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| [[File:Sum of reciprocals of primes.svg|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 purple function is a lower bound that also diverges.]]
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| The '''sum of the [[multiplicative inverse|reciprocal]]s of all [[prime number]]s [[Divergent series|diverges]]'''; that is:
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| :<math>\sum_{p\text{ prime }}\frac1p = \frac12 + \frac13 + \frac15 + \frac17 + \frac1{11} + \frac1{13} + \frac1{17} + \cdots = \infty</math>
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| This was proved by [[Leonhard Euler]] in 1737, and strengthens [[Euclid]]'s 3rd-century-BC result that [[Euclid's theorem|there are infinitely many prime number]]s.
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| There are a variety of proofs of Euler's result, including a lower bound for the partial sums stating that
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| :<math>\sum_{\scriptstyle p\text{ prime }\atop \scriptstyle p\le n}\frac1p \ge \log \log (n+1) - \log\frac{\pi^2}6</math> | |
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| for all natural numbers ''n''. The double [[natural logarithm]] indicates that the divergence might be very slow, which is indeed the case, see [[Meissel–Mertens constant]].
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| ==The harmonic series== | |
| First, we describe how Euler originally discovered the result. He was considering the [[harmonic series (mathematics)|harmonic series]]
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| : <math>
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| \sum_{n=1}^\infty \frac{1}{n} =
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| 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots | |
| </math>
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| He had already used the following "[[Riemann zeta function#Euler product formula|product formula]]" to show the existence of infinitely many primes.
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| : <math> | |
| \sum_{n=1}^\infty \frac{1}{n} = \prod_{p} \frac{1}{1-p^{-1}}
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| = \prod_{p} \left( 1+\frac{1}{p}+\frac{1}{p^2}+\cdots \right)
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| </math>
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| (Here, the product is taken over all primes ''p''; in the following, a sum or product taken over ''p'' always represents a sum or product taken over a specified set of primes, unless noted otherwise.)
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| Such infinite products are today called [[Euler product]]s. The product above is a reflection of the [[fundamental theorem of arithmetic]]. Of course, the above "equation" is not necessary because the harmonic series is known (by other means) to diverge. This type of formal manipulation was common at the time, when mathematicians were still experimenting with the new tools of [[calculus]].{{Citation needed|date=October 2010}}
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| 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. (In modern language, we now say that the existence of infinitely many primes is reflected by the fact that the [[Riemann zeta function]] has a [[pole (complex analysis)|simple pole]] at ''s'' = 1.)
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| ==Proofs==
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| ===First===
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| Euler took the above product formula and proceeded to make a sequence of audacious leaps of logic. First, he took the natural logarithm of each side, then he used the Taylor series expansion for ln(''x'') as well as the sum of a geometric series:
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| : <math>
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| \begin{align}
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| \ln \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \ln\left( \prod_p \frac{1}{1-p^{-1}}\right)
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| = -\sum_p \ln \left( 1-\frac{1}{p}\right) \\
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| & {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\
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| & {} = \left( \sum_{p}\frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( \frac{1}{2} + \frac{1}{3p} + \frac{1}{4p^2} + \cdots \right) \\
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| & {} < \left( \sum_p \frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( 1 + \frac{1}{p} + \frac{1}{p^2} + \cdots \right) \\
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| & {} = \left( \sum_p \frac{1}{p} \right) + \left( \sum_p \frac{1}{p(p-1)} \right) \\
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| & {} = \left( \sum_p \frac{1}{p} \right) + C
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| \end{align}
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| </math>
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| for a fixed constant ''C'' < 1. Since the sum of the reciprocals of the first ''n'' positive integers is asymptotic to ln(''n''), (i.e. their ratio approaches one as ''n'' approaches infinity), Euler then concluded
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| :<math>\frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \cdots = \ln \ln (+ \infty)</math>
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| It is almost certain that Euler meant that the sum of the reciprocals of the primes less than ''n'' is asymptotic to ln(ln(''n'')) as ''n'' approaches infinity. It turns out this is indeed the case; Euler had reached a correct result by questionable means.{{Citation needed|date=January 2012}}
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| ====A variation====
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| : <math>
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| \begin{align}
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| & {} \quad \log \left( \sum_{n=1}^\infty \frac{1}{n}\right) = \log \left( \prod_p \frac{1}{1-p^{-1}}\right) = \sum_p \log \left( \frac{p}{p-1}\right) = \sum_p \log\left(1+\frac{1}{p-1}\right)
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| \end{align}
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| </math>
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| Since
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| : <math> e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots</math>
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| Shows that <math>\scriptstyle e^x \,>\, 1 \,+\, x</math> therefore <math>\scriptstyle \log(e^x) \,>\, \log(1 \,+\, x)</math> , so <math> \scriptstyle x \,>\, \log(1 \,+\, x)</math>. So
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| : <math> \sum_p \frac{1}{p - 1}>\sum_p \log\left(1+\frac{1}{p-1}\right)=\quad \log \left( \sum_{n=1}^\infty \frac{1}{n}\right)</math>
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| Hence <math>\scriptstyle \sum_p \frac{1}{p - 1}</math> diverges. But <math>\scriptstyle \frac{1}{p_i - 1} \,<\, \frac{1}{p_{i - 1}}</math> (consider the '''''i''''' from 3). Where <math>\scriptstyle p_i</math> is the '''''i'''''<sup> th</sup> prime, (because <math>\scriptstyle {p_i-1} \,>\, {p_{i-1}}</math>).
