avewright/proofwiki-math huggingface dataset:
Theorem: Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial/Condition for Convergence

Consider the series:
{{:Factorial as Sum of Series of Subfactorial by Falling Factorial over Factorial}}
This converges only when $n \in \Z_{\ge 0}$, that is, when $n$ is a non-negative integer.

Proof:
First we show that this series converges when $n \in \Z_{\ge 0}$.
Consider the coefficients:
:$1, \paren {1 - \dfrac 1 {1!} }, \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} }, \ldots$
By Power Series Expansion for Exponential Function, they converge to $\dfrac 1 e$.
Starting from the $\paren {n + 1}$th term, there is a factor of $\paren {n - n}$.
In this case, all subsequent terms of the expansion equal $0$.
Therefore the series converges.
{{qed|lemma}}
Then we show that this series diverges for $n \notin \Z_{\ge 0}$.
We do this via Divergence Test.
That is, we show that $\ds \lim_{k \mathop \to \infty} \frac { {!k} \, n^{\underline k} } {k!} \ne 0$.
{{AimForCont}} $\ds \lim_{k \mathop \to \infty} \frac { {!k} \, n^{\underline k} } {k!}$ exists.
By Power Series Expansion for Exponential Function, we have that $\ds \lim_{k \mathop \to \infty} \frac {k!} {!k} = e$.
Hence $\ds \lim_{k \mathop \to \infty} n^{\underline k}$ would exist as well.
For $k > \floor {n + 2}$:
{{begin-eqn}}
{{eqn | l = n - k
      | o = <
      | r = -2 + n + 2 - \floor {n + 2}
}}
{{eqn | o = \le
      | r = -2
}}
{{eqn | ll = \leadsto
      | l = \size {n^{\underline k} }
      | r = \size {n^{\underline {k - 1} } } \size {n - k}
}}
{{eqn | o = >
      | r = 2 \size {n^{\underline {k - 1} } }
}}
{{end-eqn}}
Since $n - k \ne 0$, $\size {n^{\underline k} }$ increases without bound.
Hence its limit does not exist, a contradiction.
Therefore $\ds \lim_{k \mathop \to \infty} \frac { {!k} \, n^{\underline k} } {k!}$ does not exist for $n \notin \Z_{\ge 0}$.
By Divergence Test, our series diverges.
{{qed}}
Theorem: Bounded Subspace of Euclidean Space is Totally Bounded

Let $\struct {\R^n, \norm {\, \cdot \,} }$ be a Euclidean space, where $\norm {\, \cdot \,}$ denotes the usual metric.
Let $M$ be a metric subspace of $\struct {\R^n, \norm {\, \cdot \,} }$.
Let $M$ be bounded.
Then $M$ is totally bounded.

Proof:
As $M$ is bounded, it has a finite diameter $R$.
Consider arbitary $x, y \in \R^n$, expressed in their usual coordinates.
{{begin-eqn}}
{{eqn | l = \norm {y - x}
      | r = \sqrt {\sum_{i \mathop = 1}^n \paren {y_i - x_i}^2}
}}
{{eqn | o = \le
      | r = \sqrt {n \max_i \set {\paren {y_i - x_i}^2} }
}}
{{eqn | r = \sqrt n \max_i \size {y_i - x_i}
}}
{{end-eqn}}
Thus, up to a positive constant multiple, the inequality:
:$\ds \sup_{x, y \mathop \in M} \max_i \size {y_i - x_i} \le R$
implies that $M$ is bounded with radius $R$.
But this inequality is a characterization of an $n$-cube.
Because $R \le 2 R \sqrt n$ for all $R > 0$, $M$ can be placed inside an $n$-cube of the form $\closedint {c - R} {c + R}^n$, by diameter of $n$-cube.
For any $\epsilon > 0$, pick an integer $k > \sqrt n R \epsilon^{-1}$.
Consider a normal subdivision of $\closedint {c - R} {c + R}$ into $k$ pieces.
The length of each subinterval is then:
:$\dfrac {\paren {c + R - \paren {c - R} } } k = 2 R k^{-1}$
For any subinterval chosen this way, the product of $n$ copies of such an interval will itself be a cube, because the subintervals were chosen as to all be the same length.
By diameter of $n$-cube, each sub-cube has diameter $2 R k^{-1} \sqrt n$
By the choice of $k$, we have $2 R k^{-1} \sqrt n < 2 \epsilon$
That means that each sub-cube can fit inside an open ball $n$-ball of radius $\epsilon$.
This in turn implies that $M$ can be covered by a finite number of balls with radius $\epsilon$.
Thus $M$ is totally bounded by definition, as $\epsilon$ was arbitrary.
{{qed}}
Theorem: Difference is Rational is Equivalence Relation

Define $\sim$ as the relation on real numbers given by:
:$x \sim y \iff x - y \in \Q$
That is, that the difference between $x$ and $y$ is rational.
Then $\sim$ is an equivalence relation.

