nyuuzyou/wikis huggingface dataset:
Hardy-Littlewood Circle Method

== Theorem ==
Let $\AA$ be a subset of the non-negative integers.
Let:
:$\ds \map T s = \sum_{a \mathop \in \AA} s^a$
be the generating function for $\AA$.
For $N \in \N$, let $\map {r_{\AA, \ell} } N$ be the number of solutions $\tuple {x_1, \ldots, x_\ell} \in \AA^\ell$ to the equation:
:$x_1 + \cdots + x_\ell = N$
Then:
:$\ds \forall \rho \in \openint 0 1: \map {r_{\AA, \ell} } N = \oint_{\size s \mathop = \rho} \frac {\map T s^\ell} {s^{N + 1} } \rd s$
== Proof ==
We have:
:$\ds \map T s^\ell = \sum_{N \mathop = 0}^\infty \map {r_{\AA, \ell} } N s^N$
and:
:$\map {\dfrac {\d^N} {\d s^N} } {\map T s^\ell} = N! \cdot \map {r_{\AA, \ell} } N + \map \OO s$
so:
:$\map {r_{\AA, \ell} } N = \dfrac 1 {N!} \dfrac {\d^N} {\d s^N} \bigintlimits {\map T s^\ell} {s \mathop = 0}{}$
Now recall Cauchy's Integral Formula for Derivatives for a complex function $f$ holomorphic on a domain $D$, and a path $\gamma \subseteq D$ winding once around $a$:
:$\ds \valueat {\frac {\d^N} {\d s^N} \map f s} {s \mathop = a} = \frac {N!} {2 \pi i} \oint_\gamma \frac {\map f s} {\paren {s - a}^{N + 1} } \rd s$
Since $\map T s$ is defined by a generating function, $\map T s^\ell$ has a Taylor series about $s = 0$ which converges for all $\size s < 1$.
Applying Cauchy's formula:
where $\gamma$ a circle about zero of radius $\rho$ for any $\rho < 1$.
Category:Analytic Number Theory
Hardy-Ramanujan Number/Examples/1729

== Theorem ==
The $2$nd Hardy-Ramanujan number $\map {\operatorname {Ta}} 2$ is $1729$:
== Proof ==
We wish to demonstrate that $1729$ is the $2$nd Hardy-Ramanujan number making it the smallest positive integer which can be expressed as the sum of $2$ positive cubes in $2$ different ways.
To accomplish this, we will need to inspect $66$ sums of $a^3 + b^3$ starting with $1^3 + 1^3$ and ending with $11^3 + 11^3$
Recall from Integer Addition is Commutative, $a^3 + b^3 = b^3 + a^3$, therefore we only need to inspect one of these pairs.
, we will choose to inspect $a \ge b$ in the table below.
The reason that we don't need to inspect $a \ge 12$ is that $12^3 + 1^3 = 1729$ and adding anything larger than $1$ to $12^3$ will be larger than $1729$ and the cube of any number larger than $12$ will be larger than $1729$
The reason there are $66$ sums to inspect is that the $1$st row will have $1$ sum, the $2$nd row will have $2$ sums, all the way up to the $11$th row having $11$ sums. 
From Closed Form for Triangular Numbers, this sum is known to be:
:$\ds T_{11} = \sum_{i \mathop = 1}^{11} i = \frac {11 \paren {11 + 1} } 2 = 66$
Finally, we note that the only other strictly positive integer(s) that "almost" took this honor from $1729$ are $854$ and $855$ which are only apart by $1$.
:
Category:Hardy-Ramanujan Numbers
Category:Taxicab Numbers
Hardy-Ramanujan Number/Examples/48,988,659,276,962,496

== Theorem ==
The $5$th Hardy-Ramanujan number $\map {\operatorname {Ta} } 5$ is $48 \, 988 \, 659 \, 276 \, 962 \, 496$:
== Proof ==
== Historical Note ==
== Sources ==
* : $48,988,659,276,962,496$
* 
Category:Hardy-Ramanujan Numbers
Category:48,988,659,276,962,496
Hardy-Ramanujan Number/Examples/48,988,659,276,962,496/Historical Note

