I know that $\sum_{p \leq N} \frac{1}{p} \geq \log\log N -1 $
However, want to show that $\sum_{p \leq x} \frac{1}{p} \geq \log\log x -1 $.
If let $N=[x]$, then we get a bound for x, i.e. $N \leq x
Can't seem to be able to conclude that. As we have $\frac{1}{N+1}$