nyuuzyou/wikis huggingface dataset:
Sum of Rational Cuts is Rational Cut

== Theorem ==
Let $p \in\ Q$ and $q \in \Q$ be rational numbers.
Let $p^*$ and $q^*$ denote the rational cuts associated with $p$ and $q$.
Then:
:$p^* + q^* = \paren {p + q}^*$
Thus the operation of addition on the set of rational cuts is closed.
== Proof ==
From Sum of Cuts is Cut, $p^* + q^*$ is a cut.
Let $r \in p^* + q^*$.
Then:
:$r = s + t$
where $s < p$ and $t < q$
Thus:
:$r < p + q$
and so:
:$r \in \paren {p + q}^*$
Hence:
:$p^* + q^* \subseteq \paren {p + q}^*$
Let $r \in \paren {p + q}^*$.
Then:
:$r < p + q$
Let:
:$h = p + q - r$
:$s = p - \dfrac h 2$
:$t = q - \dfrac h 2$
Then:
:$s \in p^*$
:$t \in q^*$
Hence:
:$r = s + t$
and so:
:$r \in p^* + q^*$
So:
:$\paren {p + q}^* \subseteq p^* + q^*$
Hence the result by definition of set equality.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.28$. Theorem: $\text {(a)}$
Category:Cuts
Category:Addition
Sum of Reciprocals in Base 10 with Zeroes Removed

== Theorem ==
The infinite series
:$\ds \sum_{\map P n} \frac 1 n$
where $\map P n$ is the propositional function:
:$\forall n \in \Z_{>0}: \map P n \iff$ the decimal representation of $n$ contains no instances of the digit $0$
converges to the approximate limit $23 \cdotp 10345 \ldots$
== Proof ==
For each $k \in \N$, there are $9^k$ $k$-digit numbers containing no instances of the digit $0$.
Each of these numbers is at least $10^{k - 1}$.
Hence the reciprocals of each of these numbers is at most $\dfrac 1 {10^{k - 1}}$.
Thus:
showing that the sum converges.
== Sources ==
* 
* : $23 \cdotp 103 \, 45 \ldots$
* : $23 \cdotp 10345 \ldots$
Category:Reciprocals
Category:Series
Sum of Reciprocals of Cubes of Odd Integers Alternating in Sign

== Theorem == 
== Proof ==
By Half-Range Fourier Sine Series for $x \paren {\pi - x}$ over $\openint 0 \pi$:
:$\ds x \paren {\pi - x} = \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \paren {2 n + 1} x} {\paren {2 n + 1}^3}$
for $x \in \openint 0 \pi$.
Setting $x = \dfrac \pi 2$:
whence the result.
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.28$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.28.$
Category:Formulas for Pi
Category:Cube Numbers
Sum of Reciprocals of Cubes of Odd Integers Alternating in Sign in Pairs

== Theorem == 
== Proof ==
By Half-Range Fourier Sine Series for $x \paren {\pi - x}$ over $\openint 0 \pi$:
:$\ds x \paren {\pi - x} = \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \paren {2 n + 1} x} {\paren {2 n + 1}^3}$
for $x \in \openint 0 \pi$.
Setting $x = \dfrac \pi 4$:
Now we need to show that:
:$\ds \frac 1 {1^3} + \frac 1 {3^3} - \frac 1 {5^3} - \frac 1 {7^3} + \cdots = \sum_{n \mathop = 0}^\infty \frac {\sin \frac {n \pi} 2 + \cos \frac {n \pi} 2} {\paren {2 n + 1}^3}$
For $n \equiv 0 \pmod 4$:
:$\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \sin 2 k \pi + \cos 2 k \pi = 1$
For $n \equiv 1 \pmod 4$:
:$\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \map \sin {2 k \pi + \dfrac \pi 2} + \map \cos {2 k \pi + \dfrac \pi 2} = 1$
For $n \equiv 2 \pmod 4$:
:$\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \map \sin {2 k \pi + \pi} + \map \cos {2 k \pi + \pi} = -1$
For $n \equiv 3 \pmod 4$:
:$\sin \dfrac {n \pi} 2 + \cos \dfrac {n \pi} 2 = \map \sin {2 k \pi + \dfrac {3 \pi} 2} + \map \cos {2 k \pi + \dfrac {3 \pi} 2} = -1$
This shows that the sum is alternating in sign in pairs.
Hence the result.
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.29$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.29.$
Category:Formulas for Pi
Category:Cube Numbers
Sum of Reciprocals of Divisors equals Abundancy Index

