nyuuzyou/wikis huggingface dataset:
Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Basis for the Induction

== Theorem ==
Let $r, s, t \in \R, n \in \Z$.
Consider the equation:
:$\ds (1): \quad \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
where $\dbinom {r - t k} k$ etc. are binomial coefficients.
Then equation $(1)$ holds for the special case where $s = n - 1 - r + n t$.
== Proof ==
Substituting $n - 1 - r + n t$ for $s$ in the :
Substituting $n - 1 - r + n t$ for $s$ in the :
The two products give a polynomial of degree $n - 1$ in $k$.
Hence the sum for all $k$ is $0$.
Thus we have:
Thus the equation indeed holds for the special case where $s = n - 1 - r + n t$.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $22$
Category:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk
Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Lemma

== Theorem ==
Let this hold for $\tuple {r, s, t, n}$:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
and also for $\tuple {r, s - t, t, n - 1}$.
Then it also holds for $\tuple {r, s + 1, t, n}$.
== Proof ==
Evaluating the equation for $\tuple {r, s - t, t, n - 1}$:
Adding the equation in $\tuple {r, s, t, n}$:
Hence the equation holds for $\tuple {r, s + 1, t, n}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $23$
Category:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk
Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 2

== Theorem ==
== Proof ==
From Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$:
:$\ds \sum_k \map {A_k} {r, t} z^k = x^r$
where:
:$\map {A_n} {x, t}$ is the polynomial of degree $n$ defined as:
::$\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$
:for $x \ne n t$
:$z = x^{t + 1} - x^t$.
From Sum over $k$ of $\dbinom {r - k t} k$ by $z^k$:
:$\ds \sum_k \dbinom {r - t k} k z^k = \frac {x^{r + 1} } {\paren {t + 1} x - t}$
Thus:
Equating coefficients of $z$:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $27$
Category:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk
Sum over k of r Choose k by -1^r-k by Polynomial

== Theorem ==
Let $r \in \Z_{\ge 0}$.
Then:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$.
== Proof 1 ==
== Proof 2 ==
Category:Binomial Coefficients
Category:Sum over k of r Choose k by -1^r-k by Polynomial
Sum over k of r Choose k by -1^r-k by Polynomial/Proof 1

== Theorem ==
== Proof ==
From the corollary to Sum over $k$ of $\dbinom r k \dbinom {s + k} n \paren {-1}^{r - k}$:
:$\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \delta_{n r}$
where $\delta_{n r}$ denotes the Kronecker delta.
Thus when $n \ne r$:
:$\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = 0$
and so:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \paren {c_0 \binom k 0 + c_1 \binom k 1 + \cdots + c_m \binom k m} = c_r$
as the only term left standing is the $r$th one.
Choosing the coefficients $c_i$ as appropriate, a polynomial in $k$ can be expressed as a summation of binomial coefficients in the form:
:$c_0 \dbinom k 0 + c_1 \dbinom k 1 + \cdots + c_m \dbinom k m$
Thus we can rewrite such a polynomial in $k$ as:
:$b_0 + b_1 k + \cdots + b_r k^r$
Since each $c_m \dbinom k m$ is a polynomial of degree $m$, it follows that the only one with a non-zero degree $r$ term is $c_r \dbinom k r$.
The coefficient of $k^r$ in $c_r \dbinom k r$ must be equal to $b_r$, that is:
:$b_r = \dfrac {c_r}{r!}$
Hence the result.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(34)$
Category:Sum over k of r Choose k by -1^r-k by Polynomial
Sum over k of r Choose k by -1^r-k by Polynomial/Proof 2

== Theorem ==
== Proof ==
From Summation of Powers over Product of Differences:
:$\ds \sum_{j \mathop = 1}^n \begin {pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end {pmatrix} = \begin {cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end {cases}$
Now we have:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $29$
Category:Sum over k of r Choose k by -1^r-k by Polynomial
Sum over k of r Choose k by Minus r Choose m Minus 2k

