nyuuzyou/wikis huggingface dataset:
Sum over Integers of Cosine of n + alpha of theta over n + alpha

== Theorem ==
Let $\alpha \in \R$ be a real number which is specifically not an integer.
For $0 \le \theta < 2 \pi$:
:$\ds \dfrac 1 \alpha + \sum_{n \mathop \ge 1} \dfrac {2 \alpha} {\alpha^2 - n^2} = \sum_{n \mathop \in \Z} \dfrac {\cos \paren {n + \alpha} \theta} {n + \alpha}$
=== Corollary ===
== Proof ==
First we establish the following, as they will be needed later.
We have:
Setting $\theta = 0$:
To establish this identity for all other values of $\theta$ on the interval $0 \le \theta < 2\pi$, we will demonstrate that the sum is a constant function.
We will do this by showing that the derivative of the function is zero everywhere which by Zero Derivative implies Constant Function will complete the proof.
We have:
Then:
From Cosine Function is Even and Sine Function is Odd, we have:
:$\map \cos {-n \theta} = \map \cos {n \theta}$
and:
:$\map \sin {-n \theta} = -\map \sin {n \theta}$
Therefore: 
== Also see ==
* Sum over Integers of Sine of n + alpha of theta over n + alpha
== Sources ==
* : Exercises on Chapter $\text {II}$: $2 \ \text {(i)}$
Category:Cosine Function
Sum over Integers of Sine of n + alpha of theta over n + alpha

== Theorem ==
Let $\alpha \in \R$ be a real number which is specifically not an integer.
For $0 < \theta < 2\pi$:
:$\ds \sum_{n \mathop \in \Z} \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha} = \pi$
== Proof ==
First we establish the following, as they will be needed later.
Setting $\theta = \pi$:
Then:
To establish this identity for all other values of $\theta$ on the interval $0 < \theta < 2 \pi$, we will demonstrate that the sum is a constant function.
We will do this by showing that the derivative of the function is zero everywhere which by Zero Derivative implies Constant Function will complete the proof.
We have:
From Cosine Function is Even and Sine Function is Odd, we have:
:$\map \cos {-n \theta} = \map \cos {n \theta}$
and:
:$\map \sin {-n \theta} = -\map \sin {n \theta}$
Therefore: 
== Also see ==
* Sum over Integers of Cosine of n + alpha of theta over n + alpha
== Sources ==
* : Exercises on Chapter $\text {II}$: $1$.
Category:Sine Function
Sum over Union of Finite Sets

== Theorem ==
Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.
Let $S$ and $T$ be finite sets.
Let $f: S \cup T \to \mathbb A$ be a mapping.
Then we have the equality of summations over finite sets:
:$\ds \sum_{u \mathop \in S \mathop \cup T} \map f u = \sum_{s \mathop \in S} \map f s + \sum_{t \mathop \in T} \map f t - \sum_{v \mathop \in S \mathop \cap T} \map f v$
== Proof ==
Follows from:
:Mapping Defines Additive Function of Subalgebra of Power Set
:Power Set is Algebra of Sets
:Inclusion-Exclusion Principle
== Also see ==
* Inclusion-Exclusion Principle
Category:Summations
Sum over j, k of -1^j+k by j+j Choose k+l by r Choose j by n Choose k by s+n-j-k Choose m-j

== Theorem ==
Let $l, m, n \in \Z$ be integers such that $n \ge 0$.
Then:
:$\ds \sum_{j, \ k \mathop \in \Z} \paren {-1}^{j + k} \dbinom {j + k} {k + l} \dbinom r j \dbinom n k \dbinom {s + n - j - k} {m - j} = \paren {-1}^l \dbinom {n + r} {n + l} \dbinom {s - r} {m - n - l}$
== Proof ==
== Sources ==
* : $\S 5.1$: Basic Identities $(32)$
* : $\S 5$: Binomial Coefficients: Exercise $83$
* : $\S 5$: Binomial Coefficients: Exercise $106$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $63$
Category:Binomial Coefficients
Sum over j of Function of Floor of mj over n

== Theorem ==
Let $f$ be a real function.
Then:
:$\ds \sum_{0 \mathop \le j \mathop < n} \map f {\floor {\dfrac {m j} n} } = \sum_{0 \mathop \le r \mathop < m} \ceiling {\dfrac {r n} m} \paren {\map f {r - 1} - \map f r} + n \map f {m - 1}$
=== Corollary ===
== Proof ==
Hence:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $45$
Category:Summations
Category:Floor Function
Sum over j of Function of Floor of mj over n/Corollary