== Historical Note on Hardy-Ramanujan Number $48 \, 988 \, 659 \, 276 \, 962 \, 496$ ==
In his of $1997$, attributes this to , in the Numbers Count column of , February $1995$.
== Sources ==
* : $48,988,659,276,962,496$
Category:Hardy-Ramanujan Numbers
Category:48,988,659,276,962,496
Hardy-Ramanujan Number/Examples/6,963,472,309,248

== Theorem ==
The $4$th Hardy-Ramanujan number $\operatorname {Ta} \left({4}\right)$ is $6 \, 963 \, 472 \, 309 \, 248$:
== Historical Note ==
== Proof ==
== Sources ==
* 
* : $6,963,472,309,248$
Category:Hardy-Ramanujan Numbers
Category:6,963,472,309,248
Hardy-Ramanujan Number/Examples/6,963,472,309,248/Historical Note

== Historical Note on Hardy-Ramanujan Number $6 \, 963 \, 472 \, 309 \, 248$ ==
It is widely reported that the $4$th Hardy-Ramanujan number $\map {\operatorname {Ta} } 4$ was discovered by , and in $1991$.
However, reports in his of $1997$ that this result can be found in the Numbers Count column of , November $1989$.
== Sources ==
* : $6,963,472,309,248$
Category:Historical Notes
Hardy-Ramanujan Number/Examples/87,539,319

== Theorem ==
The $3$rd Hardy-Ramanujan number $\map {\mathrm {Ta} } 3$ is $87 \, 539 \, 319$:
== Proof ==
== Sources ==
* 
* : $87,539,319$
* : $87,539,319$
Category:Hardy-Ramanujan Numbers
Category:87,539,319
Hardy-Ramanujan Number/Examples/87,539,319/Mistake

== Source Work ==
:
:The Dictionary
::$87,539,319$
:
:The Dictionary
::$87,539,319$
== Mistake ==
:The smallest number that can be represented as the sum of $2$ cubes in $3$ different ways.
This should say:
:The smallest number that can be represented as the sum of $2$ positive cubes in $3$ different ways.
This is because we have that $4104$ is the smallest natural number which can be expressed as the sum of $2$ cubes in $3$ different ways:
== Sources ==
* : $87,539,319$
* : $87,539,319$
Category:Mistakes/Curious and Interesting Numbers
Hardy-Weinberg Law

== Theorem ==
Let $A$ and $a$ be alleles.
Let $A$ and $a$ occur in a population in proportions $p$ and $q = 1 - p$.
Then after one generation of random mating, the genotypes $AA$, $Aa$ and $aa$ are in proportions $p^2$, $2 p q$ and $q^2$.
== Proof ==
== Also known as ==
== Also see ==
* Definition:Hardy-Weinberg Equilibrium
== Historical Note ==
== Sources ==
* 
Category:Hardy-Weinberg Law
Category:Genetics
Harmonic Conjugacy is Symmetric

== Theorem ==
Let $AB$ and $PQ$ be line segments on a straight line.
Let $P$ and $Q$ be harmonic conjugates with respect to $A$ and $B$.
Then $A$ and $B$ are harmonic conjugates with respect to $P$ and $Q$ .
== Proof ==
By definition of harmonic conjugates, $\tuple {AB, PQ}$ is a harmonic range.
:
We have:
Hence, by definition, $\tuple {PQ, AB}$ is a harmonic range.
Hence the result, by definition of harmonic conjugates.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $19$. Harmonic ranges and pencils
Category:Harmonic Ranges
Harmonic Functions on Connected Riemannian Manifold with Matching Restrictions to Boundary are Identical

== Theorem ==
Let $\struct {M, g}$ be a connected Riemannian manifold with the boundary $\partial M \ne \empty$.
Let $u, v \in \map {C^\infty} M$ be smooth harmonic functions on $M$.
Suppose restrictions of $u$ and $v$ to $\partial M$ agree.
Then $u = v$ identically.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Problems
Category:Riemannian Geometry
Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum

== Theorem ==
Let $n \in \Z_{>0}$ be a positive integer.
The harmonic mean of the divisors of $n$ is given by:
:$\map H n = \dfrac {n \, \map {\sigma_0} n} {\map {\sigma_1} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
== Proof ==
== Sources ==
* 
Category:Harmonic Mean
Category:Divisor Sum Function
Category:Divisor Count Function
Harmonic Mean of two Positive Real Numbers is Between them

== Theorem ==
Let $a, b \in \R_{\gt 0}$ be (strictly) positive real numbers such that $a < b$.
Let $\map H {a, b}$ denote the harmonic mean of $a$ and $b$.
Then:
:$a < \map H {a, b} < b$
== Proof ==
By definition of harmonic mean:
:$\dfrac 1 {\map H {a, b} } := \dfrac 1 2 \paren {\dfrac 1 a + \dfrac 1 b}$
Thus:
But $\dfrac 1 {\map H {a, b} }$ is the arithmetic mean of $\dfrac 1 b$ and $\dfrac 1 a$.
Hence from Arithmetic Mean of two Real Numbers is Between them:
:$\dfrac 1 b < \dfrac 1 {\map H {a, b} } < \dfrac 1 a$
So by Reciprocal Function is Strictly Decreasing:
:$b > \map H {a, b} > a$
Hence the result.
Category:Harmonic Mean
Harmonic Number as Unsigned Stirling Number of First Kind over Factorial

== Theorem ==
:$H_n = \dfrac { {n + 1} \brack 2} {n!}$
where:
:$H_n$ denotes the $n$th harmonic number
:$n!$ denotes the $n$th factorial
:$\ds { {n + 1} \brack 2}$ denotes an unsigned Stirling number of the first kind.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$H_n = \dfrac { {n + 1} \brack 2} {n!}$
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$H_k = \dfrac { {k + 1} \brack 2} {k!}$
from which it is to be shown that:
:$H_{k + 1} = \dfrac { {k + 2} \brack 2} {\paren {k + 1}!}$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: H_n = \dfrac { {n + 1} \brack 2} {n!}$
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $6$
Category:Harmonic Numbers
Category:Stirling Numbers
Category:Factorials
Harmonic Number is Greater than Logarithm plus Gamma

== Theorem ==
:$H_n > \ln n + \gamma$
where:
:$H_n$ denotes the $n$th harmonic number
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
From Approximate Size of Sum of Harmonic Series:
:$H_n \approx \ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} - \epsilon$
where $0 < \epsilon < \dfrac 1 {252 n^6}$.
We have that:
:$\dfrac 1 {2 n} + \dfrac 1 {120 n^4} > \dfrac 1 {12 n^2} + \dfrac 1 {252 n^6}$
and the result follows.
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $4$
Category:Harmonic Numbers
Harmonic Number is not Integer

== Theorem ==
Let $H_n$ be the $n$th harmonic number.
Then $H_n$ is not an integer for $n \ge 2$.
That is, the only harmonic numbers that are integers are $H_0$ and $H_1$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $19$
Category:Harmonic Number is not Integer
Category:Harmonic Numbers
Harmonic Number is not Integer/Proof 1

