Divergence of the sum of the reciprocals of the primes

Proofs

Euler's proof

Euler's proof works by first taking the natural logarithm of each side, then using the Taylor series expansion for log x as well as the sum of a converging series:

$${\displaystyle \begin{array}{cc}\log \left(\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\right)& =\log \left(\prod _p\frac{1}{1-p^{-1}}\right)=-\sum _p\log \left(1-\frac{1}{p}\right)\\ & =\sum _p\left(\frac{1}{p}+\frac{1}{2p^2}+\frac{1}{3p^3}+\cdots \right)\\ & =\sum _p\frac{1}{p}+\frac{1}{2}\sum _p\frac{1}{p^2}+\frac{1}{3}\sum _p\frac{1}{p^3}+\frac{1}{4}\sum _p\frac{1}{p^4}+\cdots \\ & =A+\frac{1}{2}B+\frac{1}{3}C+\frac{1}{4}D+\cdots \\ & =A+K\end{array}}$$

for a fixed constant K < 1. This holds by the p-series test and the properties of the Riemann zeta function.

Then, by using the following relation:

$${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}=\log \mathrm{\infty },}$$

of which, as shown in a later 1748 work, the right hand side can be obtained by setting x = 1 in the Taylor series expansion

$${\displaystyle \log \left(\frac{1}{1-x}\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}.}$$

Thus, $${\displaystyle A=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\cdots =\log \log \mathrm{\infty }.}$$

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 n is asymptotic to log log n as n 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 p_i denote the ith prime number. Assume that the sum of the reciprocals of the primes converges.

Then there exists a smallest positive integer k such that

$${\displaystyle \sum _{i=k+1}^{\mathrm{\infty }}\frac{1}{p_i}<\frac{1}{2}(1)}$$

For a positive integer x, let M_x denote the set of those n in {1, 2, ..., x} which are not divisible by any prime greater than p_k (or equivalently all n ≤ x which are a product of powers of primes p_i ≤ p_k). We will now derive an upper and a lower estimate for |M_x|, the number of elements in M_x. For large x, these bounds will turn out to be contradictory.

Upper estimate

Every n in M_x can be written as n = m²r with positive integers m and r, where r is square-free. Since only the k primes p₁, ..., p_k can show up (with exponent 1) in the prime factorization of r, there are at most 2^k different possibilities for r. Furthermore, there are at most √x possible values for m. This gives us the upper estimate $${\displaystyle |M_x|\le 2^k\sqrt{x}(2)}$$

Lower estimate

The remaining x − |M_x| numbers in the set difference {1, 2, ..., x} \ M_x are all divisible by a prime greater than p_k. Let N_i,x denote the set of those n in {1, 2, ..., x} which are divisible by the ith prime p_i. Then $${\displaystyle \{1,2,\dots ,x\}\setminus M_x=\bigcup _{i=k+1}^{\mathrm{\infty }}{N}_{i,x}}$$

Since the number of integers in N_i,x is at most ⁠x/p_i⁠ (actually zero for p_i > x), we get $${\displaystyle x-|M_x|\le \sum _{i=k+1}^{\mathrm{\infty }}|N_{i,x}|<\sum _{i=k+1}^{\mathrm{\infty }}\frac{x}{p_i}}$$

Using (1), this implies $${\displaystyle \frac{x}{2}<|M_x|(3)}$$

This produces a contradiction: when x ≥ 2^{2k + 2}, the estimates (2) and (3) cannot both hold, because ⁠x/2⁠ ≥ 2^k√x.

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 log log n. The proof is due to Ivan Niven, adapted from the product expansion idea of Euler. In the following, a sum or product taken over p always represents a sum or product taken over a specified set of primes.

The proof rests upon the following four inequalities:

• Every positive integer i can be uniquely expressed as the product of a square-free integer and a square as a consequence of the fundamental theorem of arithmetic. To see this, start with $${\displaystyle i=q_1^{2{\alpha }_1+{\beta }_1}\cdot q_2^{2{\alpha }_2+{\beta }_2}\cdots q_r^{2{\alpha }_r+{\beta }_r},}$$ where the βs are 0 (the corresponding power of prime q is even) or 1 (the corresponding power of prime q 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: $${\displaystyle i=(p_1{p}_2\cdots p_s)\cdot b^2,}$$ where the first factor, a product of primes to the first power, is square free. Inverting all the is gives the inequality $${\displaystyle \sum _{i=1}^n\frac{1}{i}\le \left(\prod _{p\le n}\left(1+\frac{1}{p}\right)\right)\cdot \left(\sum _{k=1}^n\frac{1}{k^2}\right)=A\cdot B.}$$

To see this, note that $${\displaystyle \frac{1}{i}=\frac{1}{p_1{p}_2\cdots p_s}\cdot \frac{1}{b^2},}$$ and $${\displaystyle \begin{array}{cc}\left(1+\frac{1}{p_1}\right)\left(1+\frac{1}{p_2}\right)\dots \left(1+\frac{1}{p_s}\right)& =\left(\frac{1}{p_1}\right)\left(\frac{1}{p_2}\right)\cdots \left(\frac{1}{p_s}\right)+\dots \\ & =\frac{1}{p_1{p}_2\cdots p_s}+\dots .\end{array}}$$ That is, ${\displaystyle 1/(p_1{p}_2\cdots p_s)}$ is one of the summands in the expanded product A. And since ${\displaystyle 1/b^2}$ is one of the summands of B, every summand ${\displaystyle 1/i}$ is represented in one of the terms of AB when multiplied out. The inequality follows.

