Why Mertens could not prove the prime number theorem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T11:14:26Z http://mathoverflow.net/feeds/question/95743 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95743/why-mertens-could-not-prove-the-prime-number-theorem Why Mertens could not prove the prime number theorem? Nilotpal Sinha 2012-05-02T10:03:24Z 2012-05-04T11:39:47Z <p>We know that $$\sum_{n \le x}\frac{1}{n\ln n} = \ln\ln x + c_1 + O(1/x)$$</p> <p>where $c_1$ is a constant. Again Mertens' theorem says that the primes $p$ satisfy $$\sum_{p \le x}\frac{1}{p} = \ln\ln x + M + O(1/\ln x).$$</p> <p>Thus both these divergent series grow at the same rate. Mertens' theorem was proved without using the prime number theorem, some 25 years before PNT was proved. However from these two examples, we cannot conclude that </p> <p>$$\lim_{n \to \infty} \frac{p_n}{n\ln n} = 1$$ otherwise Mertens' would have been the first to prove PNT. <strong>My question is - based on the above two series, what are the technical difficulties that prevent us from reaching the conclusion that $p_n/n\ln n = 1$.</strong> There may be counter examples with other series, so such conclusions may not be true in general. However I am not interested in the general case. Instead I am asking only in case of the sequence $1/n\ln n$ and $1/p_n$.</p> http://mathoverflow.net/questions/95743/why-mertens-could-not-prove-the-prime-number-theorem/95763#95763 Answer by David Speyer for Why Mertens could not prove the prime number theorem? David Speyer 2012-05-02T12:36:39Z 2012-05-02T16:39:46Z <p>Here is a heuristic that I plan to make into a blog post some day. Suppose that there were only finitely many primes with first digit $9$. Is your estimate good enough to see that?</p> <p>To be more precise, suppose that there were no primes between $9 \times 10^k$ and $10^{k+1}$ for all sufficiently large $k$. And suppose that the number of primes between $a$ and $b$, for $10^k \leq a &lt; b \leq 9 \times 10^k$ is $\approx \frac{\log 10}{\log 9} \int_{a}^b dt/\log t$ (when $a$ is not too close to $b$). We'll see later where the fraction $\log 10/\log 9$ comes from.</p> <p>The first thing to note is that this would violate the prime number theorem. In this scenario, we would have $\pi(9 \times 10^k) = \pi(10 \times 10^k)$ for $k$ large. But the prime number theorem says that $$\pi(10\times 10^k) - \pi(9 \times 10^k) \sim \frac{10 \times 10^k}{k \log 10 + \log {10}} - \frac{9 \times 10^k}{k \log 10 + \log {9}} \sim \frac{10^k}{k \log 10} \to \infty.$$ So proving the prime number theorem involves disproving this ridiculous scenario.</p> <p>Now, let's see that the scenario is consistent with $\sum_{p \leq N} 1/p = \log \log N + M + O(1/\log N)$. The sum over the primes between $10^k$ and $10^{k+1}$ would be roughly $$\frac{\log 10}{\log 9} \int_{10^k}^{9 \times 10^k} \frac{dt}{t \log t} = \frac{\log 10}{\log 9} \left( \log \log (9 \times 10^k) - \log \log 10^k \right)$$ $$=\frac{\log 10}{\log 9} \left( \log( k \log 10+\log 9) - \log (k \log 10) \right) = \frac{\log 10}{\log 9} \left( \log \left( 1+\frac{\log 9}{k \log 10} \right) \right)$$ $$= \frac{\log 10}{\log 9} \frac{\log 9}{k \log 10} + O(1/k^2)= \frac{1}{k} + O(1/k^2)$$</p> <p>So $$\sum_{p \leq 10^{n+1}} \frac{1}{p} = \sum_{j=1}^n \left( \frac{1}{j} + O(1/j^2) \right)=$$ $$\log n + B + O(1/n) = \log \log 10^{n+1} + C + O(1/\log 10^{n+1}).