avewright/proofwiki-math huggingface dataset:
Theorem: Local Uniform Convergence Implies Compact Convergence

Let $X$ be a topological space.
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {f_n}$ be a sequence of mappings $f_n : X \to M$.
Let $f_n$ converge locally uniformly to $f : X \to M$.
Then $f_n$ converges compactly to $f$.

Proof:
{{Proofread}}
{{tidy}}
{{MissingLinks}}
Recall that a sequence $f_n$ converges compactly to $f$, if $f_n$ converges uniformly to $f$ over any compact subset $K \subset X$.
Consider a sequence $f_n$ which converges locally uniformly to $f : X \to M$.f
Let $K$ be an arbitrary subset $K \subset X$.
For $x \in K$ there exists an open neighborhood $U_x$ such that over $U_x$ the sequence $f_n$ converges uniformly to $f$.
Observe that $\set {U_x : x \in K}$ is an open cover of $K$.
Since $K$ is compact ({{WRT}} the induced topology), every open cover for $K$ has a finite subcover.
So there exists a finite collection of points $x_1, \ldots, x_\ell$ such that $\set {U_{x_1}, \ldots, U_{x_\ell} }$ is an open subcover of $K$. 
{{explain|What do you mean by open subcover? There is "subcover" and "open cover", but not sure what you mean by "open subcover"}}
Fix $\epsilon > 0$.
Consider $i \in \set {1, \ldots, \ell}$.
Then there exists $N_i \in \N$ such that:
:$\forall n \ge N_i: \sup \set {\map d {\map {f_n} z, \map f z} : z \in U_{x_i} } < \epsilon$
Take $N = \max \set {N_1, \ldots, N_\ell}$. 
Then for $\ds z \in \bigcup_{i \mathop = 1}^\ell U_{x_i}$:
:$\sup \set {\map d {\map {f_n} z, \map f z} : z \in U_{x_i} } < \epsilon$
But since:
:$\ds K \subset \bigcup_{i \mathop = 1}^\ell U_{x_i}$
and $\epsilon$ is arbitrary, we conclude that $f_n$ converges uniformly over $K$.
The fact that $K$ is an arbitrary compact subset finishes the proof. 
{{qed}}
Theorem: Compact Convergence Implies Local Uniform Convergence if Weakly Locally Compact

Let $T = \struct {S, \tau}$ be a weakly locally compact topological space.
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {f_n}$ be a sequence of mappings $f_n: X \to M$.
Let $f_n$ converge compactly to $f: X \to M$.
Then $f_n$ converges locally uniformly to $f$.

Proof:
Because $T$ is weakly locally compact, every point of $S$ has a compact neighborhood in $T$.
We are given $f_n$ converge compactly to $f$, so $f_n$ converges uniformly to $f$ on every compact subset of $X$
So every point of $S$ has a (compact) neighborhood on which $f_n$ converges uniformly to $f$.
Thus $f_n$ converges locally uniformly to $f$.
{{qed}}
Theorem: Smallest Pair of Quasiamicable Numbers

The smallest pair of quasiamicable numbers is $48$ and $75$.

