avewright/proofwiki-math huggingface dataset:
Theorem: Equiangular Triangle is Equilateral

Let $\triangle ABC$ be equiangular.
Then $\triangle ABC$ is an equilateral triangle.

Proof:
Let $\triangle ABC$ be equiangular.
By definition of equiangular polygon, any two of the internal angles of $\triangle ABC$ are equal.
{{WLOG}}, let $\angle ABC = \angle ACB$.
Then by Triangle with Two Equal Angles is Isosceles, $AB = AC$.
As the choice of equal angles was arbitrary, it follows that any two sides of $\triangle ABC$ are equal.
Hence all $3$ sides of $\triangle ABC$ are equal.
Hence the result by definition of equilateral triangle.
{{qed}}
Theorem: Characterization of Strictly Increasing Mapping on Woset

Let $J$ and $E$ be well-ordered sets.
Let $h: J \to E$ be a mapping.
Let $S_\alpha$ denote an initial segment determined by $\alpha$.
{{TFAE}}
:$(1): \quad h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$
:$(2): \quad \forall \alpha \in J: \map h \alpha = \map \min {E \setminus h \sqbrk {S_\alpha} }$, and $h \sqbrk {S_\alpha} = S_{\map h \alpha}$
where:
:$h \sqbrk {S_\alpha}$ denotes the image of $S_\alpha$ under $h$
:$\min$ denotes the smallest element of the set.

Proof:
=== $(1)$ implies $(2)$ ===
Suppose $h$ satisfies:
:$h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$
Then for any $x,y \in J$:
{{begin-eqn}}
{{eqn | l = x
      | o = \prec
      | r = y
}}
{{eqn | ll= \leadsto
      | l = \map h x
      | o = \prec
      | r = \map h y
      | c = {{Defof|Strictly Increasing Mapping}}
}}
{{eqn | l = h \sqbrk {S_y}
      | r = \set {\map h x \in E: \exists x \in J: \map h x \prec \map h y}
}}
{{eqn | r = S_{\map h y}
}}
{{eqn | l = \map \min {E \setminus h \sqbrk {S_y} }
      | r = \map \min {E \setminus S_{\map h y} }
}}
{{eqn | r = \map h y
      | c = {{Defof|Smallest Element}} and {{Defof|Initial Segment}}
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: No Order Isomophism Between Distinct Initial Segments of Woset

Let $E$ be a well-ordered set.
Let $S_\alpha, S_\beta$ be initial segments of $E$ that are order isomorphic.
Then $S_\alpha = S_\beta$.

Proof:
{{AimForCont}} $S_\alpha \ne S_\beta$.
Then $\alpha \ne \beta$.
By the trichotomy law, $\alpha \prec \beta$ or $\beta \prec \alpha$.
{{WLOG}} assume $\alpha \prec \beta$.
Then $S_\alpha \subsetneqq S_\beta$.
That is, $S_\alpha$ is an initial segment of $S_\beta$.
We have {{hypothesis}} that $S_\alpha$ and $S_\beta$ are order isomorphic.
Thus there is an order isomorphism between $S_\beta$ and an initial segment of $S_\beta$.
This contradicts Well-Ordered Class is not Isomorphic to Initial Segment.
Thus $S_\alpha = S_\beta$.
{{qed}}
Theorem: Perpendicular is Shortest Straight Line from Point to Straight Line

Let $AB$ be a straight line.
Let $C$ be a point which is not on $AB$.
Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$.
Then the length of $CD$ is less than the length of all other line segments that can be drawn from $C$ to $AB$.

Proof:
Let $E$ on $AB$ such that $E$ is different from $D$.
Then $CDE$ forms a right triangle where $CE$ is the hypotenuse.
By Pythagoras's Theorem:
:$CD^2 + DE^2 = CE^2$
and so $CD < CE$.
{{qed}}
Theorem: Straight Lines which make Equal Angles with Perpendicular to Straight Line are Equal

Let $AB$ be a straight line.
Let $C$ be a point which is not on $AB$.
Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$.
Let $E, F$ be points on $AB$ such that $\angle DCE = \angle DCF$.
Then $CE = CF$.