== Theorem ==
Let $n$ be a positive integer.
Let $\map {\sigma_1} n$ denote the divisor sum function of $n$.
Then:
:$\ds \sum_{d \mathop \divides n} \frac 1 d = \frac {\map {\sigma_1} n} n$
where $\dfrac {\map {\sigma_1} n} n$ is the abundancy index of $n$.
== Proof ==
Category:Number Theory
Category:Abundancy
Sum of Reciprocals of Divisors of Perfect Number is 2

== Theorem ==
Let $n$ be a perfect number.
Then:
:$\ds \sum_{d \mathop \divides n} \dfrac 1 d = 2$
That is, the sum of the reciprocals of the divisors of $n$ is equal to $2$.
== Proof ==
The result follows by definition of perfect number:
== Sources ==
* : $28$
* : $28$
Category:Perfect Numbers
Sum of Reciprocals of Even Powers of Integers Alternating in Sign

== Theorem ==
Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
where $B_{2 n}$ is the $2 n$th Bernoulli number.
=== Corollary ===
== Proof ==
== Also presented as ==
== Also see ==
* Sum of Reciprocals of Squares Alternating in Sign
* Sum of Reciprocals of Fourth Powers Alternating in Sign
* Sum of Reciprocals of Sixth Powers Alternating in Sign
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.37$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.37.$
Category:Sum of Reciprocals of Even Powers of Integers Alternating in Sign
Category:Bernoulli Numbers
Category:Formulas for Pi
Category:Sums of Sequences
Sum of Reciprocals of Even Powers of Integers Alternating in Sign/Also presented as

== Sum of Reciprocals of Even Powers of Integers Alternating in Sign: Also presented as == 
This can also be expressed using the archaic form of the Bernoulli numbers as:
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.37$
* : $\S 21$: Series of Constants: Sums of Powers of Positive Integers: $21.37.$
Category:Sum of Reciprocals of Even Powers of Integers Alternating in Sign
Sum of Reciprocals of Even Powers of Integers Alternating in Sign/Corollary

== Corollary to Sum of Reciprocals of Even Powers of Integers Alternating in Sign ==
Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
== Proof ==
== Also presented as ==
== Sources ==
* : $\S 21$: Series involving Bernoulli and Euler Numbers: $21.10$
* : $\S 23$: Bernoulli and Euler Numbers: Series involving Bernoulli and Euler Numbers: $23.10.$
Category:Sum of Reciprocals of Even Powers of Integers Alternating in Sign
Sum of Reciprocals of Even Powers of Integers Alternating in Sign/Corollary/Also presented as

== : Also presented as == 
 can also be presented in the form:
where ${B_n}^*$ denotes the archaic form of the Bernoulli numbers.
== Sources ==
* : $\S 21$: Series involving Bernoulli and Euler Numbers: $21.10$
* : $\S 23$: Bernoulli and Euler Numbers: Series involving Bernoulli and Euler Numbers: $23.10.$
Category:Sum of Reciprocals of Even Powers of Integers Alternating in Sign
Sum of Reciprocals of Even Powers of Integers Alternating in Sign/Corollary/Mistake

== Source Work ==
:Chapter $23$: Bernoulli and Euler Numbers
::Series involving Bernoulli and Euler Numbers: $23.10.$
== Mistake ==
== Correction ==
The correct equation is:
The mistake appears to have crept in since the $1$st edition, which presented this correctly.
Note that here $B_n$ denotes the archaic form of the Bernoulli numbers.
== Sources ==
* : $\S 23$: Bernoulli and Euler Numbers: Series involving Bernoulli and Euler Numbers: $23.10.$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Sum of Reciprocals of Even Powers of Odd Integers

== Theorem ==
Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
=== Corollary ===
== Proof ==
 
== Also presented as ==
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.36$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.36.$
Category:Sum of Reciprocals of Even Powers of Odd Integers
Category:Bernoulli Numbers
Category:Sums of Sequences
Sum of Reciprocals of Even Powers of Odd Integers/Also presented as