Proof:
Checking in turn each of the criteria for equivalence:
Theorem: Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta

:$\ds \prod_{n \mathop \ge 1} \dfrac {\paren {n + \alpha_1} \cdots \paren {n + \alpha_k} } {\paren {n + \beta_1} \cdots \paren {n + \beta_k} } = \dfrac {\map \Gamma {1 + \beta_1} \cdots \map \Gamma {1 + \beta_k} } {\map \Gamma {1 + \alpha_1} \cdots \map \Gamma {1 + \alpha_k} }$
where:
:$\alpha_1 + \cdots + \alpha_k = \beta_1 + \cdots + \beta_k$
:none of the $\beta$s is a negative integer.

Proof:
First we note that if any of the $\beta$s is a negative integer, the {{LHS}} would have $0$ as its denominator, and so would be undefined.
We have from the Euler form of the Gamma function that:
:$\map \Gamma {1 + \beta_i} = \ds \lim_{m \mathop \to \infty} \dfrac {m^{1 + \beta_i} m!} {\paren {1 + \beta_i} \paren {2 + \beta_i} \cdots \paren {m + 1 + \beta_i} }$
and so the {{RHS}} can be written as:
{{begin-eqn}}
{{eqn | o = 
      | r = \dfrac {\paren {\ds \lim_{m \mathop \to \infty} \dfrac {m^{1 + \beta_1} m^{1 + \beta_2} \cdots m^{1 + \beta_k} \paren {m!}^k} {\ds \prod_{1 \mathop \le n \mathop \le m + 1} \paren {n + \beta_1} \paren {n + \beta_2} \cdots \paren {n + \beta_k} } } } {\paren {\ds \lim_{m \mathop \to \infty} \dfrac {m^{1 + \alpha_1} m^{1 + \alpha_2} \cdots m^{1 + \alpha_k} \paren {m!}^k} {\ds \prod_{1 \mathop \le n \mathop \le m + 1} \paren {n + \alpha_1} \paren {n + \alpha_2} \cdots \paren {n + \alpha_k} } } }
      | c = 
}}
{{eqn | r = \lim_{m \mathop \to \infty} \dfrac {m^k m^{\beta_1 + \beta_2 + \cdots + \beta_k} \paren {m!}^k \ds \prod_{1 \mathop \le n \mathop \le m + 1} \paren {n + \alpha_1} \paren {n + \alpha_2} \cdots \paren {n + \alpha_k} } {m^k m^{\alpha_1 + \alpha_2 + \cdots + \alpha_k} \paren {m!}^k \ds \prod_{1 \mathop \le n \mathop \le m + 1} \paren {n + \beta_1} \paren {n + \beta_2} \cdots \paren {n + \beta_k} }
      | c = 
}}
{{eqn | r = \lim_{m \mathop \to \infty} \ds \prod_{1 \mathop \le n \mathop \le m + 1} \dfrac {\paren {n + \alpha_1} \paren {n + \alpha_2} \cdots \paren {n + \alpha_k} } {\paren {n + \beta_1} \paren {n + \beta_2} \cdots \paren {n + \beta_k} }
      | c = as $\alpha_1 + \cdots + \alpha_k = \beta_1 + \cdots + \beta_k$
}}
{{eqn | r = \prod_{n \mathop \ge 1} \dfrac {\paren {n + \alpha_1} \paren {n + \alpha_2} \cdots \paren {n + \alpha_k} } {\paren {n + \beta_1} \paren {n + \beta_2} \cdots \paren {n + \beta_k} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Characterization of N-Cube

Let $\struct {\R^n, d}$ be a Euclidean $n$-Space equipped with the usual metric $d$.
Let $x, y \in \R^n$, where:
:$x = \tuple {x_1, x_2, \ldots, x_n}$
:$y = \tuple {y_1, y_2, \ldots, y_n}$
Let $R > 0$ be fixed.
Let:
:$\ds Q = \set {x, y \in \R^n: \sup_{x, y} \max_i \size {y_i - x_i} \le R}$
Then $Q$ is an $n$-cube.

