## Gamma

This shows that the sum of the reciprocals of the natural numbers, e.g. 1/1 + 1/2 + 1/3 + 1/4 + 1/5... grows just like the natural logarithm.

$\gamma =\underset{\mathrm{n\to \mathrm{\infty}}}{\overset{\mathrm{lim}}{}}\left(\sum _{\mathrm{\underset{p\in \mathbb{N}}{\overset{p\le \mathrm{n}}{}}}}^{}\frac{1}{p}-\mathrm{ln}n\right)\cong 0.577215\dots $

For details look here.

## Meissel Mertens

This shows that the sum of the reciprocals of the prime numbers, e.g. 1/2 + 1/3 + 1/5 + 1/7 + 1/11 ... grows just like the natural logarithm iterated twice.

$M=\underset{\mathrm{n\to \mathrm{\infty}}}{\overset{\mathrm{lim}}{}}\left(\sum _{\mathrm{\underset{p\in \mathbb{P}}{\overset{p\le \mathrm{n}}{}}}}^{}\frac{1}{p}-\mathrm{ln}\mathrm{ln}n\right)\cong 0.261497\dots $

For details look here.

## Convergence

Note that the convergence is quite slow, as the following Mathematica output shows:

In[1]:= n=1000000; N[Total[1/Select[Range[n],True &]] - Log[ n ]] N[Total[1/Select[Range[n],PrimeQ]] - Log[Log[n]]] Out[2]= 0.577216 Out[3]= 0.261536

## A final thought

Which subset {a, b, c, d, e, ...} of the natural numbers, when summing up their reciprocals, 1/a + 1/b + 1/c + 1/d + 1/e ... grows just like log(log(log(n)))? Prime twins is not big enough as their sum converges (Cf. Brun's Theorem).