== Sum of Reciprocals of Even Powers of Odd Integers: Also presented as == 
This can also be expressed using the archaic form of the Bernoulli numbers as:
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.36$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.36.$
Category:Sum of Reciprocals of Even Powers of Odd Integers
Sum of Reciprocals of Even Powers of Odd Integers/Corollary

== Corollary to Sum of Reciprocals of Even Powers of Odd Integers ==
Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
== Proof ==
== Also presented as ==
== Sources ==
* : $\S 21$: Series involving Bernoulli and Euler Numbers: $21.9$
* : $\S 23$: Bernoulli and Euler Numbers: Series involving Bernoulli and Euler Numbers: $23.9.$
Category:Sum of Reciprocals of Even Powers of Odd Integers
Sum of Reciprocals of Even Powers of Odd Integers/Corollary/Also presented as

== : Also presented as == 
 can also be presented in the form:
where ${B_n}^*$ denotes the archaic form of the Bernoulli numbers.
== Sources ==
* : $\S 21$: Series involving Bernoulli and Euler Numbers: $21.9$
* : $\S 23$: Bernoulli and Euler Numbers: Series involving Bernoulli and Euler Numbers: $23.9.$
Category:Sum of Reciprocals of Even Powers of Odd Integers
Sum of Reciprocals of Fourth Powers Alternating in Sign

== Theorem == 
== Proof == 
== Also see ==
* Sum of Reciprocals of Even Powers of Integers Alternating in Sign
* Sum of Reciprocals of Squares Alternating in Sign
* Sum of Reciprocals of Sixth Powers Alternating in Sign
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.23$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.23.$
Category:Formulas for Pi
Category:Sums of Sequences
Category:Fourth Powers
Sum of Reciprocals of Fourth Powers of Odd Integers

== Theorem ==
== Proof ==
 
== Also presented as ==
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.26$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.26.$
Category:Sum of Reciprocals of Fourth Powers of Odd Integers
Category:Formulas for Pi
Category:Fourth Powers
Sum of Reciprocals of Fourth Powers of Odd Integers/Also presented as

== Sum of Reciprocals of Fourth Powers of Odd Integers: Also presented as == 
This result can also be presented as:
:$\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^4} = \dfrac {\pi^4} {96}$
Category:Sum of Reciprocals of Fourth Powers of Odd Integers
Sum of Reciprocals of Powers as Euler Product

== Theorem ==
Let $\zeta$ be the Riemann zeta function.
Let $s \in \C$ be a complex number with real part $\sigma > 1$.
Then:
:$\ds \map \zeta s = \prod_{\text {$p$ prime} } \frac 1 {1 - p^{-s} }$
where the infinite product runs over the prime numbers.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Historical Note ==
== Sources ==
* : $\S 3$: Appendix $\text A$: Euler
* : $0,5$
* : $0 \cdotp 5$
* : Chapter $\text {A}.32$: Riemann ()
* : $0 \cdotp 5$
* 
* 
Category:Riemann Zeta Function
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers as Euler Product/Corollary 1

== Corollary to Sum of Reciprocals of Powers as Euler Product ==
Let $\zeta$ be the Riemann zeta function.
Let $s \in \C$ be a complex number with real part $\sigma > 1$.
Then:
:$\ds\prod_{\text {$p$ prime} } \paren {1 + p^{-s} } = \dfrac {\map \zeta s} {\map \zeta {2 s} }$
where the infinite product runs over the prime numbers.
== Proof ==
== Examples ==
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers as Euler Product/Corollary 1/Examples

== Examples of Use of Sum of Reciprocals of Powers as Euler Product/Corollary 1 ==
=== Example: $\ds \prod_{\text {$p$ prime} } \paren {1 + p^{-2} }$ ===
=== Example: $\ds \prod_{\text {$p$ prime} } \paren {1 + p^{-4} }$ ===
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers as Euler Product/Corollary 1/Examples/Zeta(2) over Zeta(4)

== Example of Use of Sum of Reciprocals of Powers as Euler Product/Corollary 1 ==
:$\ds \prod_{\text {$p$ prime} } \paren {1 + p^{-2} } = \dfrac {15 } {\pi^2}$
where the infinite product runs over the prime numbers.
== Proof ==
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers as Euler Product/Corollary 1/Examples/Zeta(4) over Zeta(8)