Proof:
For ease of presentation, denote:
:$y - x = r \in \R^n$
and:
:$y_j - x_j = r_j$
for $j = 1, 2, \ldots, n$.
{{begin-eqn}}
{{eqn | l = Q
      | r = \set {r: \map {\sup_r} {\max_i \size {r_i} \le R} }
}}
{{eqn | r = \set {r: \map {\sup_r} {\map \max {\size {r_1}, \size {r_2}, \ldots, \size {r_n} } \le R} }
}}
{{eqn | r = \set {r: \map {\sup_\ell} {r_1 \in \closedint {-\ell} \ell, r_2 \in \closedint {-\ell} \ell, \ldots,  r_n \in \closedint {-\ell} \ell, \ell \le R} }
      | c = where $\ell = \map \max {\size {r_1}, \size {r_2}, \ldots, \size {r_n} }$
}}
{{eqn | r = \prod_{i \mathop = 1}^n \closedint {-R} R_i
}}
{{end-eqn}}
{{qed}}
Theorem: Factorial of Integer plus Reciprocal of Integer

Let $x \in \Z$ be a positive integer.
Then:
:$\ds \lim_{n \mathop \to \infty} \dfrac {\paren {n + x}!} {n! n^x} = 1$

Proof:
We have that:
{{begin-eqn}}
{{eqn | l = \dfrac {\paren {n + x}!} {n! n^x}
      | r = \dfrac {\paren {n + 1} \paren {n + 2} \cdots \paren {n + x} } {n^x}
      | c = 
}}
{{eqn | r = \paren {\dfrac {n + 1} n} \paren {\dfrac {n + 2} n} \cdots \paren {\dfrac {n + x} n}
      | c = 
}}
{{eqn | r = \paren {1 + \frac 1 n} \paren {1 + \frac 2 n} \cdots \paren {1 + \frac x n}
      | c = 
}}
{{end-eqn}}
As $n \to \infty$, the quantity on the {{RHS}} indeed tends to $1$.
{{qed}}
Theorem: Sum of Indices of Falling Factorial

:$x^{\underline {m + n} } = x^{\underline m} \paren {x - m}^{\underline n}$

Proof:
{{begin-eqn}}
{{eqn | l = x^{\underline {m + n} }
      | r = \prod_{j \mathop = 0}^{m + n - 1} \paren {x - j}
      | c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \paren {x - j} \prod_{j \mathop = m}^{m + n - 1} \paren {x - j}
      | c = 
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \paren {x - j} \prod_{j \mathop = 0}^{n - 1} \paren {x - {\paren {m + j} } }
      | c = Translation of Index Variable of Product
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \paren {x - j} \prod_{j \mathop = 0}^{n - 1} \paren {\paren {x - m} - j}
      | c = 
}}
{{eqn | r = x^{\underline m} \paren {x - m}^{\underline n}
      | c = {{Defof|Falling Factorial}}
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Indices of Rising Factorial

:$x^{\overline {m + n} } = x^{\overline m} \paren {x + m}^{\overline n}$

Proof:
{{begin-eqn}}
{{eqn | l = x^{\overline {m + n} }
      | r = \prod_{j \mathop = 0}^{m + n - 1} \paren {x + j}
      | c = {{Defof|Rising Factorial}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \paren {x + j} \prod_{j \mathop = m}^{m + n - 1} \paren {x + j}
      | c = 
}}
{{eqn | r = \prod_{j \mathop = 0}^{m - 1} \paren {x + j} \prod_{j \mathop = 0}^{n - 1} \paren {x + \paren {m + j} }
      | c = Translation of Index Variable of Product
}}
{{eqn | r = x^{\overline m} \paren {x + m}^{\overline n}
      | c = {{Defof|Rising Factorial}}
}}
{{end-eqn}}
{{qed}}
Theorem: Union of Indexed Family of Sets Equal to Union of Disjoint Sets

Let $\family {E_n}_{n \mathop \in \N}$ be a countable indexed family of sets where at least two $E_n$ are distinct.
Then there exists a countable indexed family of disjoint sets $\family {F_n}_{n \mathop \in \N}$ defined by:
:$\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$
satisfying:
:$\ds \bigsqcup_{n \mathop \in \N} F_n = \bigcup_{n \mathop \in \N} E_n$
where $\bigsqcup$ denotes disjoint union.