Proof:
$\triangle CDE$ and $\triangle CDF$ are right triangle where $CE$ and $CF$ are the hypotenuses.
We have:
:$\angle CDE = \angle CDF$ as both are right angles.
:$\angle DCE = \angle DCF$ by hypothesis.
:$CD$ is common.
Thus by Triangle Angle-Side-Angle Congruence, $\triangle CDE$ and $\triangle CDF$ are congruent.
Hence the result.
{{qed}}
Theorem: Distance between Two Parallel Straight Lines is Everywhere the Same

Let $AB$ and $CD$ be parallel straight lines.
Let perpendiculars $EF$ and $GH$ be drawn from $AB$ to $CD$, where $E, G$ are on $AB$ and $F, H$ are on $CD$.
Then $EF = GH$.
That is, the distance between  $AB$ and $CD$ is the same everywhere along their length.

Proof:
$\angle EFH, \angle FEG, \angle EGH, \angle FHG$ are all right angles.
Then $EF$ and $GH$ are parallel.
Thus $\Box EFHG$ is by definition a parallelogram.
From Opposite Sides and Angles of Parallelogram are Equal it follows that $EF = GH$.
{{qed}}
Theorem: Diagonals of Rhombus Bisect Each Other at Right Angles

Let $ABCD$ be a rhombus.
The diagonals $AC$ and $BD$ of $ABCD$ bisect each other at right angles.

Proof:
By the definition of a rhombus, $AB = AD = BC = DC$.
{{WLOG}}, consider the diagonal $BD$.
Thus:
: $\triangle ABD$ is an isosceles triangle whose apex is $A$ and whose base is $BD$.
By Diagonals of Rhombus Bisect Angles, $AC$ bisects $\angle BAD$.
From Bisector of Apex of Isosceles Triangle also Bisects Base, $AC$ bisects $BD$.
From Bisector of Apex of Isosceles Triangle is Perpendicular to Base, $AC$ bisects $BD$ at right angles.
Hence the result.
{{qed}}
Theorem: Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel

Let $ABCD$ be a quadrilateral.
Then:
:$ABCD$ is a parallelogram
{{iff}}:
:$AB = CD$ and $AB \parallel CD$
where $AB \parallel CD$ denotes that $AB$ is parallel to $CD$.

Proof:
=== Sufficient Condition ===
Let $ABCD$ be a parallelogram.
Then $AB \parallel CD$ by definition.
From Opposite Sides and Angles of Parallelogram are Equal it follows that $AB = CD$.
{{qed|lemma}}
Theorem: Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel

Let $ABCD$ be a quadrilateral.
Then:
:$ABCD$ is a parallelogram
{{iff}}:
:either $AB = CD$ and $AD = BC$
:or $AB \parallel CD$ and $AD \parallel BC$
where $AB \parallel CD$ denotes that $AB$ is parallel to $CD$.

Proof:
=== Sufficient Condition ===
Let $ABCD$ be a parallelogram.
Then by definition:
:$AB \parallel CD$ and $AD \parallel BC$
and from Opposite Sides and Angles of Parallelogram are Equal:
:$AB = CD$ and $AD = BC$
From Conjunction implies Disjunction, it follows that:
:either $AB = CD$ and $AD = BC$
:or $AB \parallel CD$ and $AD \parallel BC$.
{{qed|lemma}}
Theorem: Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal

Let $ABCD$ be a quadrilateral.
Then:
:$ABCD$ is a parallelogram
{{iff}}:
:$\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$.

Proof:
=== Sufficient Condition ===
Let $ABCD$ be a parallelogram.
Then by Opposite Sides and Angles of Parallelogram are Equal:
:$\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$.
{{qed|lemma}}
Theorem: Quadrilateral is Parallelogram iff Diagonals Bisect each other

Let $ABCD$ be a quadrilateral.
Then:
: $ABCD$ is a parallelogram
{{iff}}:
:both:
:: $AD$ is a bisector of $BC$
:and:
:: $BC$ is a bisector of $AD$.