Proof:
From Quasiamicable Numbers: $48$ and $75$ we have that $48$ and $75$ are quasiamicable numbers.
It remains to be demonstrated that these are indeed the smallest such.
Let $n$ be the smaller number of the quasiamicable pair.
Then we must have $\map {\sigma_1} n - n - 1 > n$.
Since $\map {\sigma_1} n > 2 n$, $n$ is abundant.
The abundant numbers for $n < 48$ are:
:$12, 18, 20, 24, 30, 36, 40, 42$
And we have:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {12} - 13
      | r = \paren {1 + 2 + 4} \paren {1 + 3} - 13
}}
{{eqn | r = 15
}}
{{eqn | l = \map {\sigma_1} {15} - 16
      | r = \paren {1 + 3} \paren {1 + 5} - 16
}}
{{eqn | r = 8
      | rr = \ne 12
}}
{{eqn | l = \map {\sigma_1} {18} - 19
      | r = \paren {1 + 2} \paren {1 + 3 + 9} - 19
}}
{{eqn | r = 20
}}
{{eqn | l = \map {\sigma_1} {20} - 21
      | r = \paren {1 + 2 + 4} \paren {1 + 5} - 21
}}
{{eqn | r = 21
      | rr = \ne 18
}}
{{eqn | l = \map {\sigma_1} {21} - 22
      | r = \paren {1 + 3} \paren {1 + 7} - 22
}}
{{eqn | r = 10
      | rr = \ne 20
}}
{{eqn | l = \map {\sigma_1} {24} - 25
      | r = \paren {1 + 2 + 4 + 8} \paren {1 + 3} - 25
}}
{{eqn | r = 35
}}
{{eqn | l = \map {\sigma_1} {35} - 36
      | r = \paren {1 + 5} \paren {1 + 7} - 36
}}
{{eqn | r = 12
      | rr = \ne 24
}}
{{eqn | l = \map {\sigma_1} {30} - 31
      | r = \paren {1 + 2} \paren {1 + 3} \paren {1 + 5} - 31
}}
{{eqn | r = 41
}}
{{eqn | l = \map {\sigma_1} {41} - 42
      | r = 0
      | rr = \ne 30, 36
}}
{{eqn | l = \map {\sigma_1} {36} - 37
      | r = \paren {1 + 2 + 4} \paren {1 + 3 + 9} - 37
}}
{{eqn | r = 41
}}
{{eqn | l = \map {\sigma_1} {40} - 41
      | r = \paren {1 + 2 + 4 + 8} \paren {1 + 5} - 41
}}
{{eqn | r = 49
}}
{{eqn | l = \map {\sigma_1} {49} - 50
      | r = \paren {1 + 7 + 49} - 50
}}
{{eqn | r = 7
      | rr = \ne 40
}}
{{eqn | l = \map {\sigma_1} {42} - 43
      | r = \paren {1 + 2} \paren {1 + 3} \paren {1 + 7} - 43
}}
{{eqn | r = 53
}}
{{eqn | l = \map {\sigma_1} {53} - 54
      | r = 0
      | rr = \ne 42
}}
{{end-eqn}}
hence $n$ is at least $48$, proving the result.
{{qed}}
Theorem: Bounds for Complex Logarithm

Let $\ln$ denote the complex logarithm.
Let $z \in \C$ with $\cmod z \le \dfrac 1 2$.
Then:
:$\dfrac 1 2 \cmod z \le \cmod {\map \ln {1 + z} } \le \dfrac 3 2 \cmod z$

Proof:
By definition of complex logarithm:
:$-\map \ln {1 + z} = \ds \sum_{n \mathop = 1}^\infty \frac {\paren {-z}^n} n$
Thus
{{begin-eqn}}
{{eqn  | l = \cmod {1 - \frac {\map \ln {1 + z} } z}
       | r = \cmod {\sum_{n \mathop = 2}^\infty \frac {\paren {-1}^n z^{n - 1} }n}
       | c = Linear Combination of Convergent Series
}}
{{eqn  | o = \le
       | r = \sum_{n \mathop = 2}^\infty \cmod {\frac {\paren {-z}^{n - 1} } n}
       | c = Triangle Inequality for Series
}}
{{eqn  | o = \le
       | r = \sum_{n \mathop = 2}^\infty \frac {\cmod z^{n - 1} } 2
       | c = $n \ge 2$
}}
{{eqn  | r = \frac {\cmod z / 2} {1 - \cmod z}
       | c = Sum of Infinite Geometric Sequence
}}
{{eqn  | o = \le
       | r = \frac 1 2
       | c = $\cmod z \le \dfrac 1 2$
}}
{{end-eqn}}
By the Triangle Inequality:
:$\dfrac 1 2 \le \cmod {\dfrac {\map \ln {1 + z} } z} \le \dfrac 3 2$
{{qed}}
Theorem: Reciprocal of Absolutely Convergent Product is Absolutely Convergent

Let $\struct {\mathbb K, \norm{\,\cdot\,}}$ be a valued field.
Let $\sequence {1 + a_n}$ be a sequence of nonzero elements of $\mathbb K$.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty \left({1 + a_n}\right)$ converge absolutely to $a \in \mathbb K \setminus \set 0$.
Then $\ds \prod_{n \mathop = 1}^\infty \frac 1 {1 + a_n}$  converges absolutely to $1 / a$.