== Theorem ==
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:
Thus we have:
Comparing the coefficients of $x^m$ on both sides yields the result.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $49$
Category:Binomial Coefficients
Sum over k of r Choose k by s+k Choose n by -1^r-k

== Theorem ==
Let $s \in \R, r \in \Z_{\ge 0}, n \in \Z$.
Then:
:$\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$
where $\dbinom r k$ etc. are binomial coefficients.
=== Corollary ===
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $\text {I} \ (23)$
Category:Binomial Coefficients
Category:Sum over k of r Choose k by s+k Choose n by -1^r-k
Sum over k of r Choose k by s+k Choose n by -1^r-k/Corollary

== Corollary to Sum over $k$ of $\dbinom r k \dbinom {s + k} n \paren {-1}^{r - k}$ ==
Let $r \in \Z_{\ge 0}, n \in \Z$.
Then:
:$\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \delta_{n r}$
where $\delta_{n r}$ is the Kronecker delta.
== Proof ==
From Sum over $k$ of $\dbinom r k \dbinom {s + k} n \paren {-1}^{r - k}$:
:$\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$
which holds for $s \in \R, r \in \Z_{\ge 0}, n \in \Z$.
Setting $s = 0$:
:$\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \binom 0 {n - r}$
We have by definition of binomial coefficient that:
:$\dbinom 0 0 = 1$
and:
:$\forall n \in \Z_{\ne 0}: \dbinom 0 n = 0$
So, using Iverson's convention:
:$\dbinom 0 {n - r} = \sqbrk {n = r}$
The result follows by definition of the Kronecker delta.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(33)$
Category:Sum over k of r Choose k by s+k Choose n by -1^r-k
Sum over k of r Choose k by s+k Choose n by -1^r-k/Proof 1

== Theorem ==
== Proof ==
The proof proceeds by induction on $r$.
For all $r \in \Z_{>0}$, let $\map P r$ be the proposition:
:$\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$
=== 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 $\map P m$ is true, where $m \ge 0$, then it logically follows that $\map P {m + 1}$ is true.
This is the induction hypothesis:
:$\ds \sum_k \binom m k \binom {s + k} n \paren {-1}^{m - k} = \binom s {n - m}$
from which it is to be shown that:
:$\ds \sum_k \binom {m + 1} k \binom {s + k} n \paren {-1}^{m + 1 - k} = \binom s {n - \paren {m + 1} }$
=== Induction Step ===
This is the induction step:
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$
for all $s \in \R, r \in \Z_{\ge 0}, n \in \Z$.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $19$
Category:Sum over k of r Choose k by s+k Choose n by -1^r-k
Sum over k of r Choose k by s-kt Choose r by -1^k

== Theorem ==
Let $r \in \Z_{\ge 0}$.
Then:
:$\ds \sum_k \binom r k \binom {s - k t} r \paren {-1}^k = t^r$
where $\dbinom r k$ etc. are binomial coefficients.
== Proof ==
From Sum over $k$ of $\dbinom r k \paren {-1}^k$ by Polynomial:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(35)$
Category:Binomial Coefficients
Sum over k of r Choose k by s Choose k by k

== Theorem ==
Let $s \in \R, r \in \Z_{\ge 0}$.
Then:
:$\ds \sum_k \binom r k \binom s k k = \binom {r + s - 1} {r - 1} s$
where $\dbinom r k$ etc. are binomial coefficients.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Example $1$
Category:Binomial Coefficients
Sum over k of r Choose m+k by s Choose n+k

== Theorem ==
Let $s \in \R, r \in \Z_{\ge 0}, m, n \in \Z$.
Then:
:$\ds \sum_k \binom r {m + k} \binom s {n + k} = \binom {r + s} {r - m + n}$
where $\dbinom r {m + k}$ etc. are binomial coefficients.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $\text {I} \ (22)$
Category:Binomial Coefficients
Sum over k of r by r+kt to the Power of k-1 over k Factorial by Power of z

