prime harmonic seriesPrimeHarmonicSeries2005-03-09 14:32:332006-10-14 15:06:51Theoremprime seriesprime sumreciprocal primeprime harmonic% This is Cosmin's PlanetMath preamble.
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The prime\footnote{\(\bbP\) denotes the set of
primes.} harmonic series (also known as series of reciprocals of primes) is the infinite sum \(\sum_{p\in\bbP}\frac{1}{p}\). The following result was originally proved by Euler (using the Euler product of the Riemann Zeta function) but the following extremely elegant proof is due to Paul Erd\H{o}s \cite{Er}.
\begin{thm}
The series \( \sum_{p\in\bbP}\frac{1}{p} \) diverges.
\end{thm}
\begin{proof}
Assume that this series is convergent. If so, then, for a certain \(
k\in\bbN \), we have:
\[ \sum_{i > k}\frac{1}{p_i} < \frac{1}{2}, \]
where \(p_i\) is the \(i^{\mathrm{th}}\) prime. Now, we define \(
\lambda_k(n) := \#\{ m\leq n : p_i \div m \Rightarrow i \leq k \} \), the
number of integers less than \(n\) divisible only by the first \(k\)
primes. Any of these numbers can be expressed as
\(m = p_1^{\omega_1} \dotsm p_k^{\omega_k} \cdot s^2,\;
\omega_i\in\{0,1\} \) (i.e. a square multiplied by a square-free number). There are \(2^k\) ways to chose the square-free part and clearly \(s\leq
\sqrt{n}\), so \( \lambda_k(n) \leq 2^k \sqrt{n}\). Now, the number of integers divisible by \(p_i\) less than \(n\) is \(\bigl\lfloor\frac{n}{p_i}\bigr\rfloor\), so the number of integers less than \(n\) divisible by primes bigger than \(p_k\) (which we shall denote by \( \Lambda_k(n) \)) is bounded above as follows:
\[\Lambda_k(n) \leq \sum_{i>k}\left\lfloor\frac{n}{p_i}\right\rfloor \leq \sum_{i>k}\frac{n}{p_i} < \frac{n}{2}.\]
However, by their definitions, \(\lambda_k(n)+\Lambda_k(n) = n\) for all \(n\in\bbN\) and so it is sufficient to find an \(n\) such that \(\lambda_k(n)\leq\frac{n}{2}\) for a contradiction and, using the previous bound for \( \lambda_k(n) \), which is \( \lambda_k(n) \leq 2^k \sqrt{n}\), we see that \( n = 2^{2k+2} \) works.
\end{proof}
The series is in some ways similar to the \PMlinkname{Harmonic series}{HarmonicSeries} \(H_n := \sum_{k = 1}^n \frac{1}{n}\). In fact, it is well known that \(H_n = \log n + \gamma + o(1) \), where \(\gamma = 0.577215664905\dots\) is Euler's constant, and this series obeys the similar asymptotic relation \(\sum_{k=1}^n\frac{1}{p_k}= \log\log n + C + o(1)\), where \(C = 0.26149721\dots\) and is sometimes called the Mertens constant. Its divergence, however, is extremely slow: for example, taking \(n\) as the biggest currently known prime, the \( 42^{\mathrm{nd}} \) Mersenne prime \(M_{25964951}\), we get \(\log\log(2^{25964951} - 1) + C \approx 16.9672\) (while \(H_{M_{25964951}}\approx 1.79975\cdot 10^{17} \) which is enormous considering \(H_n\)'s also slow divergence).
\begin{thebibliography}{2}
\bibitem{AZ} \textsc{M.~Aigner \& G.~M.~Ziegler}: \emph{Proofs from THE BOOK}, 3\(^\mathrm{rd}\) edition (2004), Springer-Verlag, 5--6.
\bibitem{Er} \textsc{P.~Erd\H{o}s}: \emph{\"Uber die Reihe \(\sum\frac{1}{p}\)}, Mathematica, Zutphen B 7 (1938).
\end{thebibliography}