Proof:
=== Sufficient Condition ===
Let $ABCD$ be a parallelogram.
Then by Diameters of Parallelogram Bisect each other:
: $AD$ is a bisector of $BC$
and
: $BC$ is a bisector of $AD$.
{{qed|lemma}}
Theorem: Parallelograms are Congruent if Two Adjacent Sides and Included Angle are respectively Equal

Let $ABCD$ and $EFGH$ be parallelograms.
Then $ABCD$ and $EFGH$ are congruent if:
: $2$ adjacent sides of $ABCD$ are equal to $2$ corresponding adjacent sides of $EFGH$
: the angle between those $2$ adjacent sides on both $ABCD$ and $EFGH$ are equal.

Proof:
{{WLOG}} let the $2$ adjacent sides of $ABCD$ be $AB$ and $BC$.
Let the $2$ corresponding adjacent sides of $EFGH$ be $EF$ and $FG$ such that $AB = EF$ and $BC = FG$.
Hence let $\angle ABC = \angle EFG$.
From Triangle Side-Angle-Side Congruence:
:$\triangle ABC = \triangle EFG$
From Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal:
:$\angle ABC = \angle ADC$
:$\angle EFG = \angle EHG$
From Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel:
:$AB = CD$
and:
:$EF = GF$
Thus again by Triangle Side-Angle-Side Congruence:
:$\triangle ADC = \triangle EHG$
So:
:$\triangle ABC + \triangle ADC = \triangle EFG + \triangle EHG$
and the result follows.
{{qed}}
Theorem: Rectangles with Equal Bases and Equal Altitudes are Congruent

Let $ABCD$ and $EFGH$ be rectangles.
Then $ABCD$ and $EFGH$ are congruent if:
: the base of $ABCD$ equals the base of $EFGH$
: the altitude of $ABCD$ equals the altitude of $EFGH$.

Proof:
A rectangle is a parallelogram whose vertices are right angles.
Thus the altitudes of $ABCD$ and of $EFGH$ coincide with the sides of $ABCD$ and $EFGH$ which are adjacent to the bases.
The result then follows from Parallelograms are Congruent if Two Adjacent Sides and Included Angle are respectively Equal.
{{qed}}
Theorem: Parallel Lines which intercept Equal Segments on Transversals

Let $3$ or more parallel lines intersect equal line segments on one transversal.
Then those same $3$ or more parallel lines intersect equal line segments on every transversal.

Proof:
{{ProofWanted|Someone who understands what this is really trying to say?}}
Theorem: Line Parallel to Side of Triangle which Bisects One Side also Bisects Other Side

Let $ABC$ be a triangle.
Let $DE$ be a straight line parallel to $BC$.
Let $DE$ bisect $AB$.
Then $DE$ also bisects $AC$.
That is, $DE$ is a midline of $\triangle ABC$.
400px

Proof:
This is a direct application of the Parallel Transversal Theorem.
{{qed}}
Theorem: Midline Theorem

The midline of a triangle is parallel to the third side of that triangle and half its length.

Proof:
400px
Let $\triangle ABC$ be a triangle.
Let $DE$ be the midline of $\triangle ABC$ through $AB$ and $AC$.
Extend $DE$ to $DF$ so $DE = EF$.
As $E$ is the midpoint of $AC$, the diagonals of the quadrilateral $ADCF$ bisect each other.
From Quadrilateral with Bisecting Diagonals is Parallelogram, $ADCF$ is a parallelogram.
By definition of a parallelogram, $AB \parallel CF$.
From Opposite Sides and Angles of Parallelogram are Equal, $AD = CF$.
But $AD = DB$ as $D$ is the midpoint of $AB$.
So $DB = CF$ and $DB \parallel CF$.
From Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel, $BCFD$ is a parallelogram.
Thus also by Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel $DF = BC$ and $DF \parallel BC$.
As $DE = EF$, $DE$ is the midpoint of $DF$ and so $DE = \dfrac 1 2 DF$.
Thus $DE = \dfrac 1 2 BC$ and $DE \parallel BC$.
Hence the result.
{{qed}}
Theorem: Midline and Median of Triangle Bisect Each Other

Let $\triangle ABC$ be a triangle.
Let $DE$ be the midline of $\triangle ABC$ which bisects $AB$ and $AC$.
Let $AF$ be the median of $ABC$ which bisects $BC$.
Then $AF$ and $DE$ bisect each other.