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| Hence <math>\scriptstyle \sum_p \frac{1}{p}</math> diverges.
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| ===Second===
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| The following [[proof by contradiction]] is due to [[Paul Erdős]].
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| Let ''p''<sub>''i''</sub> denote the ''i''<sup>''th''</sup> prime number. Assume that the [[series (mathematics)|sum]] of the reciprocals of the primes [[convergent series|converges]]; i.e.,
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| :<math>\sum_{i=1}^\infty {1\over p_i} < \infty</math>
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| Then there exists a smallest [[Positive number|positive]] [[integer]] ''k'' such that
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| :<math>\sum_{i=k+1}^\infty {1\over p_i} < {1\over 2} \qquad(1)</math>
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| For a positive integer ''x'' let ''M<sub>x</sub>'' denote the set of those ''n'' in {1, 2, . . ., ''x''} which are not [[divisible]] by any prime greater than ''p<sub>k</sub>''. We will now derive an upper and a lower estimate for the [[cardinality|number]] |''M<sub>x</sub>''| of elements in ''M<sub>x</sub>''. For large ''x'', these bounds will turn out to be contradictory. | |
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| ====Upper estimate====
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| Every ''n'' in ''M<sub>x</sub>'' can be written as ''n'' = ''r m''<sup>2</sup> with positive integers ''m'' and ''r'', where ''r'' is [[square-free integer|square-free]]. Since only the ''k'' primes ''p''<sub>1</sub>, …, ''p<sub>k</sub>'' can show up (with exponent 1) in the [[Fundamental theorem of arithmetic|prime factorization]] of ''r'', there are at most 2<sup>''k''</sup> different possibilities for ''r''. Furthermore, there are at most √''x'' possible values for ''m''. This gives us the upper estimate
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| :<math>|M_x| \le 2^k\sqrt{x} \qquad(2)</math> | |
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| ====Lower estimate====
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| The remaining ''x'' − |''M<sub>x</sub>''| numbers in the [[set difference]] {1, 2, . . ., ''x''} \ ''M<sub>x</sub>'' are all divisible by a prime greater than ''p''<sub>''k''</sub>. Let ''N<sub>i,x</sub>'' denote the set of those ''n'' in {1, 2, . . ., x} which are divisible by the ''i''<sup>th</sup> prime ''p<sub>i</sub>''. Then
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| :<math>\{1,2,\ldots,x\}\setminus M_{x}=\bigcup_{i=k+1}^\infty N_{i,x}</math>
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| Since the number of integers in ''N<sub>i,x</sub>'' is at most ''x''/''p<sub>i</sub>'' (actually zero for ''p<sub>i</sub>'' > ''x''), we get
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| :<math>x-|M_x| \le \sum_{i=k+1}^\infty |N_{i,x}|< \sum_{i=k+1}^\infty {x\over p_i}</math>
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| Using (1), this implies
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| :<math>{x\over 2} < |M_x| \qquad(3)</math>
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| ====Contradiction====
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| When ''x'' ≥ 2<sup>2''k'' + 2</sup>, the estimates (2) and (3) cannot both hold, because <math>\tfrac{x}{2}\ge 2^k\sqrt{x}</math>.
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| ===Third ===
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| 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 an adaptation of 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.