Proof:
By continuity of $x \mapsto 1 / x$:
:$\ds \lim_{N \mathop \to \infty} \prod_{n \mathop = 1}^N \frac 1 {1 + a_n} = \frac 1 a$
It remains to prove the absolute convergence.
Because $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$  converges absolutely, $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.
By Factors in Absolutely Convergent Product Converge to One, $\norm {a_n} \le \dfrac 1 2 $ for $n$ sufficiently large.
Thus $\norm {\dfrac 1 {a_n + 1} - 1} = \norm {\dfrac {a_n} {a_n + 1} } \le 2 \norm {a_n}$ for $n$ sufficiently large.
By the Comparison Test, $\ds \sum_{n \mathop = 1}^\infty \paren {\frac 1 {a_n + 1} - 1}$ converges absolutely.
Thus $\ds \prod_{n \mathop = 1}^\infty \frac 1 {1 + a_n}$  converges absolutely.
{{qed}}
Category:Infinite Products
bvj860sg8ra24s02t8zbbtpfd9gxcwc
Theorem: Equivalence of Definitions of Quasiamicable Numbers

Let $m \in \Z_{>0}$ and $n \in \Z_{>0}$ be (strictly) positive integers.
{{TFAE|def = Quasiamicable Numbers}}

Proof:
Let $\map s n$ denote the sum of the proper divisors of (strictly) positive integer $n$.
The sum of all the divisors of a (strictly) positive integer $n$ is $\map {\sigma_1} n$, where $\sigma_1$ is the divisor sum function.
The proper divisors of $n$ are the divisors $n$ with $1$ and $n$ excluded.
Thus:
:$\map s n = \map {\sigma_1} n - n - 1$
Suppose:
:$\map s n = m$
and:
:$\map s m = n$
Then:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} n - n - 1
      | r = m
      | c = {{Defof|Proper Divisor of Integer}}
}}
{{eqn | ll= \leadsto
      | l = \map {\sigma_1} n
      | r = m + n + 1
      | c = 
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} m - m - 1
      | r = n
      | c = {{Defof|Proper Divisor of Integer}}
}}
{{eqn | ll= \leadsto
      | l = \map {\sigma_1} m
      | r = m + n + 1
      | c = 
}}
{{end-eqn}}
Thus:
:$\map s n = \map s m = m + n + 1$
The argument reverses.
{{qed}}
Category:Quasiamicable Numbers
ia0aegmbjpvqohobq949bxsd7qms2tb
Theorem: Equivalence of Definitions of Absolute Convergence of Product

Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let $\sequence {a_n}$ be a sequence in $\mathbb K$.
{{TFAE|def = Absolute Convergence of Product}}
=== Definition 1 ===
{{:Definition:Absolute Convergence of Product/General Definition/Definition 1}}
=== Definition 2 ===
{{:Definition:Absolute Convergence of Product/General Definition/Definition 2}}

Proof:
=== 1 implies 2 ===
By the Monotone Convergence Theorem, it suffices to show that the partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$ are bounded.
Because $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \norm {a_n} }$ converges, its partial products are bounded.
By Bounds for Finite Product of Real Numbers:
:$\ds \sum_{n \mathop = 1}^N \norm {a_n} \le \prod_{n \mathop = 1}^N \paren {1 + \norm {a_n} }$
{{qed}}
Theorem: Factors in Convergent Product Converge to One

Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ be convergent.
Then $a_n \to 1$.

Proof:
By definition of convergent product, there exists $n_0 \in \N$ such that:
:$a_n \ne 0$ for $n \ge n_0$
:the sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty a_n$ has a nonzero limit.
Let $p_n$ denote the $n$th partial product of $\ds \prod_{n \mathop = n_0}^\infty a_n$.
For $n > n_0$:
:$a_n = \dfrac {p_n} {p_{n - 1} }$
By the Combination Theorem for Sequences:
:$a_n \to 1$
{{qed}}
Theorem: Logarithm of Convergent Product of Real Numbers

{{TFAE}}
:$(1): \quad$ The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ converges to $a \in \R_{\ne 0}$.
:$(2): \quad$ The series $\ds \sum_{n \mathop = 1}^\infty \ln a_n$ converges to $\ln a$.

