Given $h>0$, I would like to estimate the following summation by some function $f(N)$: 
$$
S\_N=\sum\_{n=2}^{N} \frac{\log n}{\log^h(\log n)}=O(f(N)).
$$


Obviously, we see that 
$$
S\_N>\sum\_{n=2}^{N} \log^\alpha n=O(N^\alpha\log^\alpha(N)), ~~\text{for every}~~0<\alpha<1,
$$
and 
$$
S\_N<\sum\_{n=2}^{N} \log n=O(N\log(N)).
$$
But there is a huge gap between $O(N^\alpha\log^\alpha(N))$ and $O(N\log(N))$. More generally, I want to estimate the summation
$$
S\_N=\sum\_{n=2}^{N} \frac{g(n)}{\log^h(g(n))},
$$
where $g(n)$ has the form $\log(\log (\dots\log n ))$.



Summation of $\log n/ \log(\log n)$
I will answer the question in the title.


Put $f(x) = \sum\_{n \leq x} \log n$. We then have, by partial summation,


$\displaystyle \sum\_{n\leq x} \frac{\log n}{\log \log n} = \frac{1}{\log \log x} f(x) + \int\_1^x \frac{1}{t (\log \log t)^2 \log t} f(t)dt.$


We have an asymptotic formula for $f(x)$, namely 


$\displaystyle f(x) = x \log x - x + O(\log x)$. 


It therefore follows that


$$\displaystyle \sum\_{n \leq x} \frac{\log n}{\log \log n} = \frac{x \log x}{\log \log x} + \int\_1^x \frac{dt}{(\log \log t)^2} + O(x).$$


The integral $\displaystyle \int\_1^x \frac{dt}{(\log \log t)^2}$ is at most $O(x)$, hence we have


$$\displaystyle \sum\_{n \leq x} \frac{\log n}{\log \log n} = \frac{x \log x}{\log \log x} + O(x).$$


