thumb|300px|The sum of the reciprocal of the primes increasing without bound. The x axis is in log scale, showing that the divergence is very slow. The blue function is a lower bound that also diverges.
The sum of the reciprocals of all prime numbers diverges; that is:
: <math>\sum_{p\text{ prime\frac1p = \frac12 + \frac13 + \frac15 + \frac17 + \frac1{11} + \frac1{13} + \frac1{17} + \cdots = \infty</math>
This was proved by Leonhard Euler in 1737, and strengthens Euclid's 3rd-century-BC result that there are infinitely many prime numbers and Nicole Oresme's 14th-century proof of the divergence of the sum of the reciprocals of the integers (harmonic series).
There are a variety of proofs of Euler's result, including a lower bound for the partial sums stating that
<math display="block">\sum_{\scriptstyle p\text{ prime}\atop \scriptstyle p\le n}\frac1p \ge \log \log (n+1) - \log\frac{\pi^2}6</math>
for all natural numbers . The double natural logarithm () indicates that the divergence might be very slow, which is indeed the case. For instance, the sum of the reciprocals of all prime numbers up to a certain integer does not exceed 3 for the first time until 5195977 and does not exceed 4 for the first time until approximately 1.8 quintillion.
The harmonic series
First, we will describe how Euler originally discovered the result. He was considering the harmonic series
<math display="block"> \sum_{n=1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots = \infty </math>
He had already used the following "product formula" to show the existence of infinitely many primes.
<math display="block"> \sum_{n=1}^\infty \frac{1}{n} = \prod_{p} \left( 1+\frac{1}{p}+\frac{1}{p^2}+\cdots \right) = \prod_{p} \frac{1}{1-p^{-1 </math>
Here the product is taken over the set of all primes.
Such infinite products are today called Euler products. The product above is a reflection of the fundamental theorem of arithmetic. Euler noted that if there were only a finite number of primes, then the product on the right would clearly converge, contradicting the divergence of the harmonic series.
Proofs
Euler's proof
Euler's proof works by first taking the natural logarithm of each side, then using the Taylor series expansion for as well as the sum of a converging series:
<math display="block">\begin{align}
\log \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \log\left( \prod_p \frac{1}{1-p^{-1\right)
= -\sum_p \log \left( 1-\frac{1}{p}\right) \\[5pt]
& = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\[5pt]
& = \sum_{p}\frac{1}{p} + \frac{1}{2}\sum_p \frac{1}{p^2} + \frac{1}{3}\sum_p \frac{1}{p^3} + \frac{1}{4}\sum_p \frac{1}{p^4}+ \cdots \\[5pt]
& = A + \frac{1}{2} B+ \frac{1}{3} C+ \frac{1}{4} D + \cdots \\[5pt]
& = A + K
\end{align}</math>
for a fixed constant . Then, by using the following relation:
<math display="block">\sum_{n=1}^\infty\frac{1}{n} = \log\infty,</math>
of which, as shown in a later 1748 work, the right hand side can be obtained by setting in the Taylor series expansion
<math display="block">\log\left(\frac1{1-x}\right)=\sum_{n=1}^\infty\frac{x^{nn.</math>
Thus,
<math display="block">A = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \cdots = \log \log \infty.</math>
It is almost certain (H. M. Edwards was less sanguine and suggested "[i]t is not clear exactly what Euler understood this equation to mean") that Euler meant that the sum of the reciprocals of the primes less than is asymptotic to as approaches infinity. It turns out this is indeed the case, and a more precise version of this fact was rigorously proved by Franz Mertens in 1874. Thus Euler obtained a correct result by questionable means.
Erdős's proof by upper and lower estimates
The following proof by contradiction comes from Paul Erdős.
Let denote the th prime number. Assume that the sum of the reciprocals of the primes converges.
Then there exists a smallest positive integer such that
<math display="block">\sum_{i=k+1}^\infty \frac 1 {p_i} < \frac12 \qquad(1)</math>
For a positive integer , let denote the set of those in which are not divisible by any prime greater than (or equivalently all which are a product of powers of primes ). We will now derive an upper and a lower estimate for , the number of elements in . For large , these bounds will turn out to be contradictory.