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| The proof rests upon the following four inequalities:
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| * Every positive integer ''i'' can be uniquely expressed as the product of a square-free integer and a square. This gives the inequality
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| ::<math> \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}}</math>
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| :where for every ''i'' between 1 and ''n'' the (expanded) product corresponds to the [[Radical of an integer|square-free part]] of ''i'' and the sum corresponds to the square part of ''i'' (see [[fundamental theorem of arithmetic]]).
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| * The upper estimate for the [[natural logarithm]]
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| ::<math>
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| \log(n+1)
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| = \int_1^{n+1}\frac{dx}x
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| = \sum_{i=1}^n\underbrace{\int_i^{i+1}\frac{dx}x}_{{} \,<\, 1/i}
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| < \sum_{i=1}^n{\frac{1}{i}}
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| </math>
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| * The lower estimate 1 + ''x'' < exp(''x'') for the [[exponential function]], which holds for all ''x'' > 0.
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| * Let n ≥ 2. The upper bound (using a [[telescoping sum]]) for the partial sums (convergence is all we really need)
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| ::<math>
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| \sum_{k=1}^n{\frac{1}{k^2}}
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| < 1 + \sum_{k=2}^n\underbrace{\left(\frac1{k - \frac{1}{2}} - \frac1{k + \frac{1}{2}}\right)}_{=\, 1/(k^2 - 1/4) \,>\, 1/k^2}
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| = 1 + \frac23 - \frac1{n + \frac{1}{2}} < \frac53
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| </math>
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| Combining all these inequalities, we see that
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| :<math>\begin{align}
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| {} & {} \log(n+1) \\
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| < &\sum_{i=1}^n\frac{1}{i} \\
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| \le &\prod_{p \le n}{\left(1 + \frac{1}{p}\right)}\sum_{k=1}^n{\frac{1}{k^2}} \\
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| < &\frac53\prod_{p \le n}{\exp\left(\frac{1}{p}\right)} \\
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| = &\frac53\exp\left(\sum_{p \le n}{\frac{1}{p}}\right)
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| \end{align}</math>
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| Dividing through by {{sfrac|5|3}} and taking the natural logarithm of both sides gives
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| :<math>\log \log(n + 1) - \log\frac53 < \sum_{p \le n}{\frac{1}{p}}</math>
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| as desired. [[Q.E.D.|∎]]
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| Using
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| :<math>\sum_{k=1}^\infty{\frac{1}{k^2}} = \frac{\pi^2}6</math>
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| (see [[Basel problem]]), the above constant ln ({{sfrac|5|3}}) = 0.51082… can be improved to ln({{sfrac|π<sup>2</sup>|6}}) = 0.4977…; in fact it turns out that
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| :<math>
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| \lim_{n \to \infty } \left(
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| \sum_{p \leq n} \frac{1}{p} - \log \log(n)
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| \right) = M
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| </math>
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| where ''M'' = 0.261497… is the [[Meissel–Mertens constant]] (somewhat analogous to the much more famous [[Euler–Mascheroni constant]]).
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| ===Fourth===
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| From [[Dusart's inequality]], we get
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| :<math> p_n < n \log n + n \log \log n \quad\mbox{for } n \ge 6</math>
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| Then
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| :<math>\begin{align}
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| \sum_{n=1}^\infty \frac1{ p_n}
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| &\ge \sum_{n=6}^\infty \frac1{ p_n} \\
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| &\ge \sum_{n=6}^\infty \frac1{ n \log n + n \log \log n} \\
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| &\ge \sum_{n=6}^\infty \frac1{2n \log n} \\
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| &= \infty
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| \end{align}</math>
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| by the [[integral test for convergence]]. This shows that the series on the left diverges.
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| ==See also==
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| *[[Euclid's theorem]] that there are infinitely many primes
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| *[[Small set (combinatorics)]]
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| *[[Brun's theorem]]
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| ==References==
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| * {{cite book | author=William Dunham | authorlink=William Dunham (mathematician) | title=Euler The Master of Us All | publisher=[[Mathematical Association of America|MAA]] | year=1999 | isbn=0-88385-328-0 | pages=61–79 }}
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| ==External links==
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| * [http://www.utm.edu/research/primes/infinity.shtml Chris K. Caldwell: There are infinitely many primes, but, how big of an infinity?]
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| {{DEFAULTSORT:Sum Of The Reciprocals Of The Primes Diverges}}
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| [[Category:Mathematical series]]
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| [[Category:Articles containing proofs]]
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| [[Category:Theorems about prime numbers]]
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