avewright/proofwiki-math huggingface dataset:
Theorem: Continuous Real Function is Darboux Integrable

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then $f$ is Darboux integrable on $\closedint a b$.

Proof:
We have the result $f$ is bounded by Continuous Real Function is Bounded.
By Condition for Darboux Integrability, it suffices to show that for all $\epsilon > 0$, there exists a subdivision $P$ of $\closedint a b$ such that:
:$\map U P - \map L P < \epsilon$
where $\map U P$ and $\map L P$ denote the upper Darboux sum and lower Darboux sum of $f$ on $\closedint a b$ belonging to the subdivision $P$.
Let $\epsilon > 0$.
We have the result Continuous Function on Closed Real Interval is Uniformly Continuous.
By the definition of uniform continuity, there exists a $\delta > 0$ such that if $x, y \in \closedint a b$ are such that $\size {x - y} < \delta$, then:
:$\size {\map f x - \map f y} < \dfrac \epsilon {b - a}$
Let $P = \set {x_0, x_1, x_2, \ldots, x_n}$ be a subdivision of $\closedint a b$ such that:
:$\ds \max_{1 \mathop \le k \mathop \le n} \paren {x_k - x_{k - 1} } < \delta$
For all integers $k$ satisfying $1 \le k \le n$, it follows from the Heine-Borel theorem that $\closedint {x_{k - 1} } {x_k}$ is compact.
So we can apply {{Corollary|Continuous Image of Compact Space is Compact|3}} to conclude that there exist $u_k, v_k \in \closedint {x_{k - 1} } {x_k}$ such that:
{{begin-eqn}}
{{eqn | l = \map f {u_k}
      | r = \sup \set {\map f x: x \in \closedint {x_{k - 1} } {x_k} }
}}
{{eqn | l = \map f {v_k}
      | r = \inf \set {\map f x: x \in \closedint {x_{k - 1} } {x_k} }
}}
{{end-eqn}}
By assumption, $x_k - x_{k - 1} < \delta$, so:
:$\size {u_k - v_k} < \delta$
It follows from the definition of $\delta$ that:
:$\ds \map f {u_k} - \map f {v_k} < \frac \epsilon {b - a}$
This gives:
{{begin-eqn}}
{{eqn | l = \map U P - \map L P
      | r = \sum_{k \mathop = 1}^n \map f {u_k} \paren {x_k - x_{k - 1} } - \sum_{k \mathop = 1}^n \map f {v_k} \paren {x_k - x_{k - 1} }
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {\map f {u_k} - \map f {v_k} } \paren {x_k - x_{k - 1} }
}}
{{eqn | o = <
      | r = \frac \epsilon {b - a} \sum_{k \mathop = 1}^n \paren {x_k - x_{k - 1} }
}}
{{eqn | r = \frac \epsilon {b - a} \paren {x_n - x_0}
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
as desired.
{{qed}}
Theorem: Definite Integral on Zero Interval

Let $f$ be a real function which is defined on the closed interval $\Bbb I := \closedint a b$, where $a < b$.
Then:
:$\ds \forall c \in \Bbb I: \int_c^c \map f t \rd t = 0$

Proof:
Follows directly from the definition of definite integral.
There is only one finite subdivision of $\closedint c c$ and that is $\set c$.
Both the lower Darboux sum and upper Darboux sum of $\map f x$ on $\closedint c c$ belonging to the finite subdivision $\set c$ are equal to zero.
Hence the result.
{{qed}}
Theorem: Linear Combination of Derivatives

Let $\map f x, \map g x$ be real functions defined on the open interval $I$.
Let $\xi \in I$ be a point in $I$ at which both $f$ and $g$ are differentiable.
Then:
:$\map D {\lambda f + \mu g} = \lambda D f + \mu D g$
at the point $\xi$ for some $\lambda \in \R$ and $\mu \in \R$.
It follows from the definition of derivative that if $f$ and $g$ are both differentiable on the interval $I$, then:
:$\forall x \in I: \map D {\lambda \map f x + \mu \map g x} = \lambda D \map f x + \mu D \map g x$

