How to find asymptotics of this sum

Question

Is there any way to find $f(n)$ in this term: $$\sum_{k=2}^n \frac1{\ln \ln(k!^{k!})} \sim f(n)?$$

The tilde symbol means that $$\lim_{n\to∞} \frac{f(n)}{\sum_{k=2}^n \frac1{\ln \ln(k!^{k!})}} = 1$$

Answer 1

We have: k! ~ $(\frac{k}{e})^k*\sqrt{2\pi*k}$ (stirling formula)

equivalence is compatible with ln, that is: $a_n$ ~ $b_n$, $a_n$,$b_n$ > 0 => $ln(a_n)$ ~ $ ln(b_n) $

we get: $ln(k!)$ ~ $k*ln(k)$ (I skipped the calculus)

$ln(ln(k!^{k!}))$ = $ln(k!) + ln(ln(k!)) $ ~ $ln(k!) $ since $\frac{ln(x)}{x} -->0$, when x -> $+\infty$

Let $V_k$ be the general term of your series:

$V_k$ ~ 1/ln(k!) ~ $\frac{1}{k*ln(k)}$ = $U_k$

The series of general term $(U_k)$ is divergent, hence your series is also divergent and we can say that the two respective partial sums are equivalent.

So what you are looking for is : $f(n)$ ~ $\sum_{k=2}^n U_k $

But this sum is also equivalent to the integral of $g(x) = \frac{1}{x*ln(x)}$ from 2 to n (comparison with integral with a decreasing function, you can prove that)

A primitive of g is : $h(x) = ln(ln(x))$

So $f(n)$ ~ $ln(ln(n))$

What do you think?