avewright/proofwiki-math huggingface dataset:
Theorem: Digamma Function of One

:$\map \psi 1 = -\gamma$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^{n - 1} \map \psi {\frac k n}
      | r = -\paren {n - 1} \gamma - n \ln n
      | c = {{Corollary|Digamma Additive Formula}}
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 1}^{6 - 1} \map \psi {\frac k 6}
      | r = -\paren {6 - 1} \gamma - 6 \ln 6
      | c = 
}}
{{eqn | ll= \leadsto
      | n = 1
      | l = \map \psi {\frac 1 6} + \map \psi {\frac 2 6} + \map \psi {\frac 3 6} + \map \psi {\frac 4 6} + \map \psi {\frac 5 6}
      | r = -5 \gamma - 6 \ln 6
      | c = 
}}
{{eqn | n = 2
      | l = \map \psi {\frac 1 6} - \map \psi {\frac 5 6}
      | r = -\pi \map \cot {\frac \pi 6}
      | c = Digamma Reflection Formula
}}
{{eqn | ll= \leadsto
      | l = 2 \map \psi {\frac 1 6} + \map \psi {\frac 2 6} + \map \psi {\frac 3 6} + \map \psi {\frac 4 6}
      | r = -5 \gamma - 6 \ln 6 - \pi \map \cot {\frac \pi 6}
      | c = adding lines $1$ and $2$
}}
{{eqn | ll= \leadsto
      | l = 2 \map \psi {\frac 1 6}
      | r = -5 \gamma - 6 \ln 6 - \pi \map \cot {\frac \pi 6} - \map \psi {\frac 1 3} - \map \psi {\frac 1 2} - \map \psi {\frac 2 3}
      | c = rearranging
}}
{{eqn | r = -5 \gamma - 6 \ln 2 - 6 \ln 3 - \pi \times \sqrt 3
      | c = Sum of Logarithms, {{cot|30}},
}}
{{eqn | o = 
      | ro= -
      | r = \paren {-\gamma - \dfrac 3 2 \ln 3 - \dfrac \pi {2 \sqrt 3} }
      | c = Digamma Function of One Third,
}}
{{eqn | o = 
      | ro= -
      | r = \paren {-\gamma - 2 \ln 2}
      | c = Digamma Function of One Half,
}}
{{eqn | o = 
      | ro= -
      | r = \paren {-\gamma - \dfrac 3 2 \ln 3 + \dfrac \pi {2 \sqrt 3} }
      | c = and Digamma Function of Two Thirds
}}
{{eqn | r = -2 \gamma - 4 \ln 2  - 3 \ln 3 - \pi \sqrt 3
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = \map \psi {\frac 1 6}
      | r = -\gamma - 2 \ln 2 - \dfrac 3 2 \ln 3 - \dfrac {\pi \sqrt 3} 2
      | c = dividing by $2$
}}
{{end-eqn}}
{{qed}}
Theorem: Digamma Function of One

:$\map \psi 1 = -\gamma$

Proof:
{{begin-eqn}}
{{eqn | l = \map \psi {\frac 1 3}
      | r = -\gamma - \ln 6 - \frac \pi 2 \map \cot {\frac 1 3 \pi} + 2 \sum_{n \mathop = 1}^{\ceiling {3 / 2} - 1} \map \cos {\frac {2 \pi n} 3} \map \ln {\map \sin {\frac {\pi n} 3} }
      | c = Gauss's Digamma Theorem
}}
{{eqn | r = -\gamma - \ln 2 - \ln 3 - \frac \pi 2 \paren {\frac 1 {\sqrt 3} } + 2 \paren {-\frac 1 2} \map \ln {\frac {\sqrt 3} 2 }
      | c = {{cot|60}}, {{cos|120}}, {{sin|60}} and Sum of Logarithms
}}
{{eqn | r = -\gamma - \ln 2 - \ln 3 - \frac \pi 2 \paren {\frac 1 {\sqrt 3} } - \paren {\frac 1 2 \ln 3 - \ln 2}
      | c = Difference of Logarithms and Logarithm of Power
}}
{{eqn | r = -\gamma - \dfrac 3 2 \ln 3 - \dfrac {\pi} {2 \sqrt 3}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Digamma Function of One

