1
$\begingroup$

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 . However, from that all I can seem to get is this $\sum_{p \leq N+1} \frac{1}{p} \geq \log\log(N+1) -1 >\log\log x-1 $

Can't seem to be able to conclude that. As we have $\frac{1}{N+1}$

  • 0
    I cleaned up the TeX code. Please don't write $log log N$ if you mean $\log\log N$. The code for the former is "log log N"; for the latter it's "\log\log N".2012-02-09

1 Answers 1

1

Well, in your other thread we got $\displaystyle \sum_{p \leq N} \frac{1}{p} \geq \log\log(N+1) -1\ \ $ (with $N \to N+1$)

and this is enough to get $\displaystyle \sum_{p \leq x} \frac{1}{p} \geq \log\log x -1\ \ $ (for $N=\lfloor x \rfloor$)

  • 0
    @simplicity: I updated my answer in the [other thread](http://math.stackexchange.com/questions/105714/analytical-number-theory/105728#105728) to get your initial inequality.2012-02-09