Proof:
Denote:
{{begin-eqn}}
{{eqn | l = E
      | r = \bigcup_{k \mathop \in \N} E_k
}}
{{eqn | l = F
      | r = \bigcup_{k \mathop \in \N} F_k
}}
{{end-eqn}}
where:
:$\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$
We first show that $E = F$.
That $x \in E \implies x \in F$ follows from the construction of $F$ from subsets of $E$.
Thus $E \subseteq F$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \bigcup_{k \mathop \in \N} F_k
}}
{{eqn | ll= \leadsto
      | q = \exists k \in \N
      | l = x
      | o = \in
      | r = F_k
}}
{{eqn | ll= \leadsto
      | q = \exists k \in \N
      | l = x
      | o = \in
      | r = E_k
}}
{{eqn | lo= \land
      | l = x
      | o = \notin
      | r = \paren {E_0 \cup E_1 \cup E_2 \cup \cdots \cup E_{k - 1} }
}}
{{eqn | ll= \leadsto
      | q = \exists k \in \N
      | l = x
      | o = \in
      | r = E_k
      | c = Rule of Simplification
}}
{{end-eqn}}
so $F \subseteq E$.
Thus $E = F$ by definition of set equality.
To show that the sets in $F$ are (pairwise) disjoint, consider an arbitrary $x \in F$. 
Then $x \in F_k$ for some $F_k$.
By the Well-Ordering Principle, there exists a smallest such $k$.
Then:
:$\forall j < k: x \notin F_j$
Choose any distinct $\ell, m \in \N$. 
We have:
If $m > \ell$, then:
{{begin-eqn}}
{{eqn | l = x \in F_\ell
      | o = \implies
      | r = x \in E_\ell
}}
{{eqn | l = x \in F_m
      | o = \implies
      | r = x \notin E_m
}}
{{end-eqn}}
If $m < \ell$, then:
{{begin-eqn}}
{{eqn | l = x \in F_m
      | o = \implies
      | r = x \in E_m
}}
{{eqn | l = x \in F_\ell
      | o = \implies
      | r = x \notin E_\ell
}}
{{end-eqn}}
So the sets $F_\ell, F_m$ are disjoint. 
Thus $F$ is the disjoint union of sets equal to $E$:
:$\ds \bigcup_{k \mathop \in \N} E_k = \bigsqcup_{k \mathop \in \N} F_k$
{{qed}}
Theorem: Borel Sigma-Algebra Generated by Closed Sets

Let $\map \BB {S, \tau}$ be a Borel $\sigma$-algebra generated by the set of open sets in $S$.
Then $\map \BB {S, \tau}$ is equivalently generated by the set of closed sets in $S$.

Proof:
By definition, a closed set is the relative complement of an open set.
The result follows from Sigma-Algebra Generated by Complements of Generators.
{{qed}}
Theorem: Stirling Number of n with n-m is Polynomial in n of Degree 2m/Unsigned First Kind

Let $m \in \Z_{\ge 0}$.
The unsigned Stirling number of the first kind $\ds {n \brack n - m}$ is a polynomial in $n$ of degree $2 m$.

Proof:
The proof proceeds by induction over $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n \brack n - m}$ is a polynomial in $n$ of degree $2 m$.
Theorem: Unsigned Stirling Number of the First Kind of n with n-2

:$\ds {n \brack n - 2} = \binom n 4 + 2 \binom {n + 1} 4$

Proof:
The proof proceeds by induction.
Theorem: Cardinality of Infinite Sigma-Algebra is at Least Cardinality of Continuum

Let $\MM$ be an infinite $\sigma$-algebra on a set $X$.
Then $\MM$ is has cardinality at least that of the cardinality of the continuum $\mathfrak c$:
:$\map \Card \MM \ge \mathfrak c$