== Example of Use of Sum of Reciprocals of Powers as Euler Product/Corollary 1 ==
:$\ds \prod_{\text {$p$ prime} } \paren {1 + p^{-4} } = \dfrac {105 } {\pi^4}$
where the infinite product runs over the prime numbers.
== Proof ==
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers as Euler Product/Corollary 2

== Corollary to Sum of Reciprocals of Powers as Euler Product ==
Let $\zeta$ be the Riemann zeta function.
Let $s \in \C$ be a complex number with real part $\sigma > 1$.
Then:
:$\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-s} } {1 - p^{-s} } } = \dfrac {\paren {\map \zeta s}^2} {\map \zeta {2 s} }$
where the infinite product runs over the prime numbers.
== Proof ==
== Examples ==
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers as Euler Product/Corollary 2/Examples

== Examples of Use of Sum of Reciprocals of Powers as Euler Product/Corollary 2 ==
=== Example: $\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-2} } {1 - p^{-2} } }$ ===
=== Example: $\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-4} } {1 - p^{-4} } }$ ===
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers as Euler Product/Corollary 2/Examples/Zeta^2(2) over Zeta(4)

== Example of Use of Sum of Reciprocals of Powers as Euler Product/Corollary 2 ==
:$\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-2} } {1 - p^{-2} } } = \dfrac 5 2$
where the infinite product runs over the prime numbers.
== Proof ==
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers as Euler Product/Corollary 2/Examples/Zeta^2(4) over Zeta(8)

== Example of Use of Sum of Reciprocals of Powers as Euler Product/Corollary 2 ==
:$\ds \prod_{\text {$p$ prime} } \paren {\frac {1 + p^{-4} } {1 - p^{-4} } } = \dfrac 7 6$
where the infinite product runs over the prime numbers.
== Proof ==
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers as Euler Product/Historical Note

== Historical Note on Sum of Reciprocals of Powers as Euler Product ==
The Riemann $\zeta$ function presented as the Sum of Reciprocals of Powers as Euler Product was proved by in $1744$.
== Sources ==
* : Chapter $\text {B}.19$: The Series $\sum 1/ p_n$ of the Reciprocals of the Primes
* 
* 
Category:Historical Notes
Sum of Reciprocals of Powers as Euler Product/Proof 1

== Theorem ==
== Proof ==
By definition of Euler product:
:$\ds \sum_{n \mathop = 1}^\infty a_n n^{-z} = \prod_p \frac 1 {1 - a_p p^{-z} }$
 $\ds \sum_{n \mathop = 1}^\infty a_n n^{-z}$ is absolutely convergent.
For all $n \in \Z_{\ge 1}$, let $a_n = 1$.
From Convergence of P-Series:
:$\ds \sum_{n \mathop = 1}^\infty n^{-z}$ is absolutely convergent
:
:$\cmod z > 1$
It then follows that:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers as Euler Product/Proof 2

== Theorem ==
== Proof ==
From Sum of Infinite Geometric Sequence:
:$\dfrac 1 {1 - p^{-z} } = 1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \cdots$
From Convergence of P-Series:
:$\ds \sum_{n \mathop = 1}^\infty n^{-z}$ is absolutely convergent
:
:$\cmod z \gt 1$
Thus:
The Fundamental Theorem of Arithmetic shows that:
:for every integer $n$ such that $n > 1$, $n$ can be expressed as the product of one or more primes, uniquely up to the order in which they appear.
== Sources ==
* : Chapter $\text {B}.19$: The Series $\sum 1/ p_n$ of the Reciprocals of the Primes
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers as Euler Product/Proof 3

== Theorem ==
== Proof ==
Multiplying both sides above by $\dfrac 1 {2^s}$
Then subtracting:
Multiplying both sides above by $\dfrac 1 {3^s}$
Then subtracting:
:Similar to the Sieve of Eratosthenes, in each subsequent step, we multiply both sides of the equation from the previous step by $\dfrac 1 {p^s}$ and then reduce both sides from the previous step by that amount.
:After infinitely many steps, we arrive at:
Category:Sum of Reciprocals of Powers as Euler Product
Sum of Reciprocals of Powers of Odd Integers Alternating in Sign

== Theorem == 
:$\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s} = \frac 1 {2 \map \Gamma s} \int_0^\infty x^{s - 1} \map \sech x \rd x$
where:
:$\map \Re s > 0$
:$\Gamma$ is the gamma function
:$\sech$ is the hyperbolic secant function.
=== Corollary ===
== Proof == 
 
