What does $\ln(\ln(\ln(n)))$ approximate? If sum of reciprocals of harmonic series approximate $\ln(n)$
$$\lim_{n \to \infty} \left( \sum_{i=1}^{n}\frac{1}{i} - \ln(n) \right) = \gamma $$
And sum of reciprocals of primes approximate $\ln(\ln(n))$ as $n$ goes to infinity
$$\lim_{n \to \infty} \left( \sum_{p \space \text{prime}}\frac{1}{p} - \ln(\ln(n)) \right) = M$$
What does $\ln(\ln(\ln(n)))$ approximate? Or is it related with some infinite sum?
 A: First, a little correction. $\sum\limits_{p \text{ prime}}\frac{1}{p}$ is infinite. The correct equality is $$\lim_{n \to \infty} \left( \sum_{p\le n}\frac{1}{p} - \log\log n \right) = M$$
There's nothing special about the harmonic series and the sum of the reciprocal of the primes. There are a lot of series that behave like that. For example, $$\lim_{n\to\infty}\left(\sum_{k=2}^n\frac{1}{k\log k}-\log\log n\right) = C$$
And the same happen for series that behave like $\log\log\log n$.
You may take, for example $$\sum_{k=2}^n\frac{1}{k\log k \log\log k}$$
or $$\sum_{p\le n}\frac{1}{p\log\log p}$$
A: Naming constants aside, you don't get just $\ln(\ln(\ln n))$. You get the whole series of iterated logarithms provided, of course, you render $n$ large enough to make all the iterated logarithms positive (the required value of $n$ increases very fast with the number of logarithm functions).
Given any sum
$S_n=\Sigma_{n=1}^\infty a_n$
that diverges sufficiently slowly, the series
$T_n=\Sigma_{i=n}^\infty (a_i/S_i)$
also diverges and, as $n\to\infty$ we have
$T_n=\ln(S_n)+c+o(1).$
The case $a_n=1$ provides sufficiently slow convergence, so simply Iterate the above scheme to get the approximations
$\Sigma_{i=1}^n(1/n)\approx \ln n +C_1$
$\Sigma_{i=1}^n[1/(n\Sigma_{j=1}^i(1/j))]\approx \ln(\ln n)+C_2$
$\Sigma_{i=1}^n \dfrac{1/(n\Sigma_{j=1}^i(1/j))}{\Sigma_{j=1}^i[1/(n\Sigma_{k=1}^j(1/k))]}\approx \ln[\ln(\ln n)]+C_3$
and so on.