;Upper estimate:
:Every in can be written as with positive integers and , where is square-free. Since only the primes can show up (with exponent 1) in the prime factorization of , there are at most different possibilities for . Furthermore, there are at most possible values for . This gives us the upper estimate <math display="block">|M_x| \le 2^k\sqrt{x} \qquad(2)</math>
;Lower estimate:
:The remaining numbers in the set difference are all divisible by a prime greater than . Let denote the set of those in which are divisible by the th prime . Then <math display="block">\{1,2,\ldots,x\}\setminus M_x = \bigcup_{i=k+1}^\infty N_{i,x}</math>
:Since the number of integers in is at most (actually zero for ), we get <math display="block">x-|M_x| \le \sum_{i=k+1}^\infty |N_{i,x}|< \sum_{i=k+1}^\infty \frac x {p_i}</math>
:Using (1), this implies <math display="block">\frac x 2 < |M_x| \qquad(3)</math>
This produces a contradiction: when , the estimates (2) and (3) cannot both hold, because .
Proof that the series exhibits log-log growth
Here is another proof that actually gives a lower estimate for the partial sums; in particular, it shows that these sums grow at least as fast as . The proof is due to Ivan Niven, adapted from the product expansion idea of Euler. In the following, a sum or product taken over always represents a sum or product taken over a specified set of primes.
The proof rests upon the following four inequalities:
- Every positive integer can be uniquely expressed as the product of a square-free integer and a square as a consequence of the fundamental theorem of arithmetic. Start with <math display="block">i = q_1^{2{\alpha}_1+{\beta}_1} \cdot q_2^{2{\alpha}_2+{\beta}_2} \cdots q_r^{2{\alpha}_r+{\beta}_r},</math> where the βs are 0 (the corresponding power of prime is even) or 1 (the corresponding power of prime is odd). Factor out one copy of all the primes whose β is 1, leaving a product of primes to even powers, itself a square. Relabeling: <math display="block">i = (p_1 p_2 \cdots p_s) \cdot b^2,</math> where the first factor, a product of primes to the first power, is square free. Inverting all the s gives the inequality <math display="block"> \sum_{i=1}^n \frac 1 i \le \left(\prod_{p \le n} \left(1 + \frac 1 p \right)\right) \cdot \left(\sum_{k=1}^n \frac 1 {k^2}\right) = A \cdot B.</math>
To see this, note that <math display="block">\frac 1 i = \frac 1 {p_1 p_2 \cdots p_s} \cdot \frac 1 {b^2},</math> and <math display="block">\begin{align}
\left(1 + \frac{1}{p_1}\right)\left(1 + \frac{1}{p_2}\right) \ldots \left(1 + \frac{1}{p_s}\right) &= \left(\frac{1}{p_1}\right)\left(\frac{1}{p_2}\right)\cdots\left(\frac{1}{p_s}\right) + \ldots\\
&= \frac 1 {p_1 p_2 \cdots p_s} + \ldots.
\end{align}</math> That is, <math>1/(p_1p_2 \cdots p_s)</math> is one of the summands in the expanded product . And since <math>1 / b^2</math> is one of the summands of , every summand <math>1/i</math> is represented in one of the terms of when multiplied out. The inequality follows.
- The upper estimate for the natural logarithm <math display="block">\begin{align}
\log(n+1) &= \int_1^{n+1} \frac{dx}x \\
&= \sum_{i=1}^n\underbrace{\int_i^{i+1}\frac{dx}x}_ - \frac1{k + \frac{1}{2\right)}_{=\, \frac{1}{k^2 - \frac14} \,>\, \frac{1}{k^2 \\
&= 1 + \frac23 - \frac1{n + \frac{1}{2 < \frac53
\end{align}</math>
Combining all these inequalities, we see that
<math display="block">\begin{align}
\log(n+1) & < \sum_{i=1}^n\frac{1}{i} \\
& \le \prod_{p \le n} \left(1 + \frac{1}{p}\right) \sum_{k=1}^n \frac{1}{k^2} \\
& < \frac53\prod_{p \le n} \exp\left(\frac{1}{p}\right) \\
& = \frac53\exp\left(\sum_{p \le n} \frac{1}{p} \right)
\end{align}</math>
Dividing through by and taking the natural logarithm of both sides gives
<math display="block">\log\log(n + 1) - \log\frac53 < \sum_{p \le n} \frac{1}{p}</math>
as desired. Q.E.D.