$$ <b>Very important exercise left for you:</b> If you redo this computation for $\sum_{p \leq 9 \times 10^k} 1/p$, you get $\log \log (9 \times 10^k) + C + O(1/\log(9 \times 10^k))$ for the same constant $C$. The point is that $\log \log 10^{k+1} - \log \log (9 \times 10^k) = O(1/\log 10^k)$, so this estitmate is consistent with $\sum_{p \leq N} 1/p$ not growing at all between $9 \times 10^k$ and $10 \times 10^k$.</p> <p>This trick is useful for refuting other simple approaches to the PNT. For example, the "primes hate to start with $9$ scenario" is also consistent with $\sum \log p/p^s = 1/(s-1) + O(1)$ as $s \to 1^{+}$, so that is also not enough to prove PNT.</p> http://mathoverflow.net/questions/95743/why-mertens-could-not-prove-the-prime-number-theorem/95822#95822 Answer by GH for Why Mertens could not prove the prime number theorem? GH 2012-05-03T02:34:26Z 2012-05-03T06:14:04Z <p>Let me supplement David Speyer's nice response by elaborating on his original comment and Greg Martin's comment. Let us write $$\sum_{p\leq x}\frac{1}{p}=\ln\ln x+M+R(x),$$ then we have, using Riemann-Stieltjes integrals, $$F(x):=\sum_{x &lt; p \leq 2x} \frac{\ln p}{p}=\int_x^{2x}\ln t\ d(\ln\ln t+M) + \int_x^{2x} \ln(t) dR(t)$$ $$= \int_x^{2x} \frac{dt}{t} + [R(t)\ln t]_x^{2x} - \int_x^{2x} \frac{R(t)}{t} dt = \ln 2 + O( \ln x \sup_{x &lt; t \leq 2x} R(t) ).$$ If $\hat F(s)$ denotes the Mellin transform of $dF(x)$, then with the notation $$S(x):=\sum_{p \leq x} \frac{\ln p}{p}$$ we have $$\hat F(s) = \int_{1-}^\infty x^{-s}dS(2x) - \int_{2-}^\infty x^{-s}dS(x) = (2^s-1)\sum_p \frac{\ln p}{p^{s+1}}, \quad \Re s>0.$$ In particular, if $\zeta(s)$ has a zero on $\Re s=\sigma\geq\frac{1}{2}$, then $\hat F(s)$ has a pole on $\Re s=\sigma -1$. Note that on the real segment $s\geq-\frac{1}{2}$, the only singularity of $\hat F(s)$ is $s=-\frac{1}{2}$, coming from the difference between $\sum (\ln p)p^{-s}$ and $\sum\Lambda(n)n^{-s}$. Hence by a well-known principle (see Theorem 11.8 in Bateman-Diamond: Analytic number theory), for a certain $c\in\mathbb{R}$ we infer the two-sided estimates $$F(x)-\ln 2-c x^{-1/2} = \Omega_\pm(x^{\sigma-1}).$$ This implies, by our initial calculation, $$R(x) = \Omega(x^{\sigma-1}/\ln x).$$ Here we can take $\sigma=\frac{1}{2}$. In the unlikely case that RH fails, we can choose a larger $\sigma$ and replace $\Omega$ by $\Omega_\pm$. Probably, with more work, we could replace $\Omega$ by $\Omega_\pm$ for $\sigma=\frac{1}{2}$ as well.</p> <p><strong>EDIT 1.</strong> I inserted the term $c x^{-1/2}$ to account for the pole at $s=-\frac{1}{2}$.</p> <p><strong>EDIT 2.</strong> I deleted the comment regarding a possible multiple zero of $\zeta(s)$, since the corresponding zero of $-\zeta'(s)/\zeta(s)=\sum\Lambda(n)n^{-s}$ is always simple.