:$\map \psi 1 = -\gamma$

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^{n - 1} \map \psi {\frac k n}
      | r = -\paren {n - 1} \gamma - n \ln n
      | c = {{Corollary|Digamma Additive Formula}}
}}
{{eqn | ll= \leadsto
      | l = \sum_{k \mathop = 1}^{3 - 1} \map \psi {\frac k 3}
      | r = -\paren {3 - 1} \gamma - 3 \ln 3
      | c = 
}}
{{eqn | ll= \leadsto
      | n = 1
      | l = \map \psi {\frac 1 3} + \map \psi {\frac 2 3}
      | r = -2 \gamma - 3 \ln 3
      | c = 
}}
{{eqn | n = 2
      | l = \map \psi {\frac 1 3} - \map \psi {\frac 2 3}
      | r = -\pi \map \cot {\frac \pi 3}
      | c = Digamma Reflection Formula
}}
{{eqn | ll= \leadsto
      | l = 2 \map \psi {\frac 1 3}
      | r = -2 \gamma - 3 \ln 3 - \pi \map \cot {\frac \pi 3}
      | c = adding lines $1$ and $2$
}}
{{eqn | ll= \leadsto
      | l = \map \psi {\frac 1 3}
      | r = -\gamma - \dfrac 3 2 \ln 3 - \dfrac \pi {2 \sqrt 3}
      | c = dividing by $2$ and {{cot|60}}
}}
{{end-eqn}}
{{qed}}
Theorem: Gamma Function of 3

:$\map \Gamma 3 = 2$

Proof:
{{begin-eqn}}
{{eqn | l = \map \Gamma 3
      | r = \map \Gamma {2 + 1}
      | c = 
}}
{{eqn | r = 2 \map \Gamma 2
      | c = Gamma Difference Equation
}}
{{eqn | r = 2 \times 1
      | c = Gamma Function of 2
}}
{{end-eqn}}
{{qed}}
Category:Examples of Gamma Function Values
gcvkn2iren6z3839y7j5v340mdy1ef0
Theorem: Gamma Function of 2

:$\map \Gamma 2 = 1$

Proof:
{{begin-eqn}}
{{eqn | l = \map \Gamma 2
      | r = \map \Gamma {1 + 1}
      | c = 
}}
{{eqn | r = 1 \map \Gamma 1
      | c = Gamma Difference Equation
}}
{{eqn | r = 1 \times 0!
      | c = Gamma Function Extends Factorial
}}
{{eqn | r = 1
      | c = {{Defof|Factorial}}: $0! = 1$
}}
{{end-eqn}}
{{qed}}
Category:Examples of Gamma Function Values
dqvzjkbdtecfye1bjvums2f9tunb48z
Theorem: Definite Integral from 0 to 1 of Zeta of 2 minus Dilogarithm of x over One minus x

:$\ds \int_{\to 0}^{\to 1} \dfrac {\paren {\map \zeta 2 - \map {\Li_2} x } } {1 - x} \rd x = 2 \map \zeta 3$
where:
:$\map {\Li_2} x$ is the Dilogarithm function of $x$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$
:$\map \zeta 3$ is Apéry's constant: the Riemann $\zeta$ function of $3$.