Proof:
{{tidy}}
Let $\xi \in I$ be a point in $I$ at which $f$ is   differentiable. Then by definition, $f$ is differentiable at the point $\xi$ implies that the limit:
:$\ds \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h$
exists.
Let $\xi \in I$ be a point in $I$ at which $g$ is   differentiable. Then by definition, $g$ is differentiable at the point $\xi$ implies that the limit:
:$\ds \lim_{h \mathop \to 0} \frac {\map g {\xi + h} - \map g \xi} h$
exists.
So the limit:
:$\ds \lim_{h \mathop \to 0} \lambda \paren {\frac {\map f {\xi + h} - \map f \xi} h} + \mu \paren {\frac {\map g {\xi + h} - \map g \xi} h}$
exists.
Since $f$ and $g$ are real functions that are differentiable at $\xi \in I$, and $\lambda, \mu \in \R$, it follows that:
:$\map{ \paren{\lambda f + \mu g} } {x}$
exists, and is differentiable at $\xi \in I$ real function.
So the limit:
:$\ds \lim_{h \mathop \to 0} \frac { \map{ \paren{\lambda f + \mu g} } {\xi + h} - \map{ \paren{\lambda f + \mu g} } \xi} h$
exists.
We have:
:$\ds \lim_{h \mathop \to 0} \frac { \map{ \paren{\lambda f + \mu g} } {\xi + h} - \map{ \paren{\lambda f + \mu g} } \xi} h = \ds \lim_{h \mathop \to 0} \lambda \paren {\frac {\map f {\xi + h} - \map f \xi} h} + \mu \paren {\frac {\map g {\xi + h} - \map g \xi} h} $
Hence, the result follows from the definition of derivative.
{{qed}}
Theorem: Linear Combination of Integrals

Let $f$ and $g$ be real functions which are integrable on the closed interval $\closedint a b$.
Let $\lambda$ and $\mu$ be real numbers.
Then the following results hold:

Proof:
Let $F$ and $G$ be primitives of $f$ and $g$ respectively on $\closedint a b$.
By Linear Combination of Derivatives, $H = \lambda F + \mu G$ is a primitive of $\lambda f + \mu g$ on $\closedint a b$.
Hence by the Fundamental Theorem of Calculus:
{{begin-eqn}}
{{eqn | l = \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t
      | r = \bigintlimits {\lambda \map F t + \mu \map G t} a b
      | c = 
}}
{{eqn | r = \lambda \bigintlimits {\map F t} a b + \mu \bigintlimits {\map G t} a b
      | c = 
}}
{{eqn | r = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Linear Combination of Integrals

Let $f$ and $g$ be real functions which are integrable on the closed interval $\closedint a b$.
Let $\lambda$ and $\mu$ be real numbers.
Then the following results hold:

Proof:
It is clear that for step functions $s$ and $t$:
{{handwaving|"It is clear that ..."}}
{{MissingLinks}}
:$\ds \int_a^b \lambda \map s x + \mu \map t x \rd x = \lambda \int_a^b \map s x \rd x + \mu \int_a^b \map t x \rd x$
Under any partition, the lower Darboux sums and upper Darboux sums of $f$ and $g$ are step functions, so the above formula relates the lower Darboux sums and upper Darboux sums of $f$ and $g$ to the lower Darboux sums and upper Darboux sums of the linear combinations of $f$ and $g$.
Because this identity is preserved for all possible partitions of $\closedint a b$, it is preserved for the supremum and infimum of all possible lower Darboux sums and upper Darboux sums, so the linear combinations of $f$ and $g$ are integrable.
{{begin-eqn}}
{{eqn | l = \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t
      | r = \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren {\lambda f + \mu g} } } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} }
      | c = 
}}
{{eqn | r = \lambda \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren f} } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} } + \mu \sup \set {\sum_{\nu \mathop = 1}^n \map {m_\nu^{\paren g} } {x_\nu - x_{\nu - 1} }: \forall \nu \in \closedint 1 n x_\nu > x_{\nu - 1} }
      | c = 
}}
{{eqn | r = \lambda \int_a^b \map f x \rd x + \mu \int_a^b \map g x \rd x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Integration by Parts

Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$.
Let $f$ and $g$ have primitives $F$ and $G$ respectively on $\closedint a b$.
Then:
=== Primitive ===
{{:Integration by Parts/Primitive}}
=== Definite Integral ===
{{:Integration by Parts/Definite Integral}}
This technique is called '''integration by parts'''.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d t} } {\map F t \map G t}
      | r = \map f t \map G t + \map F t \map g t
      | c = Product Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \paren {\map f t \map G t + \map F t \map g t} \rd t
      | r = \map F t \map G t
      | c = Fundamental Theorem of Calculus: integrating both sides {{WRT|Integration}} $t$
}}
{{eqn | ll= \leadsto
      | l = \int \map f t \map G t \rd t + \int \map F t \map g t \rd t
      | r = \map F t \map G t
      | c = Linear Combination of Primitives
}}
{{eqn | ll= \leadsto
      | l = \int \map f t \map G t \rd t
      | r = \map F t \map G t - \int \map F t \map g t \rd t
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Integration by Parts

Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$.
Let $f$ and $g$ have primitives $F$ and $G$ respectively on $\closedint a b$.
Then:
=== Primitive ===
{{:Integration by Parts/Primitive}}
=== Definite Integral ===
{{:Integration by Parts/Definite Integral}}
This technique is called '''integration by parts'''.

Proof:
Let $\map u x$ and $\map v x$ be integrable functions defined on $\closedint a b$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {u v}
      | r = u \dfrac {\d v} {\d x} + v \dfrac {\d u} {\d x}
      | c = Product Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = v \dfrac {\d u} {\d x}
      | r = \map {\dfrac \d {\d x} } {u v} - u \dfrac {\d v} {\d x}
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = \int v \dfrac {\d u} {\d x} \rd x
      | r = \int \paren {\map {\dfrac \d {\d x} } {u v} - u \dfrac {\d v} {\d x} } \rd x
      | c = integrating both sides {{WRT|Integration}} $x$
}}
{{eqn | r = \int \map {\dfrac \d {\d x} } {u v} \rd x - \int u \dfrac {\d v} {\d x} \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = u v - \int u \dfrac {\d v} {\d x} \rd x
      | c = Fundamental Theorem of Calculus
}}
{{end-eqn}}
{{qed}}
Theorem: Relative Sizes of Definite Integrals

Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$, where $a < b$.
If:
:$\forall t \in \closedint a b: \map f t \le \map g t$
then:
:$\ds \int_a^b \map f t \rd t \le \int_a^b \map g t \rd t$
Similarly, if:
:$\forall t \in \closedint a b: \map f t < \map g t$
then:
:$\ds \int_a^b \map f t \rd t < \int_a^b \map g t \rd t$

Proof:
Suppose that $\forall t \in \closedint a b: \map f t \le \map g t$.
From the Fundamental Theorem of Calculus, $g - f$ has a primitive on $\closedint a b$.
Let $H$ be such a primitive.
Then:
:$\forall t \in \closedint a b: D_t \map H t = \map g t - \map f t \ge 0$
By Derivative of Monotone Function, it follows that $H$ is increasing on $\closedint a b$.
Thus:
:$\map H b \ge \map H a$
Hence:
:$\ds \int_a^b \map g t - \map f t \rd t = \map H b - \map H a \ge 0$
The proof for the second case is similar.
{{qed}}
Theorem: Triangle Inequality for Integrals/Real

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then:
:$\ds \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$

Proof:
From Negative of Absolute Value, we have for all $a \in \closedint a b$:
:$-\size {\map f t} \le \map f t \le \size {\map f t}$
Thus from Relative Sizes of Definite Integrals:
:$\ds -\int_a^b \size {\map f t} \rd t \le \int_a^b \map f t \rd t \le \int_a^b \size {\map f t} \rd t$
Hence the result.
{{qed}}
Theorem: Cauchy Integral Test

Let $f$ be a real function which is continuous, positive and decreasing on the interval $\hointr 1 {+\infty}$.
Let the sequence $\sequence {\Delta_n}$ be defined as:
:$\ds \Delta_n = \sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x$
Then $\sequence {\Delta_n} $ is decreasing and bounded below by zero.
Hence it converges.