== Theorem ==
== Proof ==
As $H_0 = 0$ and $H_1 = 1$, they are integers.
The claim is that $H_n$ is not an integer for all $n \ge 2$.
 otherwise:
:$(\text P): \quad \exists m \in \N: H_m \in \Z$
By the definition of the harmonic numbers:
:$H_m = 1 + \dfrac 1 2 + \dfrac 1 3 + \cdots + \dfrac 1 m$
Let $2^t$ denote the highest power of two in the denominators of the summands.
Then:
Let $S$ be the set of denominators on the of $(2)$.
Then no element of $S$ can have $2$ as a factor, as follows.
Consider an arbitrary summand:
:$\dfrac {2^{t - 1} } {2^j \times k}$
for some $k \in \Z$, where $j \ge 0$ is the highest power of $2$ that divides the denominator.
For any $2$ to remain after simplification, we would need $j > t - 1$.
Were this to be so, then $2^j \times k$ would have $2^t$ as a factor, and some denominator would be a multiple of $2^t$.
By Greatest Power of Two not Divisor, the set of denominators of the of $(1)$:
:$\set {1, 2, 3, \ldots, 2^t - 1, 2^t + 1, \ldots, m}$
contains no multiple of $2^t$.
Therefore there can be no multiple of $2$ in the denominators of the of $(2)$.
Let:
:$\ell = \map \lcm S$
be the lowest common multiple of the elements of $S$.
Because $2$ is not a divisor of any of the elements of $S$, it will not be a divisor of $\ell$.
Hence $\ell$ is likewise odd.
We have:
But the of that last equation is odd, while its is even.
As this is a contradiction, it follows that our assumption $(\text P)$ that such an $m$ exists is false.
That is, there are no harmonic numbers apart from $0$ and $1$ which are integers.
Category:Harmonic Number is not Integer
Harmonic Number is not Integer/Proof 2

== Theorem ==
== Proof ==
:
:$(\text P): \quad \exists m \in \N: H_m \in \Z$
By the definition of the harmonic numbers:
:$1 + \dfrac 1 2 + \dfrac 1 3 + \cdots +\dfrac 1 m = H_m$
$m$ is either prime or composite.
If $m$ is prime, we have that:
If $k \le n$, then $k$ divides $n!$, so all terms are integers.
Every term on the is divisible by $m$ except for one, namely $\paren {m - 1}!$.
Hence, the is not divisible by $m$.
But the is divisible by $m$, so this is a contradiction.
If $m$ is composite, let $p$ be the largest prime which is less than $m$.
Then we have:
For $k < p$, we have that $m!$ is divisible by $p$ and $k$ is not.
So from Euclid's Lemma, every term $\dfrac {m!} k$ is an integer divisible by $p$.
 there exists $k \in \Z$, with $p< k< m$ such that $\dfrac {m!} k$ is not divisible by $p$.
Because $m!$ is divisible by $p$ and $\dfrac{m!} k$ is not, it follows from Euclid's Lemma that $k$ is a multiple of $p$.
Since $k$ is a multiple of $p$ which is greater than $p$:
:$2 p \le k$
Thus:
:$p < 2 p \le k < m$
From Bertrand's Postulate, there exists a prime $q$ such that $p < q < 2 p$.
Hence:
:$p < q <m $
This contradicts the fact that $p$ is the largest prime less than $m$.
Hence, the assumption that there exists $k \in \Z$ with $p < k < m$ such that $\dfrac {m!} k$ is not divisible by $p$ is false.
Therefore, every term of the , except perhaps $\dfrac{m!} p$, is a multiple of $p$.
From above:
:$p < m < 2 p$
so $p$ is the only positive integer less than $m$ which is divisible by $p$.
Since $m!$ is the product of positive integer less than or equal to $m$, it follows that $m!$ is divisible by $p$ only once.
Therefore, $\dfrac {m!} p$ is not divisible by $p$.
Because every term of the is divisible by $p$ except for one, it follows that the is not divisible by $p$.
Because the is a multiple of $m!$, and hence a multiple of $p$, this is a contradiction.
So the assumption that there exists $m \in \N: H_m \in \Z$ is false.
Category:Harmonic Number is not Integer
Harmonic Numbers/Examples

== Examples of Harmonic Numbers ==
=== Harmonic Number $H_0$ ===
=== Harmonic Number $H_1$ ===
=== Harmonic Number $H_2$ ===
=== Harmonic Number $H_3$ ===
=== Harmonic Number $H_4$ ===
=== Harmonic Number $H_5$ ===
=== Harmonic Number $H_{10000}$ ===
Category:Examples of Harmonic Numbers
Harmonic Numbers/Examples/H10000