== Corollary to Sum over $j$ of $\map f {\floor {\frac {m j} n} }$ ==
:$\ds \sum_{0 \mathop \le j \mathop < n} \binom {\floor {m j / n} + 1} k + \sum_{0 \mathop \le j \mathop < m} \ceiling {\dfrac {j n} m} \binom j {k - 1} = n \binom m k$
== Proof ==
Hence:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $45$
Category:Summations
Category:Floor Function
Sum over k from 1 to Infinity of Zeta of 2k Minus One

== Theorem ==
== Proof ==
Sum down each column, then sum across:
:$\begin{array}{r|cccccccccc}
\paren {\map \zeta {2k} - 1} & & & & & & & \\
\hline
\paren {\map \zeta 2 - 1} & \paren {\dfrac 1 2}^2 & \paren {\dfrac 1 3}^2 & \paren {\dfrac 1 4}^2 & \cdots & \paren {\dfrac 1 n}^2 & \cdots \\
\paren {\map \zeta 4 - 1} & \paren {\dfrac 1 2}^4 & \paren {\dfrac 1 3}^4 & \paren {\dfrac 1 4}^4 & \cdots & \paren {\dfrac 1 n}^4 & \cdots \\
\paren {\map \zeta 6 - 1} & \paren {\dfrac 1 2}^6 & \paren {\dfrac 1 3}^6 & \paren {\dfrac 14}^6 & \cdots & \paren {\dfrac 1 n}^6 & \cdots \\
\paren {\map \zeta 8 - 1} & \paren {\dfrac 1 2}^8 & \paren {\dfrac 1 3}^8 & \paren {\dfrac 1 4}^8 & \cdots & \paren {\dfrac 1 n}^8 & \cdots \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
\hline
\ds \sum_{k \mathop = 1}^\infty \paren {\map \zeta {2k} - 1} & \paren {\dfrac 1 1} \paren {\dfrac 1 3} & \paren {\dfrac 1 2} \paren {\dfrac 1 4} & \paren {\dfrac 1 3} \paren {\dfrac 1 5} & \cdots & \paren {\dfrac 1 {n - 1} } \paren {\dfrac 1 {n+1} } & \cdots 
\end{array}$
First, summing down the array:
Next, summing across, we note:
Therefore,
Hence:
== Also see ==
* Sum over k from 2 to Infinity of Zeta of k Minus One
* Sum over k from 1 to Infinity of Zeta of 2k Over Odd Powers of 2
== Sources ==
* 
Category:Riemann Zeta Function
Sum over k from 1 to Infinity of Zeta of 2k Over Odd Powers of 2

== Theorem ==
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Sum over k from 1 to Infinity of Zeta of 2k Minus One
* Sum over k from 2 to Infinity of Zeta of k Minus One
== Sources ==
* 
Category:Riemann Zeta Function
Category:Riemann Zeta Function at Even Integers
Category:Sum over k from 1 to Infinity of Zeta of 2k Over Odd Powers of 2
Sum over k from 1 to Infinity of Zeta of 2k Over Odd Powers of 2/Proof 3

== Theorem ==
== Proof ==
From Laurent Series Expansion for Cotangent Function, we have:
:$\ds \pi \cot \pi z = \dfrac 1 z - 2 \sum_{k \mathop = 1}^\infty \map \zeta {2 k} z^{2 k - 1}$
Setting $z = \dfrac 1 2$:
Hence:
Category:Sum over k from 1 to Infinity of Zeta of 2k Over Odd Powers of 2
Sum over k from 1 to n of n Choose k by Sine of n Theta

== Theorem ==
:$\ds \sum_{k \mathop = 1}^n \dbinom n k \sin k \theta = \paren {2 \cos \dfrac \theta 2}^n \sin \dfrac {n \theta} 2$
== Proof ==
At the same time:
== Sources ==
* : $\S 2$. Geometrical Representations: Example $2$.
Category:Sine Function
Category:Binomial Coefficients
Sum over k from 2 to Infinity of Zeta of k Minus One