Using
<math display="block">\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}6</math>
(see the Basel problem), the above constant can be improved to ; in fact it turns out that
<math display="block"> \lim_{n \to \infty } \left( \sum_{p \leq n} \frac{1}{p} - \log \log n \right) = M</math>
where is the Meissel–Mertens constant (somewhat analogous to the much more famous Euler–Mascheroni constant).
Proof from Dusart's inequality
From Dusart's inequality, we get
<math display="block"> p_n < n \log n + n \log \log n \quad\mbox{for } n \ge 6</math>
Then
<math display="block">\begin{align}
\sum_{n=1}^\infty \frac1{ p_n}
&\ge \sum_{n=6}^\infty \frac{1}{ p_n} \\
&\ge \sum_{n=6}^\infty \frac{1}{ n \log n + n \log \log n} \\
&\ge \sum_{n=6}^\infty \frac{1}{2n \log n} = \infty
\end{align}</math>
by the integral test for convergence. This shows that the series on the left diverges.
Geometric and harmonic-series proof
The following proof is modified from James A. Clarkson.
Define the k-th tail
<math display="block">x_{k} = \sum_{n = k+1} ^{\infty} \frac{1}{p_n}.</math>
Then for <math>i \geq 0</math>, the expansion of <math>(x_{k})^{i}</math> contains at least one term for each reciprocal of a positive integer with exactly <math>i</math> prime factors (counting multiplicities) only from the set <math> \{ p_{k+1}, p_{k+2}, \cdots \}</math>. It follows that the geometric series <math display="inline">\sum_{i = 0} ^{\infty} (x_{k})^{i}</math> contains at least one term for each reciprocal of a positive integer not divisible by any <math>p_{n},n\leq k</math>. But since <math>1+j(p_{1}p_{2}\cdots p_{k})</math> always satisfies this criterion,
<math display=block>\sum_{i=0}^{\infty}(x_{k})^{i}>\sum_{j=1}^{\infty} \frac{1}{1+j(p_{1}p_{2} \cdots p_{k})}>\frac{1}{1+p_{1}p_{2} \cdots p_{k \sum_{j=1}^{\infty}\frac{1}{j}=\infty</math>
by the divergence of the harmonic series. This shows that <math>x_{k}\geq 1</math> for all <math>k</math>, and since the tails of a convergent series must themselves converge to zero, this proves divergence.
Partial sums
While the partial sums of the reciprocals of the primes eventually exceed any integer value, they never equal an integer.
One proof is by induction: The first partial sum is , which has the form . If the th partial sum (for ) has the form , then the st sum is
<math display="block">\frac\text{odd}\text{even} + \frac{1}{p_{n+1 = \frac{\text{odd} \cdot p_{n+1} + \text{even{\text{even} \cdot p_{n+1 = \frac{\text{odd} + \text{even\text{even} = \frac\text{odd}\text{even}</math>
as the st prime is odd; since this sum also has an form, this partial sum cannot be an integer (because 2 divides the denominator but not the numerator), and the induction continues.
Another proof rewrites the expression for the sum of the first reciprocals of primes (or indeed the sum of the reciprocals of any finite set of primes) in terms of the least common denominator, which is the product of all these primes. Then each of these primes divides all but one of the numerator terms and hence does not divide the numerator itself; but each prime does divide the denominator. Thus the expression is irreducible and is non-integer.
See also
- Euclid's theorem that there are infinitely many primes
- Small set (combinatorics)
- Brun's theorem, on the convergent sum of reciprocals of the twin primes
- List of sums of reciprocals
References
Sources