I would like to figure out the asymptotic rate of growth for the sum $S = \sum_{j=2}^{N}{\frac{\ln \ln \ln j}{( \ln \ln j)^2}}$ in the limit of large $N$.
Ultimately, I want to know if $S(N)$ is $O(N^{2})$, $O(N \ln \ln \ln \ln \ln \ln \ln N)$, etc.
This problem occurred in calculating the expected entropy of a particular system. Really, I'm just trying to calculate the asymptotic form of the Shannon Entropy for a particular function.
I really don't know a lot of techniques for such a weird problem. I feel like there are two ways forward: Bound the sum with an integral and try to evaluate that: $S \approx \int_{j=2}^{N}{\frac{\ln \ln \ln j \ dj}{(\ln \ln j)^2}}$.
Or we can use some tricky properties of logs:
\begin{align} e^S &= \exp\left[\sum_{j}^{N}{\ln \ln \ln j} - \sum_{j}^{N}{(\ln \ln j)^2}\right] \\ &= \exp\left[\ln (\prod_{j}^{N}{\ln \ln j)} - \sum_{j}^{N}{(\ln \ln j)^2}\right] \\ &= (\prod_{j}^{N}{\ln \ln j})\ e^{- \sum_{j}^{N}{(\ln \ln j)^2}} \end{align}
Therefore $ S = \ln (\prod_{j}^{N}{\ln \ln j}) - \sum_{j}^{N}{(\ln \ln j)^2}$.
My feeling is that the function changes so slowly that the sum would be $$O(N\ln\ln\ln N/(\ln\ln N)^2)$$
For example, when $N=10^{80}$, the value of $\ln\ln\ln N/(\ln\ln N)^2=0.0607$, while for $N=10^{100}$, the value is $0.0572$. So when you sum up to $10^{100}$, the value is within $6\%$ of the end value for $99.999999999999999999\%$ of the time.