Proof:
Let $p_n$ denote the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty a_n$.
Let $s_n$ denote the $n$th partial sum of $\ds \sum_{n \mathop = 1}^\infty \ln a_n$.
By Sum of Logarithms, $s_n = \map \ln {p_n}$.
Theorem: Factors in Absolutely Convergent Product Converge to One

Let $\struct {\mathbb K, \norm {\, \cdot \,} }$ be a valued field.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ be absolutely convergent.
Then:
:$a_n \to 0$

Proof:
We have that $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ is absolutely convergent.
By the definition of absolutely convergent product, $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} }$ is convergent.
{{begin-eqn}}
{{eqn | l = \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} }
      | r = \paren {1 + \size {a_1} } \paren {1 + \size {a_2} } \paren {1 + \size {a_3} } \paren {1 + \size {a_4} } \cdots
      | c = expanding out the product
}}
{{eqn | r = 1 + \paren {\size {a_1} + \size {a_2} + \size {a_3} + \size {a_4} + \cdots} + \paren{\size {a_1} \size {a_2} + \size {a_1} \size {a_3} + \size {a_1} \size {a_4} + \cdots + \size {a_2} \size {a_3} + \size {a_2} \size {a_4} + \cdots} + \paren {\size {a_1} \size {a_2} \size {a_3} + \size {a_1} \size {a_2} \size {a_4} + \cdots + \size {a_2} \size {a_3} \size {a_4} + \cdots} + \cdots
      | c = 
}}
{{eqn | r = 1 + \sum_{i \mathop = 1}^\infty \size {a_i} + \sum_{i \mathop = 1}^\infty \sum_{j \mathop = 1}^\infty \size {a_i} \size { a_{i + j} } + \sum_{i \mathop = 1}^\infty \sum_{j \mathop = 1}^\infty \sum_{k \mathop = 1}^\infty \size {a_i} \size { a_{i + j} } \size {a_{i + j + k} } + \cdots
      | c = 
}}
{{end-eqn}}
From the above, we see that $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} } > \sum_{i \mathop = 1}^\infty \size {a_i}$
And since $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {a_n} }$ is convergent, then $\ds \sum_{i \mathop = 1}^\infty \size {a_i}$ is convergent
Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent.
By Terms in Convergent Series Converge to Zero:
:$a_n \to 0$
{{qed}}
Theorem: Factors in Absolutely Convergent Product Converge to One

Let $\struct {\mathbb K, \norm {\, \cdot \,} }$ be a valued field.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ be absolutely convergent.
Then:
:$a_n \to 0$

Proof:
We have that $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ is absolutely convergent.
Let $b_n = \paren {1 + a_n}$
Then $\ds \prod_{n \mathop = 1}^\infty b_n$ is absolutely convergent.
From Absolutely Convergent Product is Convergent, $\ds \prod_{n \mathop = 1}^\infty b_n$ is convergent.
By Factors in Convergent Product Converge to One:
:$b_n \to 1$
Thus:
:$a_n \to 0$
{{qed}}
Theorem: Absolutely Convergent Product Does not Diverge to Zero

Let $\struct {\mathbb K, \norm {\, \cdot \,} }$ be a valued field.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ be absolutely convergent.
Then it is not divergent to $0$.

Proof:
By Factors in Absolutely Convergent Product Converge to One, $\norm {a_n} < 1$ for $n \ge n_0$.
Let $n_1 \ge n_0$.
{{AimForCont}} the product diverges to $0$.
Then:
:$\ds \prod_{n \mathop = n_1}^\infty \paren {1 + a_n} = 0$
By Norm of Limit:
:$\ds \prod_{n \mathop = n_1}^\infty \norm {1 + a_n} = 0$
By the Triangle Inequality and Squeeze Theorem:
:$\ds \prod_{n \mathop = n_1}^\infty \paren {1 - \norm {a_n} } = 0$
By the Weierstrass Product Inequality, we have for $N \ge n_1$:
:$\ds \prod_{n \mathop = n_1}^N \paren {1 - \norm {a_n} } \ge 1 - \sum_{n \mathop = n_1}^N \norm{a_n}$
Taking limits:
:$0 \ge 1 - \ds \sum_{n \mathop = n_1}^\infty \norm {a_n}$
Because $\ds \prod_{n \mathop = 1}^\infty \paren  {1 + a_n}$ is absolutely convergent:
:$\ds \sum_{n \mathop = n_1}^\infty \norm {a_n} < 1$
for $n_1$ sufficiently large.
This is a contradiction.
{{qed}}
Theorem: Absolutely Convergent Product Does not Diverge to Zero

Let $\struct {\mathbb K, \norm {\, \cdot \,} }$ be a valued field.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ be absolutely convergent.
Then it is not divergent to $0$.