Proof:
From Darboux's Theorem, we have that:
:$\ds m \paren {b - a} \le \int_a^b \map f x \rd x \le M \paren {b - a}$
where:
:$M$ is the maximum
and:
:$m$ is the minimum
of $\map f x$ on $\closedint a b$.
Since $f$ decreases, $M = \map f a$ and $m = \map f b$.
Thus it follows that:
:$\ds \forall k \in \N_{>0}: \map f {k + 1} \le \int_k^{k + 1} \map f x \rd x \le \map f k$
as $\paren {k + 1} - k = 1$.
Thus:
{{begin-eqn}}
{{eqn | l = \Delta_{n + 1} - \Delta_n
      | r = \paren {\sum_{k \mathop = 1}^{n + 1} \map f k - \int_1^{n + 1} \map f x \rd x} - \paren {\sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x}
      | c = 
}}
{{eqn | r = \map f {n + 1} - \int_n^{n + 1} \map f x \rd x
      | c = 
}}
{{eqn | o = \le
      | r = \map f {n + 1} - \map f {n + 1}
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
Thus $\sequence {\Delta_n}$ is decreasing.
{{qed|lemma}}
Also:
{{begin-eqn}}
{{eqn | l = \Delta_n
      | r = \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \int_k^{k + 1} \map f x \rd x
      | c = 
}}
{{eqn | o = \ge
      | r = \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \map f k
      | c = 
}}
{{eqn | r = \map f n
      | c = 
}}
{{eqn | o = \ge
      | r = 0
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Hence the result.
{{qed}}
Theorem: Harmonic Series is Divergent

The harmonic series:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 n$ 
diverges.

Proof:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 n = \underbrace 1_{s_0} + \underbrace {\frac 1 2 + \frac 1 3}_{s_1} + \underbrace {\frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7}_{s_2} + \cdots$ 
where $\ds s_k = \sum_{i \mathop = 2^k}^{2^{k + 1} \mathop - 1} \frac 1 i$
From Ordering of Reciprocals:
:$\forall m, n \in \N_{>0}: m < n: \dfrac 1 m > \dfrac 1 n$
so each of the summands in a given $s_k$ is greater than $\dfrac 1 {2^{k + 1} }$.
The number of summands in a given $s_k$ is $2^{k + 1} - 2^k = 2 \times 2^k - 2^k = 2^k$, and so:
:$s_k > \dfrac{2^k} {2^{k + 1} } = \dfrac 1 2$
Hence the harmonic sum $H_{2^m}$ satisfies the following inequality:
{{begin-eqn}}
{{eqn | l = H_{2^m}
      | r = \sum_{n \mathop = 1}^{2^m} \frac 1 n
      | c = 
}}
{{eqn | o = >
      | r = \sum_{n \mathop = 1}^{2^m - 1} \frac 1 n
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^m \paren {s_k}
      | c = 
}}
{{eqn | o = >
      | r = 1 + \sum_{a \mathop = 0}^m \frac 1 2
      | c = 
}}
{{eqn | r = 1 + \frac m 2
      | c = 
}}
{{end-eqn}}
The {{RHS}} diverges, from the $n$th term test.
The result follows from the the Comparison Test for Divergence.
{{qed}}
Theorem: Harmonic Series is Divergent

The harmonic series:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 n$ 
diverges.

Proof:
Observe that all the terms of the harmonic series are strictly positive.
From Reciprocal Sequence is Strictly Decreasing, the terms are decreasing.
Hence the Cauchy Condensation Test can be applied, and we examine the convergence of:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 1}^\infty 2^n \frac 1 {2^n}
      | r = \sum_{n \mathop = 1}^\infty 1
}}
{{end-eqn}}
This diverges, from the $n$th term test.
Hence $\ds \sum \frac 1 n$ also diverges.
{{qed}}
Theorem: Harmonic Series is Divergent

The harmonic series:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 n$ 
diverges.

Proof:
We have that the Integral of Reciprocal is Divergent.
Hence from the Cauchy Integral Test, the harmonic series also diverges.
{{qed}}
Theorem: Harmonic Series is Divergent

The harmonic series:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 n$ 
diverges.

Proof:
For all $N \in \N$:
:$\dfrac 1 N + \dfrac 1 {N + 1} + \cdots + \dfrac 1 {2 N} > N \cdot \dfrac 1 {2 N} = \dfrac 1 2$
Hence, by Cauchy's Convergence Criterion for Series, the Harmonic series is divergent.
{{qed}}
Theorem: Sum of Reciprocals of Primes is Divergent

Let $n \in \N: n \ge 1$.
There exists a (strictly) positive real number $C \in \R_{>0}$ such that:
:$(1): \quad \ds \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \map \ln {\ln n} - C$
where $\Bbb P$ is the set of all prime numbers.
:$(2): \quad \ds \lim_{n \mathop \to \infty} \paren {\map \ln {\ln n} - C} = +\infty$