Proof:
400px
Construct the midlines $DF$ and $EF$.
Then by the Midline Theorem $DF \parallel AE$ and $EF \parallel AD$.
Thus by Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel, $\Box ADFE$ is a parallelogram.
By construction, $AF$ and $DE$ are the diagonals of $\Box ADFE$.
The result follows from Diameters of Parallelogram Bisect each other.
{{qed}}
Theorem: Median to Hypotenuse of Right Triangle equals Half Hypotenuse

Let $\triangle ABC$ be a right triangle such that $BC$ is the hypotenuse.
Let $AD$ be the median to $BC$.
Then $AD$ is half of $BC$.

Proof:
400px
Construct $BE$ and $CE$ parallel to $AC$ and $AB$ respectively.
Then by definition $ABEC$ is a parallelogram.
By construction, $BC$ is a diagonal of $ABEC$ such that $AD$ is a bisector of it.
Thus by Quadrilateral is Parallelogram iff Diagonals Bisect each other, $AE$ is also a bisector of $ABEC$.
As $\angle BAC$ is a right angle it follows that $\angle BEC$ is also a right angle.
Thus by definition $ABEC$ is a rectangle.
From Diagonals of Rectangle are Equal:
:$AE = BC$
From above, $D$ is the midpoint of both $AE$ and $BC$.
Thus $AD = BD$ and hence the result.
{{qed}}
Theorem: Diagonals of Rectangle are Equal

The diagonals of a rectangle are equal.

Proof:
:400px
Let $ABCD$ be a rectangle.
The diagonals of $ABCD$ are $AC$ and $BD$.
Then $\angle ADC = \angle DAB$ as both are right angles by definition of rectangle.
By Rectangle is Parallelogram, $ABCD$ is also a type of parallelogram.
Thus by Opposite Sides and Angles of Parallelogram are Equal $AB = DC$.
Thus we have:
:$AB = DC$
:$\angle ADC = \angle DAB$
:$AD$ is common to both $\triangle ADC$ and $\triangle DAB$
and so by Triangle Side-Angle-Side Congruence:
:$\triangle ADC = \triangle DAB$
Thus:
:$AC = BD$
{{qed}}
Category:Rectangles
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Theorem: Summation to n of Square of kth Harmonic Number

:$\ds \sum_{k \mathop = 1}^n {H_k}^2 = \paren {n + 1} {H_n}^2 - \paren {2 n + 1} H_n + 2 n$
where $H_k$ denotes the $k$th harmonic number.

Proof:
Consider:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}
      | r = \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {k + 1} - 1} {k + 1}
      | c = factorizing
}}
{{eqn | r = \sum_{k \mathop = 1}^{n - 1} 1 - \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1}
      | c = Sum of Summations equals Summation of Sum
}}
{{eqn | r = n - 1 - \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1}
      | c = Summation of Unity over Elements
}}
{{eqn | r = n - 1 - \paren {H_n - 1}
      | c = Harmonic Number and compensate for $\dfrac 1 1$ term
}}
{{eqn | n = 1
      | r = n - H_n
      | c = simplification
}}
{{end-eqn}}
{{ExtractTheorem|The above result is worth publishing on its own page as a result in its own right. I'm surprised we don't already have it.}}
Using Summation by Parts: 
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n {H_k}^2
      | r = H_n \sum_{k \mathop = 1}^n H_k - \sum_{k \mathop = 1}^{n - 1} \paren {\paren {\sum_{i \mathop = 1}^k {H_i} } \paren {H_{k + 1} - H_k} }
      | c =
}}
{{eqn | r = H_n \paren {\paren {n + 1} H_n - n} - \sum_{k \mathop = 1}^{n - 1} \paren {\paren {k + 1} H_k - k} \paren {H_{k + 1} - H_k}
      | c = Sum of Sequence of Harmonic Numbers twice
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {k + 1} H_k - k} {k + 1}
      | c = $H_{k + 1} - H_k = \dfrac 1 {k+1}$
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \sum_{k \mathop = 1}^{n - 1} H_k + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1} 
      | c = Sum of Summations equals Summation of Sum
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \paren { \sum_{k \mathop = 1}^n H_k - H_n} + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1} 
      | c = increase range of summation and compensate
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \paren {\paren {n + 1} H_n - n - H_n } + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1} 
      | c = Sum of Sequence of Harmonic Numbers
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \paren {n + 1} H_n + n + H_n + n - H_n
      | c = $(1)$
}}
{{eqn | r = \paren {n + 1} {H_n}^2 - \paren {2 n + 1} H_n + 2 n
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Summation of Odd Reciprocals in terms of Harmonic Numbers