Proof:
We have that $\ds \prod_{n \mathop = 1}^\infty \paren {1 - \norm {a_n} }$ is absolutely convergent.
By Factors in Absolutely Convergent Product Converge to One, $\norm {a_n} < 1$ for $n \ge n_0$.
Thus $\ds \sum_{n \mathop = n_0}^\infty \map \ln {1 - \norm {a_n} }$ is absolutely convergent.
{{AimForCont}} the product diverges to $0$.
Then:
:$\ds \prod_{n \mathop = n_0}^\infty \paren {1 + a_n} = 0$
By Norm of Limit:
:$\ds \prod_{n \mathop = n_0}^\infty \norm {1 + a_n} = 0$
By the Triangle Inequality and Squeeze Theorem:
:$\ds \prod_{n \mathop = n_0}^\infty \paren {1 - \norm {a_0} } = 0$
By Logarithm of Infinite Product of Real Numbers:
:$\ds \sum_{n \mathop = n_0}^\infty \map \ln {1 - \norm {a_n} }$
diverges to $-\infty$.
This is a contradiction.
{{qed}}
Theorem: Absolute Value of Absolutely Convergent Product is Absolutely Convergent

Let the infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ converge absolutely to $a \in \mathbb K$.
Then $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$ converges absolutely to $\norm a$.

Proof:
By definition of absolute convergence of $\ds \prod_{n \mathop = 1}^\infty a_n$, $\ds \sum_{n \mathop = 1}^\infty \paren {a_n - 1}$ converges absolutely.
By the Triangle Inequality:
:$\size {\norm {a_n} - 1} \le \norm {a_n - 1}$
By the Comparison Test, $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$ converges absolutely.
By Absolute Value is Continuous, its limit is $\norm a$.
{{qed}}
{{MissingLinks|limit}}
Theorem: Absolute Value of Convergent Infinite Product

Let the infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ converge to $a \in \mathbb K$.
Then $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$  converges to $\norm{a}$.

Proof:
By Absolute Value of Limit of Sequence:
:$\ds \prod_{n \mathop = 1}^\infty \norm {a_n} = \norm a$
It remains to show that the product converges.
By the convergence, there exists $n_0 \in \N$ such that $a_n \ne 0$ for $n \ge n_0$.
Then $\norm {a_n} \ne 0$ for $n \ge n_0$.
Let $P_n$ denote the $n$th partial product of $\ds \prod_{n \mathop = n_0}^\infty a_n$.
Then $\norm {P_n}$ is the $n$th partial product of $\ds \prod_{n \mathop = n_0}^\infty \norm {a_n}$.
Let $P_n$ converge to $b \in \mathbb K \setminus \set 0$.
By Convergence of Absolute Value of Sequence:
:$\norm {P_n} \to \norm b \in \mathbb K \setminus \set 0$.
Thus $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$ converges.
{{qed}}
Theorem: Absolute Value of Divergent Infinite Product

The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$ {{iff}} $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$ diverges to $0$.

Proof:
=== 1 implies 2 ===
Theorem: Logarithm of Divergent Product of Real Numbers/Infinity

{{TFAE}}:
:$(1): \quad$ The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ diverges to $+\infty$.
:$(2): \quad$ The series $\ds \sum_{n \mathop = 1}^\infty \log a_n$ diverges to $+\infty$.