Proof:
Let $n \in \N$ be a natural number.
Let $p_n$ denote the $n$th prime number.
Consider the continued product:
:$\ds \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$
By Sum of Infinite Geometric Sequence:
{{begin-eqn}}
{{eqn | l = \frac 1 {1 - \frac 1 2}
      | r = 1 + \frac 1 2 + \frac 1 {2^2} + \cdots
      | c = 
}}
{{eqn | l = \frac 1 {1 - \frac 1 3}
      | r = 1 + \frac 1 3 + \frac 1 {3^2} + \cdots
      | c = 
}}
{{eqn | l = \frac 1 {1 - \frac 1 5}
      | r = 1 + \frac 1 5 + \frac 1 {5^2} + \cdots
      | c = 
}}
{{eqn | o = \cdots
      | c = 
}}
{{eqn | l = \frac 1 {1 - \frac 1 {p_n} }
      | r = 1 + \frac 1 {p_n} + \frac 1 {p_n^2} + \cdots
      | c = 
}}
{{end-eqn}}
Consider what happens when all these series are multiplied together.
A new series will be generated whose terms consist of all possible products of one term selected from each of the series on the {{RHS}}.
This new series will converge in any order to the product of the terms on the {{LHS}}.
By the Fundamental Theorem of Arithmetic, every integer greater than $1$ is uniquely expressible as a product of powers of different primes.
Hence the product of these series is the series of reciprocals of all (strictly) positive integers whose prime factors are no greater than $p_n$.
In particular, all (strictly) positive integers up to $p_n$ have this property.
So:
:$\ds \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$
{{begin-eqn}}
{{eqn | l = \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}
      | o = \ge
      | r = \sum_{k \mathop = 1}^{p_n} \frac 1 k
      | c = 
}}
{{eqn | o = >
      | r = \int_1^{p_n + 1} \dfrac {\d x} x
      | c = 
}}
{{eqn | r = \map \ln {p_n + 1}
      | c = 
}}
{{eqn | r = \ln p_n
      | c = 
}}
{{end-eqn}}
It follows by taking reciprocals that:
:$\ds \prod_{k \mathop = 1}^n \paren {1 - \frac 1 {p_k} } < \frac 1 {\ln p_n}$
Taking logarithms of each side:
:$(1): \quad \ds \sum_{k \mathop = 1}^n \map \ln {1 - \frac 1 {p_k} } < - \ln \ln p_n$
Next, note that the line $y = 2 x$ in the cartesian plane lies below the curve $y = \map \ln {1 + x}$ on the interval $\closedint {-\frac 1 2} 0$.
Also note that all primes are greater than or equal to $2$.
Thus it follows that:
:$-\dfrac 2 {p_k} < \map \ln {1 - \dfrac 1 {p_k} }$
Applying this to $(1)$ yields:
:$\ds -2 \sum_{k \mathop = 1}^n \dfrac 1 {p_k} < -\ln \ln p_n$
and so:
:$\ds \sum_{k \mathop = 1}^n \dfrac 1 {p_k} > \dfrac 1 2 \ln \ln p_n$
But:
:$\ds \lim_{n \mathop \to \infty} \ln \ln p_n \to \infty$
and so the series:
:$\ds \sum_{p \mathop \in \Bbb P} \frac 1 p$
is divergent.
{{qed}}
Theorem: Sum of Reciprocals of Primes is Divergent

Let $n \in \N: n \ge 1$.
There exists a (strictly) positive real number $C \in \R_{>0}$ such that:
:$(1): \quad \ds \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \map \ln {\ln n} - C$
where $\Bbb P$ is the set of all prime numbers.
:$(2): \quad \ds \lim_{n \mathop \to \infty} \paren {\map \ln {\ln n} - C} = +\infty$

