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I have started to work through the course notes titled "Integers, Polynomials and Finite Fields" by Kenneth Davidson to keep me busy this summer, and there is a question in here

This is an exercise to show that the prime number theorem is plausible.

(a) Use the integral test to show that the following series converge.

$$\sum_{n=2}^{\infty}\frac{1}{n(\log n)^2}$$ and $$\sum_{n=2}^{\infty}\frac{1}{n^2}$$

(b) Show that $\pi(n) > \frac{n}{(\log n)}$ infinitely many times. $\textbf{Hint:}$ Show that the sum of the reciprocals of the primes between $2^{k-1}$ and $2^k$ is at most $2^{1-k} \pi({2^k})$. Use this to estimate the sum of the reciprocals of all primes.

part (a) is straight foreward, however I am having some trouble with (b). I came across this on my search for clues as how to solve it, but the solution assumes the prime number theorem is true, which I obviously cannot do (unless im supposed to use it to see if I end up with a reasonable answer).

Here is what I've done so far:

There are at most $(2^k - 2^{k-1})/2 = 2^{k-1}$ primes between $2^{k-1}$ and $2^{k}$ (which is an obvious over-estimate since not every odd integer is prime) and so the sum of the reciprocals of the primes cannot be more than the product of the inverse of the least possible prime within the range and number of the most possible primes from $2$ to $2^k$, or

$$ \sum_{2^{k-2}}^{2^k} \frac{1}{p_i} < \frac{\pi(2^k)}{2^{k-1}}$$ (where we simply choose $2^{k-1}$ as the least possible value.)

So to estimate the sum of the primes up to $N$ I get, $$\sum_{n=2} \frac{1}{p_n} < \sum_{n=2}\frac{\pi(2^n)}{2^{n-1}}$$

Which covers all of the primes

I feel like I should do something along the lines of $m = 2^n$ and write $\pi(2m)/m$ or $2\pi(m)/m$ but I don't see how exactly I could change the sum of $\pi(2^k)2^{1-k}$ to $2\pi(m)/m$ without double counting. My first attempt was to try and show

$$\sum_{p\;\text{prime}}\frac{1}{p} < \sum_{n\;\text{ odd}}\frac{1}{n} < \int_{2^{k-1}}^{2^{k}}\frac{1}{x}dx < \frac{\pi(2^k)}{2^{k-1}}$$

for primes and odd numbers between $2^{k-1}$ and $2^k$, but was unable to do so. Could someone just nudge me in the right direction, or even let me know if I'm going about this the right way? It would be greatly appreciated.

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Plopperzz
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  • Can you use the inequality $p_{n}<n\log^{2}(n)$ for a sufficiently large $n$? – Marco Cantarini Jun 26 '15 at 13:36
  • @MarcoCantarini Im not too sure, but its possible. All that's given in the question is exactly what I posted; and in the notes, nothing like that mentioned expect the prime number theorem. – Plopperzz Jun 26 '15 at 14:56
  • "Use this to estimate the sum of the reciprocals of all primes." But the sum of the reciprocals of the ALL primes diverges. I think there is something wrong. – Marco Cantarini Jun 26 '15 at 15:02
  • @MarcoCantarini I thought the same thing, however the author definitely knows that it diverges since the last thing before the exercises was a proof that the sum diverges implies infinite primes. I will just try and use the inequality you posted, however and see what I can come up with. – Plopperzz Jun 26 '15 at 15:19
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    To estimate primes between two consecutive powers of two I believe that is a topic related with a known inequality involving Chebishev's $\vartheta-$function defined for $x>0$, by $\vartheta(x)=\sum_{p\leq x}\log p$, it is $\vartheta(2n)-\vartheta(n)<n \log 4$, thus $\vartheta(2^{r+1})-\vartheta(2^r)<2^{r+1} \log 2$. Perhaps, now you can work for $x\geq 2$ with $$\vartheta (x)=\pi(x)\log x -\int_2^x \frac{\pi(t)}{t}dt.$$ I don't known continue your problem. As a reference for the definitions and results see pages 75-83, Chapter 4 from Tom Apostol, Analytic Number Theory. –  Jun 26 '15 at 21:08

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