== Example of Harmonic Number ==
To $15$ decimal places:
:$H_{10000} \approx 9 \cdotp 78760 \, 60360 \, 44382 \, \ldots$
where $H_{10000}$ denotes the $10 \, 000$th harmonic number.
== Proof ==
We have:
Thus for an accuracy of $15$ decimal places it is unnecessary to consider $\dfrac 1 {120 \times 10000^4}$ and smaller terms.
Then:
 2.30258 50929 94045 68401
x 4
 -------------------------
 9.21034 03719 76182 73604
+ 0.57721 56649 01532 86060
+ 0.00005
 -------------------------
 9.78760 50368 77715 59664
- 0.00000 00008 33333 33333
 -------------------------
 9.78760 50360 44382 26331
Hence the result.
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $5$
Category:Examples of Harmonic Numbers
Harmonic Properties of Schwarz Functions

== Theorem ==
Let $f, g : \R \to \C$ be Schwarz functions.
Let $\hat f$, $\hat g$ be the Fourier transforms of $f$ and $g$ respectively.
Then:
:$(1): \quad \hat f$, $\hat g$ are Schwarz functions.
:$(2): \quad \map {\widehat {\paren {\hat f} } } x = \map f {-x}$ for all $x \in \R$.
:$(3): \quad$ If $f * g$ is the convolution of $f$ and $g$, then:
::::$\widehat {f * g} = \hat f \hat g$
== Proof ==
Category:Harmonic Analysis
Harmonic Property of Complete Quadrangle

== Theorem ==
Let $Q$ be a complete quadrangle.
Let $\LL_1$ and $\LL_2$ be the two straight lines through an arbitrary diagonal point $R$ of $Q$.
Let $\LL_3$ and $\LL_4$ be straight lines constructed so as to join $R$ to the other two diagonal points of $Q$.
Then $\LL_1$, $\LL_2$, $\LL_3$ and $\LL_4$ form a harmonic pencil.
== Proof ==
== Sources ==
* 
* 
Category:Complete Quadrangles
Category:Harmonic Pencils
Harmonic Property of Complete Quadrilateral

== Theorem ==
Let $Q$ be a complete quadrilateral.
Let $D_1$ be an arbitrary diagonal of $Q$.
Let $D_2$ and $D_3$ be the other diagonals of $Q$.
Then the $4$ points at which $D_1$ intersects $Q$, $D_2$ and $D_3$ form a harmonic range.
== Proof ==
== Sources ==
* 
* 
Category:Complete Quadrilaterals
Category:Harmonic Ranges
Harmonic Property of Pole and Polar/Circle

== Theorem ==
Let $\CC$ be a circle whose radius is $r$ and whose center is at the origin of a Cartesian plane.
Let $P$ be an arbitrary point in the Cartesian plane.
Let $\LL$ be a straight line through $P$ which intersects $\CC$ at points $U$ and $V$.
Let $Q$ be the point where $\LL$ intersects the polar of $P$.
Then $\tuple {PQ, UV}$ is a harmonic range.
== Proof ==
From Equation of Circle center Origin, we have that the equation of $\CC$ is:
:$x^2 + y^2 = r^2$
Let $P = \tuple {x_1, y_1}$.
Let $Q = \tuple {x, y}$ be a point on $\LL$.
Let $V$ divide $PQ$ in the ratio $k : 1$.
Then the coordinates of $V$ are:
:$V = \tuple {\dfrac {k x + x_1} {k + 1}, \dfrac {k y + y_1} {k + 1} }$
Substituting into the equation for $\CC$:
The roots of this quadratic equation in $k$ correspond to the points $U$ and $V$.
Now let $Q$ lie on the polar of $P$.
:
Then by definition of polar:
:$x x_1 + y y_1 - r^2 = 0$
and the roots of $(1)$ are equal but of opposite sign.
That is, $U$ and $V$ divide $PQ$ internally and externally in the same ratio.
Hence $\tuple {PQ, UV}$ is a harmonic range.
== Also see ==
* Definition:Joachimsthal's Equation
== Sources ==
* : Chapter $\text {III}$. The Circle: $5$. Harmonic property of pole and polar
Category:Harmonic Property of Pole and Polar
Harmonic Range/Examples/Unity Ratio