Proof:
With a view to expressing the primitive in the form:
{{begin-eqn}}
{{eqn | l = \int u \frac {\d v} {\d x} \rd x
      | r = u v - \int v \frac {\d u} {\d x} \rd x
      | c = Integration by Parts
}}
{{end-eqn}}
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \map \zeta 2 - \map {\Li_2} x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {\map \ln {1 - x} } x
      | c = {{Defof|Spence's Function}}: $\ds \map {\Li_2} z = -\int_0^z \frac {\map \Ln {1 - t} } t \rd t$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac 1 {\paren {1 - x} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = -\paren {\map \ln {1 - x} }
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int_{\to 0}^{\to 1} \dfrac {\paren {\map \zeta 2 - \map {\Li_2} x } } {1 - x} \rd x
      | r = \bigintlimits {\paren {\map \zeta 2 - \map {\Li_2} x} \paren {-\paren {\map \ln {1 - x} } } } 0 1 + \int_0^1 \frac {\paren {\map \ln {1 - x} }^2} x \rd x 
      | c = Integration by Parts
}}
{{eqn | r = 0 + \int_0^1 \frac {\paren {\map \ln {1 - x} }^2} x \rd x 
      | c = 
}}
{{eqn | r = \int_1^0 \frac {\paren {\map \ln u}^2} {1 - u} \paren {-\d u} 
      | c = $x \to \paren {1 - u}$ and $\rd x \to -\rd u$
}}
{{eqn | r = \int_0^1 \frac {\paren {-\map \ln u} \paren {-\map \ln u} } {1 - u} \rd u  
      | c = reversing limits of integration
}}
{{eqn | r = \int_0^1 \frac {\paren {\map \ln {\dfrac 1 u} }^{3 - 1} } {1 - u} \rd u 
      | c = Logarithm of Reciprocal
}}
{{eqn | r = \map \zeta 3 \map \Gamma 3
      | c = Integral Representation of Riemann Zeta Function in terms of Gamma Function: $\ds \map \zeta s \map \Gamma s = \int_0^1 \frac {\paren {\map \ln {\frac 1 u} }^{s - 1} } {1 - u} \rd u$
}}
{{eqn | r = 2 \map \zeta 3 
      | c = Gamma Function of $3$
}}
{{end-eqn}}
{{qed}}
Category:Spence's Function
Category:Apéry's Constant
Category:Definite Integrals involving Logarithm Function
482y640np2l2a4107uc3w0bpxu14h9o
Theorem: Sum of Digamma of n plus 2 over n plus 2 squared

{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty  \dfrac {\map \psi {n + 2} } {\paren {n + 2}^2 }
      | r = \dfrac {\map \psi 2} {2^2} + \dfrac {\map \psi 3} {3^2} + \dfrac {\map \psi 4} {4^2} + \dfrac {\map \psi 5} {5^2} + \cdots
      | c = 
}}
{{eqn | r = \gamma \cdot \paren {1 - \map \zeta 2} + \map \zeta 3
      | c = 
}}
{{end-eqn}}
where:
:$\map \psi n$ is the digamma function of $n$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$
:$\map \zeta 3$ is Apéry's constant: the Riemann $\zeta$ function of $3$.
Note the use of the $\cdot$ notation for product, or you wonder what the $\gamma$ function is.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty \dfrac {\map \psi {n + 2} } {\paren {n + 2}^2}
      | r = \sum_{n \mathop = 1}^\infty \dfrac {\map \psi n} {n^2} - \map \psi 1
      | c = $n \to n - 2$
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \dfrac {\paren {\map \psi {n + 1} - \dfrac 1 n} } {n^2} - \map \psi 1
      | c = Recurrence Relation for Digamma Function: $\map \psi n = \map \psi {n + 1} - \dfrac 1 n$
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \dfrac {\map \psi {n + 1} } {n^2} - \sum_{n \mathop = 1}^\infty \dfrac 1 {n^3} - \map \psi 1
      | c = Linear Combination of Convergent Series
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \dfrac {\map \psi {n + 1} } {n^2} - \map \zeta 3 + \gamma
      | c = Digamma Function of One and {{Defof|Apéry's Constant}}
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \dfrac {-\gamma + \ds \int_{\to 0}^{\to 1} \paren {\dfrac {1 - t^n} {1 - t} } \rd t} {n^2} - \map \zeta 3 + \gamma
      | c = {{Corollary|Reciprocal times Derivative of Gamma Function}}: $\ds \map \psi {n + 1} = -\gamma + \int_{\to 0}^{\to 1} \paren {\dfrac {1 - t^n} {1 - t} } \rd t$
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \dfrac {\ds \int_{\to 0}^{\to 1} \paren {\dfrac {1 - t^n} {1 - t} } \rd t} {n^2} - \gamma \cdot \sum_{n \mathop = 1}^\infty \dfrac 1 {n^2} - \map \zeta 3 +\gamma
      | c = Linear Combination of Convergent Series
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \dfrac {\ds \int_{\to 0}^{\to 1} \paren {\dfrac {1 - t^n} {1 - t} } \rd t} {n^2} + \gamma \cdot \paren {1 - \map \zeta 2} - \map \zeta 3
      | c = {{Defof|Riemann Zeta Function}}
}}
{{eqn | r = \int_{\to 0}^{\to 1} \dfrac {\paren {\ds \sum_{n \mathop = 1}^\infty \dfrac 1 {n^2} - \sum_{n \mathop = 1}^\infty \dfrac {t^n} {n^2} } } {1 - t} \rd t + \gamma \cdot \paren {1 - \map \zeta 2} - \map \zeta 3
      | c = Tonelli's Theorem
}}
{{eqn | r = \int_{\to 0}^{\to 1} \dfrac {\map \zeta 2 - \map {\Li_2} t } {1 - t} \rd t + \gamma \cdot \paren {1 - \map \zeta 2} - \map \zeta 3
      | c = {{Defof|Riemann Zeta Function}} and Power Series Expansion for Spence's Function
}}
{{eqn | r = 2 \map \zeta 3 + \gamma \cdot \paren {1 - \map \zeta 2} - \map \zeta 3
      | c = Definite Integral from $0$ to $1$ of $\dfrac {\map \zeta 2 - \map {\Li_2} x} {1 - x}$
}}
{{eqn | r = \gamma \cdot \paren {1 - \map \zeta 2} + \map \zeta 3
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Cumulative Distribution Function for Discrete Distribution