== Theorem ==
Let $x$ be the continuous real function of $z$ which satisfies:
:$\ln x = z x^t$
where $x = 1$ when $z = 0$.
Then:
== Proof ==
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(30)$
Category:Power Series
Sum over k to n of Stirling Number of the Second Kind of k with m by m+1^n-k

== Theorem ==
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$
where $\ds {k \brace m}$ etc. 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:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$
=== 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:
:$\ds \sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} = {r + 1 \brace m + 1}$
from which it is to be shown that:
:$\ds \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k} = {r + 2 \brace m + 1}$
=== 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:
:$\ds \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(56)$
Category:Stirling Numbers
Sum over k to n of Unsigned Stirling Number of the First Kind of k with m by n factorial over k factorial

== Theorem ==
Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$
where:
:$\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
:$ n!$ denotes a factorial.
== Proof ==
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$
=== 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:
:$\ds \sum_{k \mathop \le r} {k \brack m} \frac {r!} {k!} = {r + 1 \brack m + 1}$
from which it is to be shown that:
:$\ds \sum_{k \mathop \le r + 1} {k \brack m} \frac {\paren {r + 1}!} {k!} = {r + 2 \brack m + 1}$
=== 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:
:$\ds \forall m, n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(56)$
Category:Stirling Numbers
Category:Factorials
Sum over k to n of k Choose m by kth Harmonic Number

== Theorem ==
:$\ds \sum_{k \mathop = 1}^n \binom k m H_k = \binom {n + 1} {m + 1} \paren {H_{n + 1} - \frac 1 {m + 1} }$
where:
:$\dbinom k m$ denotes a binomial coefficient
:$H_k$ denotes the $k$th harmonic number.
== Proof ==
First we note that by Pascal's Rule:
:$\dbinom k m = \dbinom {k + 1} {m + 1} - \dbinom k {m + 1}$
Thus:
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: $(9)$
Category:Harmonic Numbers
Category:Binomial Coefficients
Sum over k to p over 2 of Floor of 2kq over p

== Theorem ==
Let $p \in \Z$ be an odd prime.
Let $q \in \Z$ be an odd integer and $p \nmid q$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < p / 2} \floor {\dfrac {2 k q} p} \equiv \sum_{0 \mathop \le k \mathop < p / 2} \floor {\dfrac {k q} p} \pmod 2$
== Proof ==
When $k < \dfrac p 4$ we have:
Here it is noted that $\dfrac {\paren {2 k + 1} q} p$ is not an integer, since we have:
:$p \nmid q$
:$p > \dfrac p 2 + 1 > 2 k + 1$
Thus it is possible to replace the last terms:
:$\floor {\dfrac {\paren {p - 1} q} p}, \floor {\dfrac {\paren {p - 3} q} p}, \ldots$
by:
:$\floor {\dfrac q p}, \floor {\dfrac {3 q} p}, \ldots$
The result follows.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $47 \ \text{c)}$
Category:Prime Numbers
Sum to Infinity of 2x^2n over n by 2n Choose n

== Theorem ==
For $\cmod x < 1$:
:$\ds \frac {2 x \arcsin x} {\sqrt {1 - x^2} } = \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n} } {n \dbinom {2 n} n}$
== Proof ==
By Gregory Series:
:$\ds \arctan t = \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}$
Let $t = \dfrac x {\sqrt {1 - x^2} }$.
Let $y = \arcsin x$.
Then:
Hence $\arctan t = \arcsin x$.
We have:
It remains to show the the coefficient of $x^{2 n}$ on the is equal to $\dfrac {2^{2 n} } {n \dbinom {2 n} n}$, that is:
:$\ds 2 \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {r + n - r - 1} {r - 1} = \frac {2^{2 n} } {n \dbinom {2 n} n}$
The above is generated by picking, for each $m > 0$, the corresponding $k = n - m$ from the right sum $\ds \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}$.
We have:
Hence the result.
== Sources ==
* 
Category:Central Binomial Coefficients
Sum to Infinity of Reciprocal of n^2 by 2n Choose n