</p> http://mathoverflow.net/questions/95743/why-mertens-could-not-prove-the-prime-number-theorem/95960#95960 Answer by Nilotpal Sinha for Why Mertens could not prove the prime number theorem? Nilotpal Sinha 2012-05-04T06:27:57Z 2012-05-04T11:39:47Z <p>Thanks to David very interesting remark and GH for the calculation of R(x). I like to present my reasoning why I thought on the Merten's theorem could imply PNT. I would have put it as a comment but due to the length, I am writing as a reply. In all likely hood, I must have committed an error somewhere hence I am reaching the ridiculous conclusion that Merten's theorem implies PNT. I would appreciate if someone tell me where I went wrong. </p> <p>Let $s(n)$ be a strictly increasing sequence, $s(1) \ge 2$, such that $$S_s(x) = \sum_{s(r) \le x}\frac{1}{s(r)} = \ln\ln x + C_s + R_s(x).$$ where $C_s$ is a constant that depends only on $s$ and $R_s(x) = o(1)$ is the error term which depends on $s$ and $x$. Let $A={[s(1)],[s(2)], [s(3)], \ldots }$ where [.] denotes the greatest integer function. Let $a(r) = 1$ if $r \in A$ and $a(r) = 0$ if $r \notin A$. Then $N_s(x) = \sum_{r \le x} a(r)$ is the number of elements of $A$ in the interval $(0,x)$. Using Abel's summation formula we obtain $$\sum_{s(r) \le x}\frac{1}{s(r)} = \frac{N_s(x)}{x} + \int_{s(1)}^{x} \frac{N_s(t)}{t^2}dt = \ln\ln x + C_s + R_s(x)$$ Differentiating under the integral sign with respect to $x$, we obtain $$\frac{N_s'(x)}{x} - \frac{N_s(x)}{x^2} + \frac{N_s(x)}{x^2} = \frac{1}{x\ln x} + R_s'(x)$$ Simplifying the above equation and integrating both sides, we obtain $$N_s(x) = \int_{s(1)}^{x}\frac{dt}{\ln t} + \int_{s(1)}^{x} tR_s'(t)dt$$ Without loss of generality, we can define $s(1) =2$ as this will only effect the constant term $C_s$ and not $R_s(x)$. Hence we have $$N_s(x) = \int_{2}^{x}\frac{dt}{\ln t} + E(x) = Li(x) + E(x).$$ Thus any sequence $s(n)$ satisfying $S_s(x) = \ln\ln x + C_s + R_s(x)$ must have a density function which is asymptotic to the logarithmic integral. We can verify this with the sequences $n\ln n, n\ln n + n\ln\ln n, nH_n$ etc; where $H_n$ is the harmonic number. Also we have found the explicit relationship between the error term in $S_s(x)$ and that of $N_s(x)$ which is $$E(x) = \int_{2}^{x} tR_s'(t)dt$$</p> <p>In case of prime numbers, $N_s(x) = \pi (x)$ and Merten's theorem shows that primes satisfy the condition on $S_s(x)$. This gives $$\pi(x) = \int_{2}^{x}\frac{dt}{\ln t} + E(x).$$</p> <p>This shows that the fact that $\pi(x) \sim Li(x)$ is a consequence of $S_p(x) \sim \ln\ln x$ and it is not a unique property of primes. In fact it is a common attribute of a general family of sequences which grow at the same asymptotic rate in their dominant term. Primes happen to be just one sequence in this family. What differentiates primes and other members of this family of sequences form each other is difference in their respective error terms $E(x)$ or in terms of $S_s(x)$ the difference in their respective constant terms $C_s$ and their error terms $R_s(x)$. Therefore the correct way to interpret Merten's theorem is that for prime numbers, the constant term in $S_p(x)$ is $$C_p = M = \gamma + \sum_{k=2}^{\infty} \frac{\mu (k)\ln \zeta(k)}{k} \simeq 0.2614972128$$ because if it were any other constant, then we know that we are dealing with some other sequence and not the sequence of primes even though the dominant term in asymptotic expansion of the sum of the reciprocal of the sequence would still be $\ln\ln x$ and its density would still be asymptotic to $Li(x)$.</p>