# Prime Number Theorem and sum of reciprocals of primes

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.

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 $y$.