== Theorem ==
== Proof ==
By Sum to Infinity of 2x^2n over n by 2n Choose n, for $0 < \size x < 1$:
Substituting $x = 0$ gives:
:$0 = 0 + C$
so $C = 0$.
Hence:
== Also see ==
* Sum to Infinity of Reciprocal of n by 2n Choose n
== Sources ==
* 
* 
Category:Central Binomial Coefficients
Sum to Infinity of Reciprocal of n^4 by 2n Choose n

== Theorem ==
It is conjectured that:
:$\map \zeta 4 = \ds \dfrac {36} {17} \sum_{n \mathop = 1}^\infty \dfrac 1 {n^4 \dbinom {2 n} n}$
== Proof ==
== Historical Note ==
== Sources ==
* : $1,08232 3237 \ldots$
* 
* 
Category:Central Binomial Coefficients
Category:Riemann Zeta Function
Sum to Infinity of Reciprocal of n^4 by 2n Choose n/Historical Note

== Historical Note on Sum to Infinity of Reciprocal of $\dfrac 1 {n^4 \dbinom {2 n} n}$ ==
On the cited MathWorld page, this result is stated as a truth.
However, according to and in their of $1983$, this was merely a conjecture.
== Sources ==
* : $1,08232 3237 \ldots$
* 
Category:Historical Notes
Sum to Infinity of Reciprocal of n by 2n Choose n

== Theorem ==
== Proof ==
By Sum to Infinity of 2x^2n over n by 2n Choose n:
== Also see ==
* Sum to Infinity of Reciprocals of Central Binomial Coefficients
== Sources ==
* 
* : $0,60459 97881 \ldots$
* 
Category:Central Binomial Coefficients
Sum to Infinity of Reciprocals of Central Binomial Coefficients

== Theorem ==
== Proof ==
By Sum to Infinity of 2x^2n over n by 2n Choose n:
== Also see ==
* Sum to Infinity of Reciprocal of n by 2n Choose n
== Sources ==
* 
* : $0,73639 98587 \ldots$
* 
Category:Central Binomial Coefficients
Sum with Maximum is Maximum of Sum

== Theorem ==
Let $a, b, c \in \R$ be real numbers.
Then:
:$a + \max \set {b, c} = \max \set {a + b, a + c}$
== Proof ==
, there are two cases to consider:
:$(1): \quad b \ge c$
:$(2): \quad b < c$
First let $b \ge c$.
We have:
Then:
Now let $b < c$.
We have:
Then:
Thus the result holds in both cases.
Hence the result, by Proof by Cases.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $35$
Category:Max Operation
Sum with One is Immediate Successor in Naturally Ordered Semigroup

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Let $1$ be the one of $S$.
Let $n \in S$.
Then $n \circ 1$ is the immediate successor of $n$.
That is, for all $m \in S$:
:$n \prec m \iff n \circ 1 \preceq m$
== Proof ==
By Zero Strictly Precedes One, $0 \prec 1$, where $0$ is the zero of $S$.
Hence from Strict Ordering of Naturally Ordered Semigroup is Strongly Compatible:
:$n \circ 0 \prec n \circ 1$
and by Zero is Identity in Naturally Ordered Semigroup, $n \circ 0 = n$.
Now suppose that $n \prec m$.
Then by , there exists $p \in S$ such that:
:$n \circ p = m$
Moreover, since $n \ne m$, it follows that $p \ne 0$.
Hence $0 \prec p$ by definition of zero.
Therefore, by definition of one:
:$1 \preceq p$
Now by compatibility of $\preceq$ with $\circ$:
:$n \circ 1 \preceq n \circ p = m$
as desired.
Conversely, if $n \circ 1 \preceq m$, it is immediate from:
:$n \prec n \circ 1$
that $n \prec m$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.4$
Category:Naturally Ordered Semigroup
Summary of Topology on P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\tau_p$ be the topology induced by the non-Archimedean norm $\norm {\,\cdot\,}_p$.
Then $\struct{\Q_p, \tau_p}$ is:
:$(1): \quad$ Hausdorff
:$(2): \quad$ second-countable
:$(3): \quad$ totally disconnected
:$(4): \quad$ locally compact
== Proof ==
Follows from:
:P-adic Numbers is Hausdorff Topological Space
:P-adic Numbers is Second Countable Topological Space
:P-adic Numbers is Totally Disconnected Topological Space
:P-adic Numbers is Locally Compact Topological Space
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.7$
Category:P-adic Number Theory
Category:Topology of P-adic Numbers
Summation/Inequality/Examples/From 1 to Pi