• The upper estimate for the natural logarithm $${\displaystyle \begin{array}{cc}\log (n+1)& ={\int }_1^{n+1}\frac{dx}{x}\\ & =\sum _{i=1}^n\underset{<\frac{1}{i}}{\underbrace{{\int }_i^{i+1}\frac{dx}{x}}}\\ & <\sum _{i=1}^n\frac{1}{i}\end{array}}$$
• The lower estimate 1 + x < exp(x) for the exponential function, which holds for all x > 0.
• Let n ≥ 2. The upper bound (using a telescoping sum) for the partial sums (convergence is all we really need) $${\displaystyle \begin{array}{cc}\sum _{k=1}^n\frac{1}{k^2}& <1+\sum _{k=2}^n\underset{=\frac{1}{k^2-\frac{1}{4}}>\frac{1}{k^2}}{\underbrace{\left(\frac{1}{k-\frac{1}{2}}-\frac{1}{k+\frac{1}{2}}\right)}}\\ & =1+\frac{2}{3}-\frac{1}{n+\frac{1}{2}}<\frac{5}{3}\end{array}}$$

Combining all these inequalities, we see that $${\displaystyle \begin{array}{cc}\log (n+1)& <\sum _{i=1}^n\frac{1}{i}\\ & \le \prod _{p\le n}\left(1+\frac{1}{p}\right)\sum _{k=1}^n\frac{1}{k^2}\\ & <\frac{5}{3}\prod _{p\le n}\exp \left(\frac{1}{p}\right)\\ & =\frac{5}{3}\exp \left(\sum _{p\le n}\frac{1}{p}\right)\end{array}}$$

Dividing through by ⁠5/3⁠ and taking the natural logarithm of both sides gives $${\displaystyle \log \log (n+1)-\log \frac{5}{3}<\sum _{p\le n}\frac{1}{p}}$$

as desired. Q.E.D.

Using

$${\displaystyle \sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}=\frac{{\pi }^2}{6}}$$

(see the Basel problem), the above constant log ⁠5/3⁠ = 0.51082... can be improved to log ⁠π²/6⁠ = 0.4977...; in fact it turns out that $${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left(\sum _{p\le n}\frac{1}{p}-\log \log n\right)=M}$$

where M = 0.261497... 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 $${\displaystyle p_n<n\log n+n\log \log n\text{for }n\ge 6}$$

Then $${\displaystyle \begin{array}{cc}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{p_n}& \ge \sum _{n=6}^{\mathrm{\infty }}\frac{1}{p_n}\\ & \ge \sum _{n=6}^{\mathrm{\infty }}\frac{1}{n\log n+n\log \log n}\\ & \ge \sum _{n=6}^{\mathrm{\infty }}\frac{1}{2n\log n}=\mathrm{\infty }\end{array}}$$ 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

$${\displaystyle x_k=\sum _{n=k+1}^{\mathrm{\infty }}\frac{1}{p_n}.}$$

Then for ${\displaystyle i\ge 0}$, the expansion of ${\displaystyle (x_k{)}^i}$ contains at least one term for each reciprocal of a positive integer with exactly ${\displaystyle i}$ prime factors (counting multiplicities) only from the set ${\displaystyle \{p_{k+1},p_{k+2},\cdots \}}$. It follows that the geometric series $\sum _{i=0}^{\mathrm{\infty }}(x_k{)}^i$ contains at least one term for each reciprocal of a positive integer not divisible by any ${\displaystyle p_n,n\le k}$. But since ${\displaystyle 1+j(p_1{p}_2\cdots p_k)}$ always satisfies this criterion,

$${\displaystyle \sum _{i=0}^{\mathrm{\infty }}(x_k{)}^i>\sum _{j=1}^{\mathrm{\infty }}\frac{1}{1+j(p_1{p}_2\cdots p_k)}>\frac{1}{1+p_1{p}_2\cdots p_k}\sum _{j=1}^{\mathrm{\infty }}\frac{1}{j}=\mathrm{\infty }}$$

by the divergence of the harmonic series. This shows that ${\displaystyle x_k\ge 1}$ for all ${\displaystyle k}$, 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 ⁠1/2⁠, which has the form ⁠odd/even⁠. If the nth partial sum (for n ≥ 1) has the form ⁠odd/even⁠, then the (n + 1)st sum is

$${\displaystyle \frac{\text{odd}}{\text{even}}+\frac{1}{p_{n+1}}=\frac{\text{odd}\cdot p_{n+1}+\text{even}}{\text{even}\cdot p_{n+1}}=\frac{\text{odd}+\text{even}}{\text{even}}=\frac{\text{odd}}{\text{even}}}$$

as the (n + 1)st prime p_{n + 1} is odd; since this sum also has an ⁠odd/even⁠ 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 n 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.