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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.

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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 $y$.

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  • $\begingroup$ Thank you very much. This looks great. $\endgroup$ – user152804 May 23 '14 at 22:37

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