So: 
:$\ds \frac 1 {2 \map \Gamma s} \int_0^\infty x^{s - 1} \map \sech x \rd x = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s}$
Category:Hyperbolic Secant Function
Category:Definite Integrals
Category:Sums of Sequences
Sum of Reciprocals of Powers of Odd Integers Alternating in Sign/Corollary

== Corollary to Sum of Reciprocals of Powers of Odd Integers Alternating in Sign == 
:$\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}^s} = \frac 1 {\map \Gamma s} \int_1^\infty \frac {\ln^{s - 1} x} {x^2 + 1} \rd x$
== Proof ==
Category:Definite Integrals
Sum of Reciprocals of Primes is Divergent

== Theorem ==
Let $n \in \N: n \ge 1$.
There exists a (strictly) positive real number $C \in \R_{>0}$ such that:
:$(1): \quad \ds \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \map \ln {\ln n} - C$
where $\Bbb P$ is the set of all prime numbers.
:$(2): \quad \ds \lim_{n \mathop \to \infty} \paren {\map \ln {\ln n} - C} = +\infty$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* : $\S 3$: Appendix $\text A$: Euler
* : $1 \cdotp 90195 \ldots$
* : $1 \cdotp 90216 \, 054 \ldots$
Category:Analytic Number Theory
Category:Sum of Reciprocals of Primes is Divergent
Sum of Reciprocals of Primes is Divergent/Lemma

== Theorem ==
Let $C \in \R_{>0}$ be a (strictly) positive real number.
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {\map \ln {\ln n} - C} = + \infty$
== Proof ==
Fix $c \in \R$.
It is sufficient to show there exists $N \in \N$, such that:
:$(1): \quad n \ge N \implies \map \ln {\ln n} - C > c$
Proceed as follows:
Let $N \in \N$ such that $N > \map \exp {\map \exp {c + C} }$.
By Logarithm is Strictly Increasing it follows that $N$ satisfies condition $(1)$.
Hence the result.
Category:Number Theory
Sum of Reciprocals of Primes is Divergent/Proof 1

== Theorem ==
== Proof ==
By Sum of Reciprocals of Primes is Divergent: Lemma:
:$\ds \lim_{n \mathop \to \infty} \paren {\map \ln {\map \ln n} - \frac 1 2} = +\infty$
It remains to be proved that:
:$\ds \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \map \ln {\ln n} - \frac 1 2$
Assume all sums and product over $p$ are over the set of prime numbers.
Let $n \ge 1$.
Then:
Category:Sum of Reciprocals of Primes is Divergent
Sum of Reciprocals of Primes is Divergent/Proof 2

== Theorem ==
The series:
:$\ds \sum_{p \mathop \in \Bbb P} \frac 1 p$
where:
:$\Bbb P$ is the set of all prime numbers
is divergent.
== Proof ==
Let $n \in \N$ be a natural number.
Let $p_n$ denote the $n$th prime number.
Consider the continued product:
:$\ds \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$
By Sum of Infinite Geometric Sequence:
Consider what happens when all these series are multiplied together.
A new series will be generated whose terms consist of all possible products of one term selected from each of the series on the .
This new series will converge in any order to the product of the terms on the .
By the Fundamental Theorem of Arithmetic, every integer greater than $1$ is uniquely expressible as a product of powers of different primes.
Hence the product of these series is the series of reciprocals of all (strictly) positive integers whose prime factors are no greater than $p_n$.
In particular, all (strictly) positive integers up to $p_n$ have this property.
So:
:$\ds \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$
It follows by taking reciprocals that:
:$\ds \prod_{k \mathop = 1}^n \paren {1 - \frac 1 {p_k} } < \frac 1 {\ln p_n}$
Taking logarithms of each side:
:$(1): \quad \ds \sum_{k \mathop = 1}^n \map \ln {1 - \frac 1 {p_k} } < - \ln \ln p_n$
Next, note that the line $y = 2 x$ in the cartesian plane lies below the curve $y = \map \ln {1 + x}$ on the interval $\closedint {-\frac 1 2} 0$.
Also note that all primes are greater than or equal to $2$.
Thus it follows that:
:$-\dfrac 2 {p_k} < \map \ln {1 - \dfrac 1 {p_k} }$
Applying this to $(1)$ yields:
:$\ds -2 \sum_{k \mathop = 1}^n \dfrac 1 {p_k} < -\ln \ln p_n$
and so:
:$\ds \sum_{k \mathop = 1}^n \dfrac 1 {p_k} > \dfrac 1 2 \ln \ln p_n$
But:
:$\ds \lim_{n \mathop \to \infty} \ln \ln p_n \to \infty$
and so the series:
:$\ds \sum_{p \mathop \in \Bbb P} \frac 1 p$
is divergent.
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.19$: The Series $\sum 1/ p_n$ of the Reciprocals of the Primes
Category:Sum of Reciprocals of Primes is Divergent
Sum of Reciprocals of Primes is Divergent/Proof 2/Historical Note