== Theorem ==
== Proof ==
Sum down each column, then sum across:
:$\begin{array}{r|cccccccccc}
\paren {\map \zeta k - 1} & & & & & & & \\
\hline
\paren {\map \zeta 2 - 1} & \paren {\dfrac 1 2}^2 & \paren {\dfrac 1 3}^2 & \paren {\dfrac 1 4}^2 & \cdots & \paren {\dfrac 1 n}^2 & \cdots \\
\paren {\map \zeta 3 - 1} & \paren {\dfrac 1 2}^3 & \paren {\dfrac 1 3}^3 & \paren {\dfrac 1 4}^3 & \cdots & \paren {\dfrac 1 n}^3 & \cdots \\
\paren {\map \zeta 4 - 1} & \paren {\dfrac 1 2}^4 & \paren {\dfrac 1 3}^4 & \paren {\dfrac 14}^4 & \cdots & \paren {\dfrac 1 n}^4 & \cdots \\
\paren {\map \zeta 5 - 1} & \paren {\dfrac 1 2}^5 & \paren {\dfrac 1 3}^5 & \paren {\dfrac 1 4}^5 & \cdots & \paren {\dfrac 1 n}^5 & \cdots \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
\hline
\ds \sum_{k \mathop = 2}^\infty \paren {\map \zeta k - 1} & \paren {\dfrac 1 1} \paren {\dfrac 1 2} & \paren {\dfrac 1 2} \paren {\dfrac 1 3} & \paren {\dfrac 1 3} \paren {\dfrac 1 4} & \cdots & \paren {\dfrac 1 {n - 1} } \paren {\dfrac 1 n} & \cdots 
\end{array}$
First, summing down the array:
Next, summing across, we note:
Therefore,
Hence:
== Also see ==
* Sum over k from 1 to Infinity of Zeta of 2k Minus One
* Sum over k from 1 to Infinity of Zeta of 2k Over Odd Powers of 2
== Sources ==
* 
Category:Riemann Zeta Function
Sum over k from 2 to Infinity of Zeta of k Over k Alternating in Sign

== Theorem ==
== Proof 1 ==
== Proof 2 ==
== Sources ==
* https://www.instagram.com/p/CzgNh31ttAF/
Category:Sum over k from 2 to Infinity of Zeta of k Over k Alternating in Sign
Category:Euler-Mascheroni Constant
Category:Riemann Zeta Function
Sum over k of -1^k by n choose k by r-kt choose n by r over r-kt

== Theorem ==
:$\ds \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
where $\delta_{n 0}$ is the Kronecker delta.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
=== Basis for the Induction ===
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P j$ is true, where $j \ge 0$, then it logically follows that $\map P {j + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_k \paren {-1}^k \dbinom j k \dbinom {r - k t} j \dfrac r {r - k t} = \delta_{j 0}$
from which it is to be shown that:
:$\ds \sum_k \paren {-1}^k \dbinom {j + 1} k \dbinom {r - k t} {j + 1} \dfrac r {r - k t} = \delta_{\paren {j + 1} 0}$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 0} \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
Category:Binomial Coefficients
Sum over k of -2 Choose k

== Theorem ==
:$\ds \sum_{k \mathop = 0}^n \binom {-2} k = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$
where:
:$\dbinom {-2} k$ is a binomial coefficient
:$\ceiling x$ denotes the ceiling of $x$.
== Proof ==
When $n$ is even, we have:
As $n$ is even, $n + 1$ is odd, and so:
:$\dfrac {n + 2} 2 = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$
When $n$ is odd, we have:
As $n$ is odd, $n + 1$ is even, and so $\dfrac {n + 1} 2$ is an integer.
Thus from Real Number is Integer iff equals Ceiling:
:$\paren {-1} \dfrac {n + 1} 2 = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$
Thus:
:$\ds \sum_{k \mathop = 0}^n \binom {-2} k = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$
whether $n$ is odd or even.
Hence the result.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(37)$
Category:Binomial Coefficients
Sum over k of Floor of Log base b of k

== Theorem ==
Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $b \in \Z$ such that $b \ge 2$.
Then:
:$\ds \sum_{k \mathop = 1}^n \floor {\log_b k} = \paren {n + 1} \floor {\log_b n} - \dfrac {b^{\floor {\log_b n} + 1} - b} {b - 1}$
== Proof ==
From Sum of Sequence as Summation of Difference of Adjacent Terms:
:$(1): \quad \ds \sum_{k \mathop = 1}^n \floor {\log_b k} = n \floor {\log_b n} - \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\map {\log_b} {k + 1} } - \floor {\log_b k} }$
Let $S$ be defined as:
:$\ds S := \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\map {\log_b} {k + 1} } - \floor {\log_b k} }$
As $b \ge 2$, we have that:
:$\map {\log_b} {k + 1} - \log_b k < 1$
As $b$ is an integer:
:$\lfloor {\map {\log_b} {k + 1} } - \floor {\log_b k} = 1$
 $k + 1$ is a power of $b$.
So:
The result follows by substituting $S$ back into $(1)$ and factoring out $\floor {\log_b n}$.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $42 \ \text{(b)}$
Category:Summations
Category:Floor Function
Category:Logarithms
Sum over k of Floor of Root k

