Best approximation for $\displaystyle \sum_{k=2}^n\ln\ln k$? I need the best approximation for 
$\displaystyle{\sum_{k = 2}^{n}\ln\left(\ln\left(k\right)\right)}$. Any suggestion or hint is welcomed.
I derived $n\ln\left(\ln\left(n!\right) \over n\right)$ so is there any better one ?

Euler-Maclaurin series:
$$ \sum_{k=2}^n \ln(\ln(k)) = C + \int_2^n \ln(\ln(t))\ dt + \dfrac{1}{2} \ln(\ln(n)) + \dfrac{1}{12 n \ln(n)} - \dfrac{1}{360 n^3 \ln(n)} - \dfrac{1}{240 n^3 \ln(n)^2} - \dfrac{1}{360 n^3 \ln(n)^3} + O(1/n^5) $$
for some constant $C$,
where $$\eqalign{\int_2^n \ln(\ln(t))\ dt &= n \ln(\ln(n)) - 2 \ln(\ln(2)) - \int_2^n \dfrac{dt}{\ln(t)}\cr &= n \ln(\ln(n)) - Li(n) - 2 \ln(\ln(2)) + Li(2)\cr}$$
Numerically it appears $C \approx -.2412388$.