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.
Then the cumulative distribution function of $X$ is given by:
:$\map F x = \ds \sum_{x_i \mathop \le x} \map \Pr {X = x_i}$

Proof:
{{ProofWanted|Not sure whether this really ought to be a theorem or a definition}}
Theorem: Divisibility by 6

An integer $N$ expressed in decimal notation is divisible by $6$ {{iff}}
:$N$ is even
and:
:$N$ is divisible by $3$.

Proof:
We have that:
:$6 \divides N \iff 2 \divides N \land 3 \divides N$
The result follows by definition of even integer.
{{qed}}
Theorem: Definite Integral over Unit Square of Logarithm of x minus Logarithm of y over x minus y

:$\ds \int_0^1 \int_0^1 \frac {\ln x - \ln y} {x - y} \rd x \rd y = 2 \map \zeta 2$

Proof:
{{begin-eqn}}
{{eqn	| l = \int_0^1 \int_0^1 \dfrac {\ln x - \ln y} {x - y} \rd x \rd y
	    | r = \int_0^1 \int_0^1 \dfrac {\map \ln {\dfrac x y} } {x - y} \rd x \rd y
        | c = Difference of Logarithms
}}
{{eqn	| r = 2 \int_0^1 \int_0^x \dfrac {\map \ln {\dfrac x y} } {x - y} \rd x \rd y
	    | c = symmetry about the line $y = x$
}}
{{eqn	| r = 2 \int_0^1 \int_0^1 \dfrac {\map \ln {\dfrac x {x t} } } {x - x t} \rd x \paren {x \rd t}
        | c = $y \to x t$ and $\rd y \to x \rd t$
}}
{{eqn	| r = 2 \int_0^1 \int_0^1 \dfrac {\map \ln {\dfrac 1 t } } {\paren {1 - t} } \rd x \rd t
        | c = canceling the $x$
}}
{{eqn	| r = 2 \int_0^1 \dfrac {\map \ln {\dfrac 1 t } } {\paren {1 - t} } \rd t \bigintlimits x 0 1 
        | c = 
}}
{{eqn	| r = -2 \int_0^1 \dfrac {\map \ln t } {\paren {1 - t} } \rd t
        | c = Logarithm of Reciprocal
}}
{{eqn	| r = -2 \int_1^0 \dfrac {\map \ln {1 - x} } x \paren {-\rd x}
        | c = $\paren {1 - t} \to x$ and $-\rd t \to \rd x$
}}
{{eqn	| r = -2 \int_0^1 \dfrac {\map \ln {1 - x} } x \rd x
	    | c = reversing limits of integration
}}
{{eqn	| r = 2 \map {\Li_2} 1
	    | c = {{Defof|Spence's Function}}
}}
{{eqn	| r = 2 \sum_{n \mathop = 1}^\infty \frac {1^n} {n^2}
	    | c = Power Series Expansion for Spence's Function
}}
{{eqn	| r = 2 \map \zeta 2
	    | c = {{Defof|Riemann Zeta Function}}
}}
{{end-eqn}}
{{qed}}
Theorem: Power Series Expansion for Harmonic Numbers