:$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = H_{2 n} - \dfrac {H_n} 2$
where $H_n$ denotes the $n$th harmonic number.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1}
      | r = \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} + \sum_{k \mathop = 1}^n \dfrac 1 {2 k} - \sum_{k \mathop = 1}^n \dfrac 1 {2 k}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^{2 n} \dfrac 1 k - \dfrac 1 2 \sum_{k \mathop = 1}^n \dfrac 1 k
      | c = 
}}
{{eqn | r = H_{2 n} - \dfrac {H_n} 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Numerator of p-1th Harmonic Number is Divisible by Prime p

Let $p$ be an odd prime.
Consider the harmonic number $H_{p - 1}$ expressed in canonical form.
The numerator of $H_{p - 1}$ is divisible by $p$.

Proof:
Add the terms of $H_{p - 1}$ using the definition of rational addition to obtain $\dfrac m n$.
Do not cancel common prime factors from $m$ and $n$.
It is seen that $n = \paren {p - 1}!$
Hence $p$ is not a divisor of $n$.
The numerator $m$ is seen to be:
:$m = \dfrac {\paren {p - 1}!} 1 + \dfrac {\paren {p - 1}!} 2 + \cdots + \dfrac {\paren {p - 1}!} {p - 1}$
Thus it is sufficient to show that $m$ is a multiple of $p$.
Each term in this sum is an integer of the form $\dfrac {\paren {p - 1}!} k$.
For each $k \in \set {1, 2, \ldots, p - 1}$, define $k'= - \dfrac {\paren {p - 1}!} k \bmod p$.
By Wilson's Theorem:
:$k k' \equiv -\paren {p - 1}! \equiv 1 \pmod p$
Therefore:
:$k' \equiv k^{-1} \pmod p$
{{Explain|$k^{-1}$}}
From {{Corollary|Reduced Residue System under Multiplication forms Abelian Group}}:
:$\struct {\Z'_p, \times}$ is an abelian group.
Since Inverse in Group is Unique, the set:
:$\set {1', 2', \ldots, \paren {p - 1}'}$
is merely the set:
:$\set {1, 2, \ldots, p - 1}$
in a different order.
Thus 
{{begin-eqn}}
{{eqn | l = m
      | r = \dfrac {\paren {p - 1}!} 1 + \dfrac {\paren {p - 1}!} 2 + \cdots + \dfrac {\paren {p - 1}!} {p - 1}
}}
{{eqn | o = \equiv
      | r = 1 + 2 + \cdots + p - 1 
      | rr= \pmod p
}}
{{eqn | o = \equiv
      | r = p \cdot \frac {p - 1} 2
      | rr= \pmod p
      | c = Closed Form for Triangular Numbers
}}
{{eqn | o = \equiv
      | r = 0
      | rr= \pmod p
      | c = Congruent to Zero iff Modulo is Divisor
}}
{{end-eqn}}
{{qed}}
Theorem: Numerator of p-1th Harmonic Number is Divisible by Prime p

Let $p$ be an odd prime.
Consider the harmonic number $H_{p - 1}$ expressed in canonical form.
The numerator of $H_{p - 1}$ is divisible by $p$.

