# I want to see more of this constant

$\Large \lim_{n \to \infty} \left( \sum_{p\leq n}\frac{1}{p}-\ln(\ln(n))\right)$

If the above limit equals $$M$$, find $$\lfloor100M \rfloor$$.

Clarification: The sum is over all primes less than or equal to $$n$$.

×