Proof:
Let:
:$\mathbb M_\infty := \set {A \in \MM : \map \Card {\MM_A} = \infty}$
where:
:$\MM_A$ denotes the trace $\sigma$-algebra of $A$ in $\MM$
Let $A \in \mathbb M_\infty$.
Observe:
{{begin-eqn}}
{{eqn | l = \map \Card {\MM_B} + \map \Card {\MM_{A \mathop \setminus B} }
      | o = \ge
      | r = \map \Card {\MM_B \cup \MM_{A \mathop \setminus B} }
}}
{{eqn | o = \ge
      | r = \map \Card {\MM_A}
}}
{{eqn | r = \infty
}}
{{end-eqn}}
Thus:
:$\forall B \in \MM_A : B \in {\mathbb M}_\infty \lor A \setminus B \in {\mathbb M}_\infty$
In particular:
:$\forall A \in {\mathbb M}_\infty : \exists B \in {\mathbb M}_\infty : B \subsetneq A$
That is:
:$\O \not \in \mathbb S$
where:
:$\mathbb S := \set {\set {\tuple {A, B} : B \in \mathbb M_\infty, B \subsetneq A} : A \in \mathbb M_\infty}$
By Axiom of Choice, there is a choice function:
:$\tilde f : \mathbb S \to \bigcup \mathbb S$
In particular, it satisfies:
:$\forall A \in \mathbb M_\infty : \map {\tilde f} {\set {\tuple {A, B} : B \in \mathbb M_\infty, B \subsetneq A} } \in \set {\tuple {A, B} : B \in \mathbb M_\infty, B \subsetneq A}$
Hence there exists a mapping:
:$f : \mathbb M_\infty \to \mathbb M_\infty$
such that:
:$\forall A \in \mathbb M_\infty : \map f A \subsetneq A$
More specifically, we can define:
:$\map f A := \map {\pr_2} {\map {\tilde f} {\set {\tuple {A, B} : B \in \mathbb M_\infty, B \subsetneq A} } }$
where $\pr_2$ denotes the second projection of $\paren {\mathbb M_\infty}^2$.
{{improve|This boilerplate for applying AoC should be a theorem}}
Let:
:$F_i := \map {f^i} X \setminus \map {f^{i + 1} } X$
for $i \in \N$.
Then $\sequence {F_i}_{i \mathop \in \N}$ are non-empty pairwise disjoint sets in $\MM$.
We can define an injection $\powerset \N \to \MM$ by:
:$\ds N \mapsto \bigsqcup_{i \mathop \in N} F_i$
That is:
:$\map \Card \MM \ge \map \Card {\powerset \N}$
But by Power Set of Natural Numbers has Cardinality of Continuum:
:$\map \Card {\powerset \N} = \mathfrak c$
Hence the result.
{{qed}}
{{AoC}}
Theorem: Stirling Number of the Second Kind of n with n-2

:$\ds {n \brace n - 2} = \binom {n + 1} 4 + 2 \binom n 4$

Proof:
The proof proceeds by induction.
Theorem: Unsigned Stirling Number of the First Kind of n with n-3

:$\ds {n \brack n - 3} = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$

Proof:
The proof proceeds by induction.
Theorem: Stirling Number of the Second Kind of n with n-3

:$\ds {n \brace n - 3} = \binom {n + 2} 6 + 8 \binom {n + 1} 6 + 6 \binom n 6$

Proof:
The proof proceeds by induction.
Theorem: Duality Law for Stirling Numbers

For all integers $n, m \in \Z$:
:$\ds {n \brace m} = {-m \brack -n}$
where:
:$\ds {n \brace m}$ denotes a Stirling number of the second kind
:$\ds {n \brack m}$ denotes an unsigned Stirling number of the first kind.

Proof:
{{ProofWanted|It is unclear exactly how the Stirling numbers are extended to the negative integers. {{AuthorRef|Donald E. Knuth|Knuth}}'s exposition is uncharacteristically non-explicit.}}
Theorem: Power of Complex Number as Summation of Stirling Numbers of Second Kind

Let $z \in \C$ be a complex number whose real part is positive.
Then:
:$z^r = \ds \sum_{k \mathop \in \Z} {r \brace r - k} z^{\underline {r - k} }$
where:
:$\ds {r \brace r - k}$ denotes the extension of the Stirling numbers of the second kind to the complex plane
:$z^{\underline {r - k} }$ denotes $z$ to the $r - k$ falling.

Proof:
{{ProofWanted|It is unclear exactly how the Stirling numbers are extended to the complex plane. {{AuthorRef|Donald E. Knuth|Knuth}}'s exposition is uncharacteristically non-explicit.}}
Theorem: Falling Factorial of Complex Number as Summation of Unsigned Stirling Numbers of First Kind

Let $z \in \C$ be a complex number whose real part is positive.
Then:
:$z^{\underline r} = \ds \sum_{k \mathop = 0}^m{r \brack r - k} \paren {-1}^k z^{r - k} + \map \OO {z^{r - m - 1} }$
where:
:$\ds {r \brack r - k}$ denotes the extension of the unsigned Stirling numbers of the first kind to the complex plane
:$z^{\underline r}$ denotes $z$ to the $r$ falling
:$\map \OO {z^{r - m - 1} }$ denotes big-$\OO$ notation.