== Example of Summation by Inequality ==
Let $n = 3 \cdotp 1 4$.
Then:
:$\ds \sum_{1 \mathop \le j \mathop \le n} a_j = a_1 + a_2 + a_3$
== Proof ==
:$\ds \sum_{1 \mathop \le j \mathop \le n} a_j$
means:
:The sum of all $a_j$ where $j \in \Z$ and $1 \le j \le n$
from which it follows that $j \in \set {1, 2, 3}$.
Hence the result.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $1$
Category:Examples of Summations
Summation Formula (Complex Analysis)

== Theorem == 
Let $N \in \N$ be an arbitrary natural number.
Let $C_N$ be the square embedded in the complex plane $\C$ with vertices $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$.
Let $f$ be a meromorphic function on $\C$ with finitely many poles. 
Suppose that: 
:$\ds \int_{C_N} \paren {\pi \cot \pi z} \map f z \rd z \to 0$
as $N \to \infty$. 
Let $X$ be the set of poles of $f$.
Then: 
 
:$\ds \sum_{n \mathop \in \Z \mathop \setminus X} \map f n = - \sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
 
If $X \cap \Z = \O$, this becomes: 
:$\ds \sum_{n \mathop = -\infty}^\infty \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
== Proof == 
By Summation Formula: Lemma, there exists a constant $A$ such that: 
:$\cmod {\map \cot {\pi z} } < A$ 
for all $z$ on $C_N$.
Since $f$ has only finitely many poles, we can take $N$ large enough so that no poles of $f$ lie on $C_N$. 
Let $X_N$ be the set of poles of $f$ contained in the region bounded by $C_N$. 
From Poles of Cotangent Function, $\map \cot {\pi z}$ has poles at $z \in \Z$. 
Let $A_N = \set {n \in \Z : -N \le n \le N}$
We then have: 
We then have, for each integer $n$: 
Note that :
:$\ds \int_{C_N} \paren {\pi \cot \pi z} \map f z \rd z \to 0$
So, taking $N \to \infty$: 
:$\ds 0 = 2 \pi i \paren {\sum_{n \mathop \in \Z \mathop \setminus X} \map f n + \sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0} }$
which gives:
:$\ds \sum_{n \mathop \in \Z \mathop \setminus X} \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
== Sources == 
* : $4.4.1$: Summation Theorem 
Category:Complex Analysis
Summation Formula (Complex Analysis)/Lemma

== Theorem ==
Let $N \in \N$ be an arbitrary natural number.
Let $C_N$ be the square embedded in the complex plane with vertices $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$.
Then there exists a constant real number $A$ independent of $N$ such that: 
:$\cmod {\map \cot {\pi z} } < A$ 
for all $z \in C_N$.
== Proof == 
Let $z = x + iy$ for real $x, y$. 
=== Case 1: $y > \frac 1 2$ === 
We have: 
=== Case 2: $y < -\frac 1 2$ === 
Similarly: 
=== Case 3: $-\frac 1 2 \le y \le \frac 1 2$ === 
First consider $z = N + \frac 1 2 + iy$. 
Then: 
Similarly in the case of $z = -N - \frac 1 2 + iy$, we have: 
Picking $A = \map \max {A_1, A_2}$ gives the desired bound. 
== Sources == 
* : $7.24$: Miscellaneous Definite Integrals
Category:Complex Analysis
Summation Formula for Alternating Series