== Historical Note on Sum of Reciprocals of Primes is Divergent ==
This proof of the divergence of the reciprocals of the primes was discovered by .
== Sources ==
* : Chapter $\text {B}.19$: The Series $\sum 1/ p_n$ of the Reciprocals of the Primes
Category:Historical Notes
Sum of Reciprocals of Primes is Divergent/Proof 3

== Theorem ==
== Proof ==
 the contrary.
If the prime reciprocal series converges then there must exist some $k \in \N$ such that:
:$\ds \sum_{n \mathop = k \mathop + 1}^{\infty} \frac 1 {p_n} < \frac 1 2$
Let:
:$\ds Q = \prod_{i \mathop = 1}^k {p_i}$
and:
:$\ds \map S r = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q}$
Let $\map S {r, j}$ be the sum of all of the terms from $\map S r$ for which $1 + i Q$ has exactly $j$ prime factors.
Notice that $1 + i Q$ is coprime with every prime factor in $Q$.
Thus every prime factor of $1 + i Q$ where $i = 1, \ldots, r$ falls into some finite sequence of consecutive primes:
:$\map P r = \set {p_{k + 1}, p_{k + 2}, \ldots, p_{\map m r} }$
Notice again that each term of $\map S {r, j}$ occurs at least once in the expansion of:
:$\ds \paren {\sum_{n \mathop = k \mathop + 1}^{\map m r} \frac 1 {p_n} }^j < \paren {\sum_{n \mathop = k \mathop + 1}^\infty \frac 1 {p_n} }^j < \paren {\frac 1 2}^j$
and also by Sum of Infinite Geometric Sequence:
:$\ds \map S r = \sum_{j \mathop = 1}^r \map S {r, j} < \sum_{j \mathop = 1}^r \paren {\frac 1 2}^j < 1$
for every $r$.
Finally notice that:
:$\ds \map S r = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q} > \frac 1 {1 + Q} \sum_{n \mathop = 1}^r \frac 1 n$
which implies that $\map S r$ diverges towards $+\infty$ by Harmonic Series is Divergent, a contradiction.
== Historical Note ==
Category:Sum of Reciprocals of Primes is Divergent
Sum of Reciprocals of Primes is Divergent/Proof 4

== Theorem ==
== Proof ==
 the contrary.
If the prime reciprocal series converges then there must exist some $k \in \N$ such that:
:$\ds \sum_{n \mathop = k + 1}^\infty \frac 1 {p_n} < \frac 1 2$
Let:
:$M_x = \set {n \in \N: 1 \le n \le x \text { and } n \text { is not divisible by any primes greater than } p_k}$.
Notice that:
:$\size {M_x} \le {2^k} \sqrt x$
because if:
:$n = m^2r$
then
:$m \le \sqrt x$
and there are at most $2^k$ distinct non-square prime compositions using primes less than $p_k$.
Now let:
:$N_{i, x} = \set {n \in \N: 1 \le n \le x \text{ and } n \text { is divisible by } p_i}$
Notice:
:$\ds \size {\set {1, \dots, x} \setminus M_x} \le \sum_{i \mathop = k + 1}^\infty \size {N_{i, x} } < \sum_{i \mathop = k + 1}^\infty \dfrac x {p_i} \implies \dfrac x 2 < \size {M_x}$
Finally notice that whenever:
:$x \ge 2^{2 k + 2}$
then it cannot be the case that both:
:$\dfrac x 2 < \size {M_x} \le {2^k} \sqrt x$
== Historical Note ==
Category:Sum of Reciprocals of Primes is Divergent