Proof:
{{ProofWanted|It is unclear exactly how the Stirling numbers are extended to the complex plane. {{AuthorRef|Donald E. Knuth|Knuth}}'s exposition is uncharacteristically non-explicit.}}
Theorem: Non-Divisbility of Binomial Coefficients of n by Prime

Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $p$ be a prime number.
Then:
:$\dbinom n k$ is not divisible by $p$ for any $k \in \Z_{\ge 0}$ where $0 \le k \le n$
{{iff}}:
:$n = a p^m - 1$ where $1 \le a < p$
for some $m \in \Z_{\ge 0}$.

Proof:
The statement:
:$\dbinom n k$ is not divisible by $p$
is equivalent to:
:$\dbinom n k \not \equiv 0 \pmod p$
The corollary to Lucas' Theorem gives:
:$\ds \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$
where:
:$n, k \in \Z_{\ge 0}$ and $p$ is prime
:the representations of $n$ and $k$ to the base $p$ are given by:
::$n = a_r p^r + \cdots + a_1 p + a_0$
::$k = b_r p^r + \cdots + b_1 p + b_0$
Consider the form of $n = a p^m - 1$ when represented to the base $p$:
{{begin-eqn}}
{{eqn | l = n
      | r = a p^m - 1
      | c = 
}}
{{eqn | r = a \paren {p^m - 1} + \paren {a - 1}
      | c = 
}}
{{eqn | r = a \paren {p - 1} \sum_{j \mathop = 0}^{m - 1} p^i + \paren {a - 1}
      | c = Sum of Geometric Sequence: Corollary 1
}}
{{eqn | r = \paren {a - 1} p^m + \sum_{j \mathop = 0}^{m - 1} \paren {p - 1} p^i
      | c = after algebra
}}
{{end-eqn}}
That is, all of the digits of $n$ are $p - 1$ except perhaps the first one.
For $k$ such that $0 \le k \le n$, the digits of $k$ will all range over $0$ to $p - 1$.
Theorem: Summation Formula for Reciprocal of Binomial Coefficient

{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom n k \dfrac {\paren {-1}^k} {k + x}
      | r = \dfrac 1 {x \binom {n + x} n}
      | c = 
}}
{{eqn | r = \dfrac {n!} {x \paren {x + 1} \cdots \paren {x + n} }
      | c = 
}}
{{end-eqn}}
as long as the denominators are not zero.

Proof:
Consider the value of $\map \Beta {x, n + 1}$, where $\Beta$ is the beta function.
We have:
{{begin-eqn}}
{{eqn | l = \map \Beta {x, n + 1}
      | r = \int_0^1 t^{x - 1} \paren {1 - t}^n \rd t
      | c = {{Defof|Beta Function|index = 1}}
}}
{{eqn | r = \int_0^1 t^{x - 1} \sum_{k \mathop \ge 0} \binom n k \paren {-t}^k 1^{n - k} \rd t
      | c = Binomial Theorem
}}
{{eqn | r = \sum_{k \mathop \ge 0} \binom n k \paren {-1}^k \int_0^1 t^{x - 1} t^k \rd t
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = \sum_{k \mathop \ge 0} \binom n k \paren {-1}^k \int_0^1 t^{x + k - 1} \rd t
      | c = Index Laws/Sum of Indices
}}
{{eqn | r = \sum_{k \mathop \ge 0} \binom n k \paren {-1}^k \frac 1 {x + k}
      | c = 
}}
{{end-eqn}}
which equates to the {{LHS}} of the identity.
We also have:
{{begin-eqn}}
{{eqn | l = \map \Beta {x, n + 1}
      | r = \frac {\map \Gamma x \map \Gamma {n + 1} } {\map \Gamma {x + n + 1} }
      | c = {{Defof|Beta Function|index = 3}}
}}
{{eqn | r = \frac {n! \map \Gamma x } {\map \Gamma {x + n + 1} }
      | c = Gamma Function Extends Factorial
}}
{{eqn | r = \frac {n! \map \Gamma x } {\paren {x + n} \map \Gamma {x + n} }
      | c = Gamma Difference Equation
}}
{{eqn | r = \frac {n! \map \Gamma x } {\paren {x + n - 1} \paren {x + n} \map \Gamma {x + n - 1} }
      | c = Gamma Difference Equation
}}
{{eqn | o = :}}
{{eqn | r = \frac {n! \map \Gamma x } {x \paren {x + 1} \cdots \paren {x + n} \map \Gamma x}
      | c = Gamma Difference Equation
}}
{{eqn | r = \frac {n!} {x \paren {x + 1} \cdots \paren {x + n} }
      | c =
}}
{{end-eqn}}
which equates to the {{RHS}} of the identity.
Hence the result.
{{qed}}
Theorem: Sum over k of r Choose k by Minus r Choose m Minus 2k