== Examples of Harmonic Ranges ==
Let $A$, $B$, $P$ and $Q$ be points on a straight line.
Let $\tuple {AB, PQ}$ be a harmonic range such that $P$ is the midpoint of $AB$.
Then $Q$ is the point at infinity.
== Proof ==
 $AQ$ is of finite length.
Let $p := AP$ and $q := AQ$ be the (undirected) lengths of $AP$ and $AQ$ respectively.
By construction, $AP = PB$.
Then:
That is, the length of $AP$ is zero.
which contradicts the definition of $AP$.
Hence $AQ$ can not be of finite length.
Hence the result.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $19$. Harmonic ranges and pencils
Category:Harmonic Ranges
Harmonic Series is Divergent

== Theorem ==
The harmonic series:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 n$ 
diverges.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also see ==
* Definition:Harmonic Numbers
* Definition:Riemann Zeta Function
== Historical Note ==
== Sources ==
* : $23 \cdotp 103 \, 45 \ldots$
* : $\S 1.2.2$: Summary of convergence tests
* : Chapter $\text {B}.19$: The Series $\sum 1/ p_n$ of the Reciprocals of the Primes
* : $23 \cdotp 10345 \ldots$
* 
* 
* 
* 
Category:Harmonic Series is Divergent
Category:Real Analysis
Category:Harmonic Series
Category:Examples of Divergent Series
Harmonic Series is Divergent/Historical Note

== Historical Note on Harmonic Series is Divergent ==
The proof that the Harmonic Series is Divergent was discovered by .
However, it was lost for centuries, before being rediscovered by in $1647$.
It was discovered yet again in $1687$ by , and a short time after that by , after whom it is usually (erroneously) attributed.
Some sources attribute its rediscovery to .
== Sources ==
* : Chapter $\text {A}.20$: The Bernoulli Brothers
Category:Historical Notes
Harmonic Series is Divergent/Proof 1

== Theorem ==
== Proof ==
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 n = \underbrace 1_{s_0} + \underbrace {\frac 1 2 + \frac 1 3}_{s_1} + \underbrace {\frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7}_{s_2} + \cdots$ 
where $\ds s_k = \sum_{i \mathop = 2^k}^{2^{k + 1} \mathop - 1} \frac 1 i$
From Ordering of Reciprocals:
:$\forall m, n \in \N_{>0}: m < n: \dfrac 1 m > \dfrac 1 n$
so each of the summands in a given $s_k$ is greater than $\dfrac 1 {2^{k + 1} }$.
The number of summands in a given $s_k$ is $2^{k + 1} - 2^k = 2 \times 2^k - 2^k = 2^k$, and so:
:$s_k > \dfrac{2^k} {2^{k + 1} } = \dfrac 1 2$
Hence the harmonic sum $H_{2^m}$ satisfies the following inequality:
The diverges, from the $n$th term test.
The result follows from the the Comparison Test for Divergence.
== Historical Note ==
== Sources ==
* : $\S 6.5$
* : $\S 1.2.7$: Harmonic Numbers: $(2)$
* : $\S 9.3$
* 
Category:Harmonic Series is Divergent
Harmonic Series is Divergent/Proof 2

== Theorem ==
== Proof ==
Observe that all the terms of the harmonic series are strictly positive.
From Reciprocal Sequence is Strictly Decreasing, the terms are decreasing.
Hence the Cauchy Condensation Test can be applied, and we examine the convergence of:
This diverges, from the $n$th term test.
Hence $\ds \sum \frac 1 n$ also diverges.
== Historical Note ==
Category:Harmonic Series is Divergent
Harmonic Series is Divergent/Proof 3