Proof:
Note that for any integer $x$:
{{begin-eqn}}
{{eqn | l = x^p - x
      | o = \equiv
      | r = 0
      | rr= \pmod p
      | c = Corollary $1$ to Fermat's Little Theorem
}}
{{eqn | o = \equiv
      | r = x^{\overline p}
      | rr= \pmod p
      | c = Divisibility of Product of Consecutive Integers
}}
{{eqn | o = \equiv
      | r = \sum_k {p \brack k} x^k
      | rr= \pmod p
      | c = Sum over k of Unsigned Stirling Numbers of First Kind by x^k
}}
{{end-eqn}}
By comparing coefficients:
:$\ds {p \brack p} \equiv 1 \pmod p$
:$\ds {p \brack 1} \equiv -1 \pmod p$
:$\ds {p \brack k} \equiv 0 \pmod p$ for $k \ne 1, p$
or in a more compact form:
:$\ds {p \brack k} \equiv \delta_{k p} - \delta _{k 1} \pmod p$
where:
:$\ds {p \brack k}$ denotes an unsigned Stirling number of the first kind
:$\delta$ is the Kronecker delta.
From Harmonic Number as Unsigned Stirling Number of First Kind over Factorial:
:$\ds H_{p - 1} = \frac {p \brack 2} {\paren {p - 1}!}$
From the above we have:
:$\ds p \divides {p \brack 2}$
By Prime iff Coprime to all Smaller Positive Integers we also have:
:$p \nmid \paren {p - 1}!$
Hence the numerator of $H_{p - 1}$, when expressed in canonical form, is divisible by $p$.
{{qed}}
Theorem: Numerator of p-1th Harmonic Number is Divisible by p^2 for Prime Greater than 3

Let $p$ be a prime number such that $p > 3$.
Consider the harmonic number $H_{p - 1}$ expressed in canonical form.
The numerator of $H_{p - 1}$ is divisible by $p^2$.

Proof:
{{ProofWanted|research and invoke Wolstenholme's Theorem}}
Theorem: Highest Power of 2 Dividing Numerator of Sum of Odd Reciprocals

Let:
: $S = \dfrac p q = \ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1}$
where $\dfrac p q$ is the canonical form of $S$.
Let $n = 2^k m$ where $m$ is odd.
Then the largest power of $2$ that divides $p$ is $2^{2 k}$.

Proof:
We have that:
:$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = \sum_{i \mathop = 0}^{M - 1} \paren {\dfrac 1 {i \times 2^{r + 1} + 1} + \dfrac 1 {i \times 2^{r + 1} + 3} + \cdots + \dfrac 1 {\paren {i + 1} \times 2^{r + 1} - 1} }$
where $k = 2^r M$ where $M$ is odd.
Collect the $r$ terms in the parenthesis on the {{RHS}} of the $i$th term under a single common denominator $P_i$.
Then the resulting numerators will each consist of $2^r$ terms, each of the form $\dfrac {P_i} {R_i}$, where $R_i$ is a distinct odd residue of $2^{r + 1}$.
These $\dfrac {P_i} {R_i}$ must themselves also be the distinct odd residue of $2^{r + 1}$, in some order.
The odd residues of $2^{r + 1}$ are:
:$1, 3, 5, \dots, 2^{r + 1} - 1$
Their sum is:
{{begin-eqn}}
{{eqn | o =
      | r = 1 + 3 + \cdots + \paren {2^{r + 1} - 1}
}}
{{eqn | r = \frac {\paren {1 + 2^{r + 1} - 1} 2^r} 2
      | c = Sum of Arithmetic Sequence
}}
{{eqn | r = 2^{2 r}
}}
{{end-eqn}}
Thus each of the $M$ numerators is of the form $2^{2 r} M_i$, where $M_i$ is odd.
Thus the numerator of $S$ is in the form $2^{2 r} Q$ where $Q$ is the sum of an odd number of odd terms.
Therefore $Q$ is itself odd.
Hence, if $2^r$ is the largest power of $2$ which divides $k$, then $2^{2 r}$ is the largest power of $2$ dividing the numerator of $S$.
{{qed}}
Theorem: Summation of Power Series by Harmonic Sequence

Consider the power series:
:$\map f x = \ds \sum_{k \mathop \ge 0} a_k x^k$
Let $\map f x$ converge for $x = x_0$.
Then:
:$\ds \sum_{k \mathop \ge 0} a_k {x_0}^k H_k = \int_0^1 \dfrac {\map f {x_0} - \map f {x_0 y} } {1 - y} \rd y$
where $H_n$ denotes the $n$th harmonic number.

