TITLE: Prime Number Theorem and sum of reciprocals of primes
QUESTION [7 upvotes]: This is not a homework problem.  I am a mathematician (group representations and classical analysis) who never studied number theory and am beginning with Niven’s book.  
My question concerns the second part of a problem from Chapter $1, \S 3$ of Niven’s Introduction to Number Theory:
I've done the first part.  Here it is:
With $\pi(x)$ = number of primes $\leq  x,$ show that the sum of the reciprocals of primes  $\leq x$ is equal to
$$\frac{\pi(x)}{x}   + \int_{2}^{x} \frac{\pi(u)}{u^{2}}du,$$ 
that is,
$$\sum_{p\leq x } \frac{1}{p}  = \frac{\pi(x)}{x}   + \int_{2}^{x} \frac{\pi(u)}{u^{2}}du $$
I am hunting for a hint on how do the second part: Use theorem $1.19$ (below) to prove that 
$$\limsup_{x\rightarrow \infty}~\frac{\pi (x)}{x/\log x}  \geq  1$$
Theorem $1.19$ says:
For every real $y\geq 2,$ the sum of the (reciprocals of primes  $\leq y$)
$$\sum_{p\leq y} \frac{1}{p}  >  \log \log y - 1 $$    
I have no idea how to begin this.  I realize that this problem is asking for what looks like a partial proof of the Prime Number Theorem, but it appears in Niven’s book far before he fully addresses the PNT.  I've tried to use MathJax, with great difficulty.

REPLY [4 votes]: Assume the contrary, i.e., that $\pi(x)\le q\frac{x}{\log x}$ for all sufficiently large $x$ with $q<1$. Then $$\sum_{p\le y}\frac 1p=\frac{\pi(y)}{y+1}+\sum_{x\le y}\frac{\pi(x)}{x(x+1)}\le \text{Const}+q\sum_{3\le x\le y}\frac{x}{x(x+1)\log x}\\ \le\text{Const}+q\int_2^y\frac{1}{x\log x}\,dx=\text{Const}+q\log\log y<\log\log y-1$$
for large $y$.