The harmonic numbers have the power series expansion:
{{begin-eqn}}
{{eqn | l = \map H x
      | r = \sum_{k \mathop = 2}^{\infty} \paren {-1}^k \map \zeta k x^{k - 1}
      | c = 
}}
{{eqn | r = \map \zeta 2 x - \map \zeta 3 x^2 + \map \zeta 4 x^3 - \map \zeta 5 x^4 + \cdots
      | c = 
}}
{{end-eqn}}
where $\zeta$ denotes the Riemann zeta function.
This is valid for all $x \in \C$ such that $\size x < 1$.

Proof:
{{begin-eqn}}
{{eqn | l = \map H x
      | r = \gamma + \frac {\map {\Gamma'} {x + 1} } {\map \Gamma {x + 1} }
      | c = Extension of Harmonic Number to Non-Integer Argument
}}
{{eqn | r = \gamma - \gamma + \sum_{m \mathop = 1}^\infty \paren {\frac 1 m - \frac 1 {x + m} }
      | c = Reciprocal times Derivative of Gamma Function
}}
{{eqn | r = \sum_{m \mathop = 1}^\infty \paren {\frac 1 m - \frac 1 {x + m} }
      | c = 
}}
{{eqn | r = \sum_{m \mathop = 1}^\infty \paren {\frac x m - \frac x {x + m} } \dfrac 1 x
      | c = multiplying by $1$
}}
{{eqn | r = \sum_{m \mathop = 1}^\infty \paren {\frac x m + \paren {\frac m {x + m} - \frac {x + m} {x + m} } } \dfrac 1 x
      | c = 
}}
{{eqn | r = \sum_{m \mathop = 1}^\infty \paren {\frac x m + \paren {\frac m {x + m} - 1} } \dfrac 1 x
      | c = 
}}
{{eqn | r = \sum_{m \mathop = 1}^\infty \paren {\frac x m + \paren {\frac 1 {1 - \paren {-\dfrac x m } } } - 1} \dfrac 1 x
      | c = 
}}
{{eqn | r = \sum_{m \mathop = 1}^\infty \paren {\frac x m + \paren {\sum_{k \mathop = 0}^\infty \paren {-\dfrac x m}^k } - 1} \dfrac 1 x
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = \sum_{m \mathop = 1}^\infty \paren {\sum_{k \mathop = 2}^\infty \paren {-\dfrac x m}^k } \dfrac 1 x
      | c = reindexing the sum
}}
{{eqn | r = \sum_{k \mathop = 2}^{\infty} \paren {-1}^k \paren {\sum_{m \mathop = 1}^\infty \dfrac 1 {m^k} } x^{k - 1}
      | c = Fubini's Theorem
}}
{{eqn | r = \sum_{k \mathop = 2}^{\infty} \paren {-1}^k \map \zeta k x^{k - 1}
      | c = {{Defof|Riemann Zeta Function}}
}}
{{end-eqn}}
Our reliance upon the Sum of Infinite Geometric Sequence above where $m = 1$ implies $\size x < 1$, where $\size x$ denotes the complex modulus of $x$.
When $m = 1$ and $x = 1$, we have Grandi's Series, a well known divergent series.
{{qed}}
Theorem: Power Series Expansion for Logarithm of Gamma Function