Proof:
{{begin-eqn}}
{{eqn | l = \int_0^1 \dfrac {\map f {x_0} - \map f {x_0 y} } {1 - y} \rd y
      | r = \int_0^1 \dfrac {\sum_{k \mathop \ge 0} a_k {x_0}^k - \sum_{k \mathop \ge 0} a_k {x_0}^k y^k} {1 - y} \rd y
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k \int_0^1 \dfrac {1 - y^k} {1 - y} \rd y
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k \int_0^1 \paren {1 + y + y^2 + \cdots + y^{k - 1} } \rd y
      | c = Sum of Geometric Sequence
}}
{{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k \intlimits {y + \dfrac {y^2} 2 + \dfrac {y^3} 3 + \cdots + \dfrac {y^k} k} 0 1
      | c = Primitive of Power
}}
{{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k \paren {\paren {1 + \dfrac {1^2} 2 + \dfrac {1^3} 3 + \cdots + \dfrac {1^k} k} - \paren {0 + \dfrac {0^2} 2 + \dfrac {0^3} 3 + \cdots + \dfrac {0^k} k} }
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k H_k
      | c = {{Defof|Harmonic Numbers}}
}}
{{end-eqn}}
{{qed}}
Theorem: Summation over k to n of Harmonic Numbers over n+1-k

:$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} {n + 1 - k} = {H_{n + 1} }^2 - H_{n + 1}^{\paren 2}$
where:
:$H_n$ denotes the $n$th harmonic number
:$H_n^{\paren 2}$ denotes a general harmonic number.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} {n + 1 - k}
      | r = \sum_{k \mathop = 1}^n \dfrac {H_{n + 1 - k} } k
      | c = Permutation of Indices of Summation
}}
{{eqn | r = \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^k \dfrac 1 {j \paren {n + 1 - k} }
      | c = {{Defof|Harmonic Number}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \sum_{j \mathop = k}^n \dfrac 1 {\paren {n + 1 - j} k}
      | c = {{Defof|Harmonic Number}}
}}
{{end-eqn}}
{{ProofWanted|all at sea}}
Theorem: Extension of Harmonic Number to Non-Integer Argument

Let $\map H x$ be the real function defined as:
:$\map H x = \gamma + \dfrac {\map {\Gamma'} {x + 1} } {\map \Gamma {x + 1} }$
where:
:$\map \Gamma x$ denotes the Gamma function
:$\map {\Gamma'} x$ denotes the derivative of the Gamma function
:$\gamma$ denotes the Euler-Mascheroni constant.
Then $H$ is an extension of the mapping $H: \N \to \Q$ defined as:
:$\forall n \in \N: \map H n = H_n$
where $H_n$ denotes the $n$th harmonic number.

Proof:
For $n \in \N$:
{{begin-eqn}}
{{eqn | l = \map H n
      | r = \gamma + \frac {\map {\Gamma'} {n + 1} } {\map \Gamma {n + 1} }
}}
{{eqn | r = \gamma - \gamma + \sum_{m \mathop = 1}^\infty \paren {\frac 1 m - \frac 1 {n + m} }
      | c = Reciprocal times Derivative of Gamma Function
}}
{{eqn | r = \lim_{k \mathop \to \infty} \sum_{m \mathop = 1}^k \paren {\frac 1 m - \frac 1 {n + m} }
}}
{{eqn | r = \lim_{k \mathop \to \infty} \paren {\sum_{m \mathop = 1}^k \paren {\frac 1 m} - \sum_{m \mathop = 1}^k \paren {\frac 1 {n + m} } }
      | c = Linear Combination of Convergent Series
}}
{{eqn | r = \lim_{k \mathop \to \infty} \paren {\sum_{m \mathop = 1}^k \paren {\frac 1 m} - \sum_{m \mathop = n + 1}^{k + n} \paren {\frac 1 m} }
      | c = reindexing sum
}}
{{eqn | r = \lim_{k \mathop \to \infty} \paren {\sum_{m \mathop = 1}^n \paren {\frac 1 m} - \sum_{m \mathop = k + 1}^{k + n} \paren {\frac 1 m} }
}}
{{eqn | r = \sum_{m \mathop = 1}^n \paren {\frac 1 m} - \lim_{k \mathop \to \infty} \sum_{m \mathop = k + 1}^{k + n} \paren {\frac 1 m}
      | c = this limit will vanish since it is positive but less than $\dfrac n {k + 1}$
}}
{{eqn | r = H_n - 0
      | c = {{Defof|Harmonic Number}}
}}
{{eqn | r = H_n
}}
{{end-eqn}}
{{qed}}