Divergence of the sum of the reciprocals of the primes

File:Sum of reciprocals of primes.svg

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 display="block">\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,<ref>Template:Cite journal</ref> 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 Template:Mvar. The double natural logarithm (Template:Math) indicates that the divergence might be very slow, which is indeed the case. See Meissel–Mertens constant.

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 Template:Math 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 Template:Math. 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,<ref>Template:Cite book</ref> the right hand side can be obtained by setting Template:Math in the Taylor series expansion

<math display="block">\log\left(\frac1{1-x}\right)=\sum_{n=1}^\infty\frac{x^{n}}n.</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 that Euler meant that the sum of the reciprocals of the primes less than Template:Mvar is asymptotic to Template:Math as Template:Mvar 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.<ref>Template:Cite journal</ref> 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 Template:Mvar denote the Template:Mvarth prime number. Assume that the sum of the reciprocals of the primes converges.

Then there exists a smallest positive integer Template:Mvar such that

<math display="block">\sum_{i=k+1}^\infty \frac 1 {p_i} < \frac12 \qquad(1)</math>

For a positive integer Template:Mvar, let Template:Mvar denote the set of those Template:Mvar in Template:Math which are not divisible by any prime greater than Template:Mvar (or equivalently all Template:Math which are a product of powers of primes Template:Math). We will now derive an upper and a lower estimate for Template:Math, the number of elements in Template:Mvar. For large Template:Mvar, these bounds will turn out to be contradictory.

Upper estimate

Every Template:Mvar in Template:Mvar can be written as Template:Math with positive integers Template:Mvar and Template:Mvar, where Template:Mvar is square-free. Since only the Template:Mvar primes Template:Math can show up (with exponent 1) in the prime factorization of Template:Mvar, there are at most Template:Math different possibilities for Template:Mvar. Furthermore, there are at most Template:Math possible values for Template:Mvar. This gives us the upper estimate <math display="block">|M_x| \le 2^k\sqrt{x} \qquad(2)</math>

Lower estimate

The remaining Template:Math numbers in the set difference Template:Math are all divisible by a prime greater than Template:Mvar. Let Template:Math denote the set of those Template:Mvar in Template:Math which are divisible by the Template:Mvarth prime Template:Mvar. 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 Template:Math is at most Template:Math (actually zero for Template:Math), 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 Template:Math, the estimates (2) and (3) cannot both hold, because Template:Math.

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 Template:Math. The proof is due to Ivan Niven,<ref>Niven, Ivan, "A Proof of the Divergence of Σ 1/Template:Mvar", The American Mathematical Monthly, Vol. 78, No. 3 (Mar. 1971), pp. 272-273. The half-page proof is expanded by William Dunham in Euler: The Master of Us All, pp. 74-76.</ref> adapted from the product expansion idea of Euler. In the following, a sum or product taken over Template:Mvar always represents a sum or product taken over a specified set of primes.

The proof rests upon the following four inequalities:

• Every positive integer Template:Mvar 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 Template:Mvar is even) or 1 (the corresponding power of prime Template:Mvar 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 Template:Mvars 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 Template:Mvar. And since <math>1 / b^2</math> is one of the summands of Template:Mvar, every summand <math>1/i</math> is represented in one of the terms of Template:Mvar 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}_{{} \,<\, \frac1i} \\
&< \sum_{i=1}^n \frac 1 i

\end{align}</math>

• The lower estimate Template:Math for the exponential function, which holds for all Template:Math.
• Let Template:Math. The upper bound (using a telescoping sum) for the partial sums (convergence is all we really need) <math display="block">\begin{align}

\sum_{k=1}^n \frac 1 {k^2} 
&< 1 + \sum_{k=2}^n \underbrace{\left(\frac1{k - \frac{1}{2}} - \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 Template:Sfrac 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 Template:Math can be improved to Template:Math; 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 Template:Math 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.<ref>Template:Cite journal</ref>

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<ref>Template:Cite journal</ref> is by induction: The first partial sum is Template:Sfrac, which has the form Template:Sfrac. If the Template:Mvarth partial sum (for Template:Math) has the form Template:Sfrac, then the Template:Mathst 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 Template:Mathst prime Template:Math is odd; since this sum also has an Template:Sfrac 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 Template:Mvar 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

• Template:Cite book

External links

• {{#invoke:citation/CS1|citation

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