== Theorem ==
Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $b \in \Z$ such that $b \ge 2$.
Then:
:$\ds \sum_{k \mathop = 1}^n \floor {\sqrt k} = \floor {\sqrt n} \paren {n - \dfrac {\paren {2 \floor {\sqrt n} + 5} \paren {\floor {\sqrt n} - 1} } 6}$
== Proof ==
From Sum of Sequence as Summation of Difference of Adjacent Terms:
:$\ds \sum_{k \mathop = 1}^n \floor {\sqrt k} = n \floor {\sqrt n} - \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\sqrt {k + 1} } - \floor {\sqrt k} }$
Let $S$ be defined as:
:$\ds S := \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\sqrt {k + 1} } - \floor {\sqrt k} }$
We have that:
:$\sqrt {k + 1} - \sqrt k < 1$
and so:
:$\floor {\sqrt {k + 1} } - \floor {\sqrt k} = 1$
 $k + 1$ is a square number.
So:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $43$
Category:Summations
Category:Floor Function
Sum over k of Stirling Number of the Second Kind of n+1 with k+1 by Unsigned Stirling Number of the First Kind of k with m by -1^k-m

== Theorem ==
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom n m$
where:
:$\dbinom n m$ denotes a binomial coefficient
:$\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
:$\ds {n + 1 \brace k + 1}$ denotes a Stirling number of the second kind.
== Proof ==
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom n m$
=== Basis for the Induction ===
$\map P 0$ is the case:
So $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_k {r + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom r m$
from which it is to be shown that:
:$\ds \sum_k {r + 2 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom {r + 1} m$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m, n \in \Z_{\ge 0}: \sum_k {n + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom n m$
== Also see ==
* Sum over $k$ of $\ds {n + 1 \brack k + 1}$ by $\ds {k \brace m}$ by $\paren {-1}^{k - m}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(55)$
Category:Stirling Numbers
Category:Binomial Coefficients
Sum over k of Stirling Numbers of the Second Kind of k+1 with m+1 by n choose k by -1^k-m

== Theorem ==
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$
where:
:$\ds {k + 1 \brace m + 1}$ and so on denotes a Stirling number of the second kind
:$\dbinom n k$ denotes a binomial coefficient.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$
=== Basis for the Induction ===
$\map P 0$ is the case:
So $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_k {k + 1 \brace m + 1} \binom r k \paren {-1}^{r - k} = {r \brace m}$
from which it is to be shown that:
:$\ds \sum_k {k + 1 \brace m + 1} \binom {r + 1} k \paren {-1}^{r + 1 - k} = {r + 1 \brace m}$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m, n \in \Z_{\ge 0}: \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(52)$
Category:Stirling Numbers
Category:Binomial Coefficients
Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k

== Theorem ==
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$
where:
:$\ds {k \brace m}$ denotes a Stirling number of the second kind
:$\dbinom n k$ denotes a binomial coefficient.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$
=== Basis for the Induction ===
$\map P 0$ is the case:
So $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_k {k \brace m} \binom r k = {r + 1 \brace m + 1}$
from which it is to be shown that:
:$\ds \sum_k {k \brace m} \binom {r + 1} k = {r + 2 \brace m + 1}$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m, n \in \Z_{\ge 0}: \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(52)$
Category:Stirling Numbers
Category:Binomial Coefficients
Sum over k of Sum over j of Floor of n + jb^k over b^k+1

== Theorem ==
Let $n, b \in \Z$ such that $n \ge 0$ and $b \ge 2$.
Then:
:$\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } } = n$
where $\floor {\, \cdot \,}$ denotes the floor function.
=== Corollary ===
== Proof ==
We have that $\floor {\dfrac {n + j b^k} {b^{k + 1} } }$ is in the form $\floor {\dfrac {m k + x} n}$ so that:
Thus:
Hence the result.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $44$
Category:Summations
Category:Floor Function
Sum over k of Sum over j of Floor of n + jb^k over b^k+1/Corollary