The logarithm of the Gamma function has the power series expansion:
{{begin-eqn}}
{{eqn | l = \map \ln {\map \Gamma {x + 1} }
      | r = -\gamma x + \sum_{k \mathop = 2}^\infty \dfrac {\map \zeta k \paren {-x}^k} k
      | c = 
}}
{{eqn | r = -\gamma x + \dfrac {\map \zeta 2 x^2} 2 - \dfrac {\map \zeta 3 x^3} 3 + \dfrac {\map \zeta 4 x^4} 4 - \dfrac {\map \zeta 5 x^5} 5 + \cdots
      | c = 
}}
{{end-eqn}}
valid for all $x \in \C$ such that $\size x < 1$ and $x = 1$.

Proof:
{{begin-eqn}}
{{eqn | l = \gamma + \frac {\map {\Gamma'} {x + 1} } {\map \Gamma {x + 1} }
      | r = \map H x
      | c = Extension of Harmonic Number to Non-Integer Argument
}}
{{eqn | ll= \leadsto 
      | l = \frac {\map {\Gamma'} {x + 1} } {\map \Gamma {x + 1} }
      | r = -\gamma + \map H x
      | c = subtracting $\gamma$ from both sides
}}
{{eqn | ll= \leadsto 
      | l = \int_0^x \frac {\map {\Gamma'} {x + 1} } {\map \Gamma {x + 1} } \rd x
      | r = \int_0^x \paren {-\gamma + \map H x} \rd x
      | c = integrating both sides
}}
{{eqn | ll= \leadsto 
      | l = \map \ln {\map \Gamma {x + 1} }
      | r = \int_0^x \paren {-\gamma + \map H x} \rd x
      | c = {{Defof|Digamma Function}} and the Fundamental Theorem of Calculus
}}
{{eqn | r = \int_0^x \paren {-\gamma + \sum_{k \mathop = 2}^\infty \paren {-1}^k \map \zeta k x^{k - 1} } \rd x
      | c = Power Series Expansion for Harmonic Numbers
}}
{{eqn | r = \intlimits {-\gamma x + \sum_{k \mathop = 2}^\infty \dfrac {\map \zeta k \paren {-x}^k} k} 0 x 
      | c = Primitive of Power and Primitive of Constant
}}
{{eqn | r = -\gamma x + \sum_{k \mathop = 2}^\infty \dfrac {\map \zeta k \paren {-x}^k} k
      | c = 
}}
{{end-eqn}}
When $x = 1$, we have $\sequence {a_k = \dfrac {\map \zeta k} k}_{k \mathop \ge 0}$ a decreasing sequence of positive terms in $\R$ which converges with a limit of zero.
By the Alternating Series Test, this alternating series converges at $x = 1$.
{{qed}}
Theorem: Upper Bound for Abundancy Index

Let $n \in \N_{>0}$ be a non-zero natural number.
Let $\map h n$ be the abundancy index of $n$.
Then:
:$\ds \map h n \le \prod_{p \mathop \divides n} \dfrac p {p - 1}$
such that equality holds {{iff}} $n = 1$.