Let $r \in \R$, $m \in \Z$.
:$\ds \sum_{k \mathop \in \Z} \binom r k \binom {-r} {m - 2 k} \paren {-1}^{m + k} = \binom r m$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \paren {1 - x^2}
      | r = \paren {1 - x} \paren {1 + x}
      | c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
      | l = \paren {1 - x}^r
      | r = \dfrac {\paren {1 - x^2}^r} {\paren {1 + x}^r}
      | c = 
}}
{{eqn | r = \paren {1 - x^2}^r \paren {1 + x}^{-r}
      | c = 
}}
{{end-eqn}}
Thus we have:
{{begin-eqn}}
{{eqn | l = \sum_{m \mathop \in \Z} \binom r m x^m
      | r = \paren {1 - x}^r
      | c = Binomial Theorem
}}
{{eqn | r = \paren {1 - x^2}^r \paren {1 + x}^{-r}
      | c = 
}}
{{eqn | r = \sum_{k \mathop \in \Z} \binom r k \paren {-x^2}^k \sum_{j \mathop \in \Z} \binom {-r} j x^j
      | c = 
}}
{{end-eqn}}
Comparing the coefficients of $x^m$ on both sides yields the result.
{{qed}}
Theorem: Sigma-Algebra Contains Generated Sigma-Algebra of Subset

Let  $\sigma_\FF$ be a  be a $\sigma$-algebra on a set $\FF$.
Let $\sigma_\FF$ contain a set of sets $\EE$.
Let $\map \sigma \EE$ be the $\sigma$-algebra generated by $\EE$.
Then $\map \sigma \EE \subseteq \sigma_\FF$

Proof:
{{MissingLinks}}
$\sigma_\FF$ is a $\sigma$-algebra containing $\EE$.
$\map \sigma \EE$ is a subset of ''all'' $\sigma$-algebras containing $\EE$, by definition of a generated $\sigma$-algebra.
Therefore it contains $\map \sigma \EE$.
{{qed}}
Theorem: Generated Sigma-Algebra Contains Generated Sigma-Algebra of Subset

Let $\map \sigma \FF$ be the $\sigma$-algebra generated by $\EE$.
Let $\map \sigma \FF$ contain a set of sets $\EE$.
Let $\map \sigma \EE$ be the $\sigma$-algebra generated by $\EE$.
Then $\map \sigma \EE \subseteq \map \sigma \FF$.

Proof:
Apply Sigma-Algebra Contains Generated Sigma-Algebra of Subset to  $\map \sigma \FF$.
{{qed}}
Theorem: Binomial Theorem/Abel's Generalisation/x+y = 0

Consider Abel's Generalisation of Binomial Theorem:
{{:Abel's Generalisation of Binomial Theorem}}
This holds in the special case where $x + y = 0$.

Proof:
As $x + y = 0$, we can substitute $y = -x$, and so:
:$\ds \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k} = 0$
is to be proved.
So:
{{begin-eqn}}
{{eqn | l = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k}
      | r = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-1}^{n - k} \paren {x - k z}^{n - k}
      | c = 
}}
{{eqn | r = \sum_k \binom n k \paren {-1}^{n - k} x \paren {x - k z}^{n - 1}
      | c = 
}}
{{end-eqn}}
Then Sum over $k$ of $\dbinom r k$ by $\paren {-1}^{r - k}$ by Polynomial can be applied:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$.
As the coefficient of $k^n = 0$, we have:
:$\ds \sum_k \binom n k \paren {-1}^{n - k} \paren {b_0 + b_1 k + \cdots + b_{n - 1} k^n + b_n 0} = 0$
Hence the result.
{{qed}}