Proof:
{{AimForCont}} the contrary.
If the prime reciprocal series converges then there must exist some $k \in \N$ such that:
:$\ds \sum_{n \mathop = k \mathop + 1}^{\infty} \frac 1 {p_n} < \frac 1 2$
Fixing $k$ from the series above, we now let:
:$\ds Q = \prod_{i \mathop = 1}^k {p_i}$
and:
:$\ds \map S r = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q}$
Let $\map S {r, j}$ be the sum of all of the terms from $\map S r$ for which $1 + i Q$ has exactly $j$ distinct prime factors.
Notice that $1 + i Q$ is coprime with every prime factor in $Q$.
Thus every prime factor of $1 + i Q$ where $i = 1, \ldots, r$ falls into some finite sequence of consecutive primes separated by $i Q$:
:$\map P r = \set {p_{k + 1}, p_{k + 2}, \ldots, p_{\map m r} }$
where $\map m r$ is the total number of different prime factors formed from all $1 + i Q \paren {i = 1, 2, \cdots, r}$
Notice again that each term of $\map S {r, j}$ occurs at least once in the expansion of:
:$\ds \paren {\sum_{n \mathop = k \mathop + 1}^{\map m r} \frac 1 {p_n} }^j < \paren {\sum_{n \mathop = k \mathop + 1}^\infty \frac 1 {p_n} }^j < \paren {\frac 1 2}^j$
and also by Sum of Infinite Geometric Sequence:
:$\ds \map S r = \sum_{j \mathop = 1}^r \map S {r, j} < \sum_{j \mathop = 1}^r \paren {\frac 1 2}^j < 1$
We now have an upper bound on our sum :$\ds \map S r < 1$
Finally notice that:
{{begin-eqn}}
{{eqn | l = \lim_{r \mathop \to \infty} \map S r 
      | r = \sum_{i \mathop = 1}^\infty \frac 1 {1 + i Q}
      | c = 
}}
{{eqn | o = \gt
      | r = \sum_{i \mathop = 1}^\infty \frac 1 {i + i Q}
      | c = 
}}
{{eqn | r = \frac 1 {1 + Q} \sum_{i \mathop = 1}^\infty \frac 1 i
      | c = 
}}
{{eqn | ll = \leadsto
      | l = \lim_{r \mathop \to \infty} \map S r 
      | o = \gt
      | r = + \infty
      | c = Harmonic Series is Divergent
}}
{{end-eqn}}
We now have a lower bound on our sum:
:$\ds \map S r > \infty$
Our upper bound and lower bound cannot both be correct.
We are forced to abandon our assumption that the prime reciprocal series converges.
Therefore, from Proof by Contradiction, the prime reciprocal series diverges.
{{qed}}
Theorem: Measure of Interval is Length

Let $I$ be a real interval whose endpoints are $a$ and $b$.
Then $I$ is Lebesgue measurable, and the value of the measure is the length of the interval $b - a$.

Proof:
{{link wanted|Insert here: Some stuff proving intervals are measurable}}
Let $L \subset \R$ be a real interval.
Then $L$ has two distinct endpoints: $a$ and $b$.
Let $\ds \set {I_n}_{n \mathop = 1}^\infty$ be a set of open real intervals satisfying:
:$\ds L \subseteq \bigcup_{n \mathop = 1}^\infty I_n$
Theorem: Classification of Compact Three-Manifolds Supporting Zero-Curvature Geometry

Every closed, orientable, path connected $3$-dimensional Riemannian manifold which supports a geometry of zero curvature is homeomorphic to one of the following:
* Torus $\mathbb T^3$
* Half-Twist Cube
* Quarter-Twist Cube
* Hantschze-Wendt Manifold
* $\frac 1 6$-Twist Hexagonal Prism
* $\frac 1 3$-Twist Hexagonal Prism
The $3$-torus is described on the torus page.
The other manifolds can be described using quotient spaces on familiar prisms, with the equivalence relations described below.
<gallery perRow="5">
File:Halftwistcube.JPG|The Half-Twist Cube
File:Hantschzewendt.JPG|The Hantschze-Wendt Manifold
File:Quartertwistcube.JPG|The Quarter-Twist Cube
File:Sixthtwisthexagon.JPG|The $\frac 1 6$-Twist Hexagonal Prism
File:Thirdtwisthexagon.JPG|The $\frac 1 3$-Twist Hexagonal Prism
</gallery>

Proof:
{{ProofWanted}}
Category:Riemannian Manifolds
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Theorem: Heine-Borel Theorem/Real Line

Let $\R$ be the real number line considered as a (real) Euclidean space.
Let $C \subseteq \R$.
Then $C$ is closed and bounded in $\R$ {{iff}} $C$ is compact.

Proof:
=== Necessary Condition ===
Let $C$ be closed and bounded in $\R$.
Then, by Closed Bounded Subset of Real Numbers is Compact, $C$ is compact.
{{qed|lemma}}