== Theorem == 
Let $N \in \N$ be an arbitrary natural number.
Let $C_N$ be the square embedded in the complex plane $\C$ with vertices $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$.
Let $f$ be a meromorphic function on $\C$ with finitely many poles. 
Suppose that: 
:$\ds \int_{C_N} \paren {\pi \csc \pi z} \map f z \rd z \to 0$
as $N \to \infty$. 
Let $X$ be the set of poles of $f$.
Then: 
:$\ds \sum_{n \mathop \in \Z \mathop \setminus X} \paren {-1}^n \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \map \csc {\pi z} \map f z} {z_0}$
If $X \cap \Z = \O$, this becomes: 
:$\ds \sum_{n \mathop = -\infty}^\infty \paren {-1}^n \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \map \csc {\pi z} \map f z} {z_0}$
</onlyinclude> 
== Proof == 
== Sources == 
* : $7.8$: Summation of Series
Category:Complex Analysis
Summation Formula for Alternating Series over Half-Integers

== Theorem == 
Let $N \in \N$ be an arbitrary natural number.
Let $C_N$ be the square embedded in the complex plane $\C$ with vertices $N \paren {\pm 1 \pm i}$.
Let $f$ be a meromorphic function on $\C$ with finitely many poles. 
Suppose that:
:$\ds \int_{C_N} \paren {\pi \sec \pi z} \map f z \rd z \to 0$
as $N \to \infty$. 
Let $X$ be the set of poles of $f$.
Let $Y$ be the set of poles of $\map f {\dfrac {2 z + 1} 2}$.
Then: 
:$\ds \sum_{n \mathop \in \Z \mathop \setminus Y} \paren {-1}^n \map f {\frac {2 n + 1} 2} = \sum_{z_0 \mathop \in X} \Res {\pi \sec \paren {\pi z} \map f z} {z_0}$
If $Y \cap \Z = \O$, this becomes: 
:$\ds \sum_{n \mathop = -\infty}^\infty \paren {-1}^n \map f {\frac {2 n + 1} 2} = \sum_{z_0 \mathop \in X} \Res {\pi \sec \paren {\pi z} \map f z} {z_0}$
== Proof == 
== Sources == 
* : $7.8$: Summation of Series
Category:Complex Analysis
Summation Formula for Polygonal Numbers

== Theorem ==
Let $\map P {k, n}$ be the $n$th $k$-gonal number.
Then:
:$\ds \map P {k, n} = \sum_{j \mathop = 1}^n \paren {\paren {k - 2} \paren {j - 1} + 1}$
== Proof ==
We have that:
$\map P {k, n} = \begin{cases}
0 & : n = 0 \\
\map P {k, n - 1} + \paren {k - 2} \paren {n - 1} + 1 & : n > 0
\end{cases}$
Proof by induction:
For all $n \in \N_{>0}$, let $\map \Pi n$ be the proposition:
:$\ds \map P {k, n} = \sum_{j \mathop = 1}^n \paren {\paren {k - 2} \paren {j - 1} + 1}$
=== Basis for the Induction ===
$\map \Pi 1$ is the statement that $\map P {k, 1} = 1$.
This follows directly from:
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map \Pi r$ is true, where $r \ge 1$, then it logically follows that $\map \Pi {r + 1}$ is true.
So this is our induction hypothesis:
:$\ds \map P {k, r} = \sum_{j \mathop = 1}^r \paren {\paren {k - 2} \paren {j - 1} + 1}$
Then we need to show:
:$\ds \map P {k, r + 1} = \sum_{j \mathop = 1}^{r + 1} \paren {\paren {k - 2} \paren {j - 1} + 1}$
=== Induction Step ===
This is our induction step:
So $\map \Pi r \implies \map \Pi {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N: \map P {k, n} = \sum_{j \mathop = 1}^n \paren {\paren {k - 2} \paren {j - 1} + 1}$
Category:Polygonal Numbers
Category:Proofs by Induction