== Corollary to Sum over $k$ of Sum over $j$ of $\floor {\frac {n + j b^k} {b^{k + 1} } }$ ==
Let $n, b \in \Z$ such that $n < 0$ and $b \ge 2$.
Then:
:$\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }= n + 1$
where $\floor {\, \cdot \,}$ denotes the floor function.
== Proof ==
From the working of Sum over $k$ of Sum over $j$ of $\floor {\dfrac {n + j b^k} {b^{k + 1} } }$:
:$\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } } = \lim_{k \mathop \to \infty} \floor {\dfrac n 1} - \floor {\dfrac n {b^{k + 1} } }$
But:
:$\forall k \in \Z_{> 0}: \dfrac n {b^{k + 1} } < 0$
and so:
:$\ds \lim_{k \mathop \to \infty} \floor {\dfrac n 1} - \floor {\dfrac n {b^{k + 1} } } = n - \paren {-1}$
Hence the result.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $44$
Category:Summations
Category:Floor Function
Sum over k of Unsigned Stirling Number of the First Kind of n+1 with k+1 by Stirling Number of the Second Kind of k with m by -1^k-m

== Theorem ==
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$
where:
:$\sqbrk {n \ge m}$ is Iverson's convention
:$\ds {n + 1 \brack k + 1}$ denotes an unsigned Stirling number of the first kind
:$\ds {k \brace m}$ denotes a Stirling number of the second kind.
== Proof ==
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$
=== Basis for the Induction ===
$\map P 0$ is the case:
So $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_k \sqbrk {r + 1 \atop k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {r \ge m} \dfrac {r!} {m!}$
from which it is to be shown that:
:$\ds \sum_k \sqbrk {r + 2 \atop k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {r + 1 \ge m} \dfrac {\paren {r + 1}!} {m!}$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m, n \in \Z_{\ge 0}: \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$
== Also see ==
* Sum over $k$ of $\ds {n + 1 \brace k + 1}$ by $\ds {k \brack m}$ by $\paren {-1}^{k - m}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $61$
Category:Stirling Numbers
Sum over k of Unsigned Stirling Numbers of First Kind by x^k

== Theorem ==
:$\ds \sum_k {n \brack k} x^k = x^{\overline n}$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$x^{\overline n}$ denotes $x$ to the $n$ rising.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $32$
* : $\S 1.2.9$: Generating Functions: $(27)$
Category:Stirling Numbers
Category:Rising Factorials
Category:Examples of Power Series
Sum over k of Unsigned Stirling Numbers of the First Kind of n+1 with k+1 by k choose m by -1^k-m

== Theorem ==
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brack k + 1} \binom k m \paren {-1}^{k - m} = {n \brack m}$
where:
:$\ds {n + 1 \brack k + 1}$ etc. denotes an unsigned Stirling number of the first kind
:$\dbinom k m$ denotes a binomial coefficient.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n + 1 \brack k + 1} \binom k m \paren {-1}^{k - m} = {n \brack m}$
=== Basis for the Induction ===
$\map P 0$ is the case:
But when $m \ne 0$ we have that:
:$\ds {0 \brack m} \paren {-1}^{- m} = 0 = {0 \brack m}$
and when $m = 0$ we have that:
:$\ds {0 \brack m} \paren {-1}^{- 0} = 1 = {0 \brack m}$
So $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_k {r + 1 \brack k + 1} \binom k m \paren {-1}^{k - m} = {r \brack m}$
from which it is to be shown that:
:$\ds \sum_k {r + 2 \brack k + 1} \binom k m \paren {-1}^{k - m} = {r + 1 \brack m}$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m, n \in \Z_{\ge 0}: \sum_k {n + 1 \brack k + 1} \binom k m \paren {-1}^{k - m} = {n \brack m}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(51)$
Category:Stirling Numbers
Category:Binomial Coefficients
Sum over k of Unsigned Stirling Numbers of the First Kind of n with k by k choose m

== Theorem ==
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n \brack k} \binom k m = {n + 1 \brack m + 1}$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$\dbinom k m$ denotes a binomial coefficient.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brack k} \binom k m = {n + 1 \brack m + 1}$
=== Basis for the Induction ===
$\map P 0$ is the case:
So $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_k {r \brack k} \binom k m = {r + 1 \brack m + 1}$
from which it is to be shown that:
:$\ds \sum_k {r + 1 \brack k} \binom k m = {r + 2 \brack m + 1}$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m, n \in \Z_{\ge 0}: \sum_k {n \brack k} \binom k m = {n + 1 \brack m + 1}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(51)$
Category:Stirling Numbers
Category:Binomial Coefficients