Proof:
First we note that $\map h 1 = 1$ demonstrating that equality holds if $n = 1$.
Let $n$ be expressed as its prime decomposition:
{{begin-eqn}}
{{eqn | l = n
      | r = \prod_{p_k \mathop \divides n} {p_k}^{a_k}
      | c = 
}}
{{eqn | r = {p_1}^{a_1} {p_2}^{a_2} \cdots {p_k}^{a_k}
      | c = 
}}
{{end-eqn}}
where $p_1, p_2, \ldots, p_k$ are distinct primes.
By definition of abundancy index:
:$\map h n = \dfrac {\map {\sigma_1} n} n$
where $\map {\sigma_1} n$ denotes the divisor sum function of $n$.
We have:
{{begin-eqn}}
{{eqn | l = \dfrac {\map {\sigma_1} n} n
      | r = \dfrac {\ds \map {\sigma_1} {\prod_{p_k \mathop \divides n} {p_k}^{a_k} } } {\ds \prod_{p_k \mathop \divides n} {p_k}^{a_k} }
      | c = 
}}
{{eqn | r = \dfrac {\ds \prod_{p_k \mathop \divides n} \map {\sigma_1} { {p_k}^{a_k} } } {\ds \prod_{p_k \mathop \divides n} {p_k}^{a_k} }
      | c = Divisor Sum Function is Multiplicative
}}
{{eqn | r = \prod_{p_k \mathop \divides n} \dfrac {\map {\sigma_1} { {p_k}^{a_k} } } { {p_k}^{a_k} }
      | c = 
}}
{{eqn | r = \prod_{p_k \mathop \divides n} \map h { {p_k}^{a_k} }
      | c = {{Defof|Abundancy Index}}
}}
{{end-eqn}}
It remains to be shown that:
:$\map h {p^a} < \dfrac p {p - 1}$
for prime $p$.
We have that:
{{begin-eqn}}
{{eqn | l = m
      | r = p^a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map h m
      | r = \dfrac {\map {\sigma_1} m} m
      | c = 
}}
{{eqn | r = \dfrac {p^{a + 1} - 1} {p - 1} / p^a
      | c = Divisor Sum of Power of Prime
}}
{{eqn | r = \dfrac {p^{a + 1} - 1} {p^a \paren {p - 1} }
      | c = 
}}
{{eqn | r = \dfrac {p^{a + 1} } {p^a \paren {p - 1} } - \dfrac 1 {p^a \paren {p - 1} }
      | c = 
}}
{{eqn | o = <
      | r = \dfrac p {p - 1}
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Abundancy
1b159yx0hgss4m5wz5bgsv4kkwaiozi
Theorem: Perfect Number has at least Two Distinct Prime Factors

Let $n \in \N$ be a perfect number.
Then $n$ has at least two distinct prime factors.

Proof:
{{AimForCont}} the contrary: that $n$ is a perfect number with exactly $1$ prime factor.
Hence let $n = p^k$ where $p$ is prime.
By definition of perfect number:
:$\map {\sigma_1} n = 2 n$
where $\map {\sigma_1} n$ denotes the divisor sum of $n$.
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} n
      | r = 2 n
      | c = {{Defof|Perfect Number}}
}}
{{eqn | l = \map {\sigma_1} {p^k}
      | r = \dfrac {p^{k + 1} - 1} {p - 1}
      | c = Divisor Sum of Power of Prime
}}
{{eqn | r = 1 + p + \cdots + p^k
      | c = Sum of Geometric Sequence
}}
{{eqn | r = 2 p^k
      | c = Definition of $n$
}}
{{eqn | ll= \leadsto
      | l = 1
      | r = 2 p^k - \paren {p + \cdots + p^k}
      | c = 
}}
{{eqn | r = p \paren {2 p^{k - 1} - \paren {1 + p + \cdots + p^{k - 1} } }
      | c = simplification
}}
{{eqn | ll= \leadsto
      | l = p
      | o = \divides
      | r = 1
      | c = 
}}
{{end-eqn}}
That is, $p$ is a divisor of $1$ which is a contradiction.
{{explain|To be rigorous, we need to specify explicitly what it contradicts: that $p > 1$ but $p \divides 1 \implies p \le 1$}}
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Perfect Number has at least Two Distinct Prime Factors

Let $n \in \N$ be a perfect number.
Then $n$ has at least two distinct prime factors.

