\limits_{k = 1}^n {\log (f)} \)
On the other hand \({\left( {\log (f)} \right)^n} = \prod\limits_{k = 1}^n {\log (f)} \)