== Theorem ==
== Proof ==
We have that the Integral of Reciprocal is Divergent.
Hence from the Cauchy Integral Test, the harmonic series also diverges.
== Note == 
The Cauchy Integral Test works in both directions.
That is, it can also be used to show that Integral of Reciprocal is Divergent based on Harmonic Series is Divergent.
This could lead to a circular proof.
Hence, if the Cauchy Integral Test is used here, it should not be used to prove Integral of Reciprocal is Divergent.
Instead, use for instance the definition of the natural logarithm as integral of reciprocal.
== Historical Note ==
== Sources ==
* : $\S 9.3$
Category:Harmonic Series is Divergent
Harmonic Series is Divergent/Proof 4

== Theorem ==
== Proof ==
For all $N \in \N$:
:$\dfrac 1 N + \dfrac 1 {N + 1} + \cdots + \dfrac 1 {2 N} > N \cdot \dfrac 1 {2 N} = \dfrac 1 2$
Hence, by Cauchy's Convergence Criterion for Series, the Harmonic series is divergent.
== Historical Note ==
== Sources ==
* 
Category:Harmonic Series is Divergent
Hartogs' Lemma (Set Theory)

== Theorem ==
Let $S$ be a set.
Then there exists an ordinal $\alpha$ such that there is no injection from $\alpha$ to $S$.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
This result is also known as Hartogs' Theorem.
Category:Set Theory
Category:Ordinals
Category:Hartogs' Lemma (Set Theory)
Hartogs' Lemma (Set Theory)/Proof 1

== Theorem ==
== Proof == 
Define $\alpha = \set {\beta: \text{$\beta$ is an ordinal and there is an injection $\beta \to S$}}$.
First of all, it is to be shown that $\alpha$ is a set.
To this end, define the set $W$ by:
:$W = \set {\paren {S', \preceq}: \text{$S' \subseteq S$ and $\preceq$ well-orders $S'$}}$
By the Counting Theorem, each $w \in W$ corresponds to a unique ordinal $\beta_w$.
Thus by the Axiom of Replacement, the following is a set:
:$\set {\beta: \exists w \in W: \beta = \beta_w}$
It follows from Injection Induces Well-Ordering that this set coincides with $\alpha$.
In particular, then, $\alpha$ is a set.
Next, to establish $\alpha$ is an ordinal.
Suppose $\beta \in \alpha$ and $\gamma < \beta$.
Let $i: \beta \to S$ be an injection, and let $\iota: \gamma \to \beta$ be the inclusion of $\gamma$ in $\beta$.
Then by Composite of Injections is Injection, $i \circ \iota: \gamma \to S$ is an injection.
Hence $\gamma \in \alpha$, and therefore $\alpha$ is an ordinal.
Finally, by Ordinal is not Element of Itself, $\alpha \notin \alpha$.
That is to say, there is no injection $\alpha \to S$.
== Notes == 
* The construction in the proof in fact defines the least such ordinal, which is called the Hartogs number of $S$.
* Hartogs' lemma forms the basis for transfinite induction: There can be no strictly increasing mapping from the class of ordinals into a poset $S$, because if there were then it would be an injection from the Hartogs number of $S$ into $S$
Category:Hartogs' Lemma (Set Theory)
Hartogs' Lemma (Set Theory)/Proof 2

== Theorem ==
== Proof ==
Let $W$ be the set of all well-orderings on subsets of $S$.
By the Counting Theorem, there exists a mapping $F: W \to \On$ defined by letting $\map F s$ be the ordinal which is isomorphic to $s$.
By Mapping from Set to Class of All Ordinals is Bounded Above, $F \sqbrk W$ has an upper bound $\alpha_0$.
Then if $\alpha$ is any ordinal strictly greater than $\alpha_0$, then $\alpha \notin F \sqbrk W$.
 there is an injection $g: \alpha \to S$.
Then by Injection to Image is Bijection, there is a bijection from $\alpha$ onto $g \sqbrk \alpha$.
But this induces a well-ordering on $g \sqbrk \alpha \subseteq S$ which is isomorphic to $\alpha$, contradicting the fact that $\alpha \notin F \sqbrk W$.
== Sources ==
* : Theorem $9.5.1$
Category:Hartogs' Lemma (Set Theory)