Proof:
{{AimForCont}} the contrary: that $n$ is a perfect number with exactly $1$ prime factor.
Hence let $n = p^k$ where $p$ is prime.
By definition of perfect number:
:$\map {\sigma_1} n = 2 n$
where $\map {\sigma_1} n$ denotes the divisor sum of $n$.
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} n
      | r = 2 n
      | c = {{Defof|Perfect Number}}
}}
{{eqn | l = \map {\sigma_1} {p^k}
      | r = \dfrac {p^{k + 1} - 1} {p - 1}
      | c = Divisor Sum of Power of Prime
}}
{{eqn | r = 1 + p + \cdots + p^k
      | c = Sum of Geometric Sequence
}}
{{eqn | r = 2 p^k
      | c = Definition of $n$
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\map {\sigma_1} {p^k} } p^k
      | r = 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map h n
      | r = 2
      | c = where $\map h n$ denotes the abundancy index of $n$
}}
{{eqn | o = <
      | r = \dfrac p {p - 1}
      | c = Upper Bound for Abundancy Index
}}
{{eqn | o = <
      | r = \dfrac 3 2
      | c = as $\dfrac x {x - 1}$ is decreasing on $x > 1$
}}
{{eqn | ll= \leadsto
      | l = 2 
      | o = <
      | r = \dfrac 3 2
      | c = which contradicts the fact that $2 > \dfrac 3 2$
}}
{{end-eqn}}
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Left Zero Divisor of Commutative Ring is Right Zero Divisor

Let $\struct {R, +, \circ}$ be a commutative ring.
Let $x \in R$ be a left zero divisor of $R$.
Then $x$ is also a right zero divisor of $R$.

Proof:
Let $x \in R$ be a left zero divisor of $R$.
Let $0_R$ denote the zero of $R$.
Then:
{{begin-eqn}}
{{eqn | q = \exists y \in R^*
      | l = x \circ y
      | r = 0_R
      | c = {{Defof|Left Zero Divisor}}
}}
{{eqn | ll= \leadsto
      | l = y \circ x
      | r = 0_R
      | c = {{Defof|Commutative Ring}}
}}
{{end-eqn}}
Hence $x$ is a right zero divisor of $R$ by definition.
{{qed}}
Theorem: Right Zero Divisor of Commutative Ring is Left Zero Divisor

Let $\struct {R, +, \circ}$ be a commutative ring.
Let $x \in R$ be a right zero divisor of $R$.
Then $x$ is also a left zero divisor of $R$.

Proof:
Let $x \in R$ be a right zero divisor of $R$.
Let $0_R$ denote the zero of $R$.
Then:
{{begin-eqn}}
{{eqn | q = \exists y \in R^*
      | l = x \circ y
      | r = 0_R
      | c = {{Defof|Left Zero Divisor}}
}}
{{eqn | ll= \leadsto
      | l = y \circ x
      | r = 0_R
      | c = {{Defof|Commutative Ring}}
}}
{{end-eqn}}
Hence $x$ is a left zero divisor of $R$ by definition.
{{qed}}
Theorem: Odd Perfect Number has at least Three Distinct Prime Factors

Let $n \in \N$ be an odd perfect number.
Then $n$ has at least $3$ distinct prime factors.

Proof:
{{AimForCont}} the contrary: that $n$ is an odd perfect number with at most $2$ distinct prime factors.
By Perfect Number has at least Two Distinct Prime Factors, $n$ must have exactly $2$ distinct prime factors.
Hence let $n = p^a q^b$ where $p$ and $q$ are distinct primes and where $a, b \in \N_{>0}$.
As $n$ is odd, both $p$ and $q$ are also odd.
By definition of perfect number:
:$\map {\sigma_1} n = 2 n$
where $\map {\sigma_1} n$ denotes the divisor sum of $n$.
Hence:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} n
      | r = 2 n
      | c = {{Defof|Perfect Number}}
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\map {\sigma_1} n} n
      | r = 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map h n
      | r = 2
      | c = where $\map h n$ denotes the abundancy index of $n$
}}
{{eqn | o = <
      | r = \dfrac p {p - 1} \dfrac q {q - 1}
      | c = Upper Bound for Abundancy Index
}}
{{eqn | o = <
      | r = \dfrac 3 2 \dfrac 5 4
      | c = as $\dfrac x {x - 1}$ is decreasing on $x > 1$, and $p$ and $q$ are distinct odd primes
}}
{{eqn | ll= \leadsto
      | l = 2 
      | o = <
      | r = \dfrac {15} 8
      | c = contradicting $2 > \dfrac {15} 8$
}}
{{end-eqn}}
Hence the result by Proof by Contradiction.
{{qed}}