Proof:
Let $p_n$ denote the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty a_n$.
Let $s_n$ denote the $n$th partial sum of $\ds \sum_{n \mathop = 1}^\infty\log a_n$.
By Sum of Logarithms:
:$s_n = \map \log {p_n}$
{{ProofWanted}}
Theorem: Logarithm of Divergent Product of Real Numbers/Zero

{{TFAE}}
{{begin-itemize}}
{{item|*|The infinite product $\ds \prod_{n \mathop {{=}} 1}^\infty a_n$ diverges to $0$.}}
{{item|*|The series $\ds \sum_{n \mathop {{=}} 1}^\infty \log a_n$ diverges to $-\infty$.}}
{{end-itemize}}

Proof:
Let $p_n$ denote the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty a_n$.
Let $s_n$ denote the $n$th partial sum of $\ds \sum_{n \mathop = 1}^\infty \log a_n$.
By Sum of Logarithms:
:$s_n = \log p_n$
{{ProofWanted}}
Theorem: Product of Convergent Products is Convergent

Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let $\ds \prod_{n \mathop = 1}^\infty a_n$ converge to $a$.
Let $\ds \prod_{n \mathop = 1}^\infty b_n$ converge to $b$.
Then $\ds \prod_{n \mathop = 1}^\infty a_n b_n$ converges to $ab$.

Proof:
Let $n_0 \in \N$ such that $a_n \ne 0$ for $n> n_0$.
Let $n_1 \in \N$ such that $b_n \ne 0$ for $n> n_1$.
Then $a_n b_n \ne 0$ for $n > n_2 = \max \set {n_0, n_1}$.
Let $p_n$ be the $n$th partial product of $\ds \prod_{n \mathop = n_2 + 1}^\infty a_n$.
Let $q_n$ be the $n$th partial product of $\ds \prod_{n \mathop = n_2 + 1}^\infty b_n$.
Then $p_n q_n$ is the $n$th partial product of $\ds \prod_{n \mathop = n_2 + 1}^\infty a_n b_n$.
Because $p_n$ and $q_n$ converge to a nonzero limit, so does $p_n q_n$.
Thus $\ds \prod_{n \mathop = 1}^\infty a_n b_n$ converges.
Let $P_n$ be the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty a_n$.
Let $Q_n$ be the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty b_n$.
Then $P_n Q_n$ is the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty a_n b_n$.
By Limit of Product of Sequences, $P_n Q_n\to ab$.
Thus $\ds \prod_{n \mathop = 1}^\infty a_n b_n$ converges to $a b$.
{{qed}}
Theorem: Trimorphic Number is not necessarily Automorphic

Let $n \in \Z_{>0}$ be a trimorphic number.
Then it is not necessarily the case that $n$ is also an automorphic number.

Proof:
Take as an example $n = 49$.
We have that:
:$49^3 = 117 \, 6 \mathbf{49}$
demonstrating that $49$ is trimorphic.
However, we also have that:
:$49^2 = 2401$
demonstrating that $49$ is not automorphic.
{{qed}}
Theorem: Reciprocal of 49 shows Powers of 2 in Decimal Expansion

The decimal expansion of the reciprocal of $49$ contains the powers of $2$:
:$\dfrac 1 {49} = 0 \cdotp \dot 02040 \, 81632 \, 65306 \, 12244 \, 89795 \, 91836 \, 73469 \, 38775 \, 5 \dot 1$

Proof:
From Reciprocal of $49$:
{{:Reciprocal of 49}}
Adding up powers of $2$, shifted appropriately to the right:
<pre>
02
  04
    08
      16
        32
          64
           128
             256
               512
                1024
                  2048
                    4096
                      8192
                       16384
                         32768
                           65536
                            131072
                              262144
                                524288
                                 1048576
                                   2097152
                                     4194304
                                       8388608
                                        16777216
                                          33554432
                                            67108864
                                             134217728
                                               268435456
                                                 536870912
                                                  1073741824
                                                    2147483648
                                                      4294967296
                                                        8589934592
                                                         ...........
--------------------------------------------------------------------
020408163265306122448979591836734693877551020408163265305947144192
--------------------------------------------------------------------
                     1 1 1111 11  11212122222112222222322222111
</pre>
As can be seen, the decimal expansion of $\dfrac 1 {49}$ matches the sum of the shifted powers of $2$ (to the limits of calculation) and the pattern is apparent.
This is what is to be expected, because:
:$0 \cdotp 02 + 0 \cdotp 0004 + 0 \cdotp 000008 + 0 \cdotp 00000016 + \cdots$
is nothing else but:
:$\dfrac 1 {50} + \paren {\dfrac 1 {50} }^2 + \paren {\dfrac 1 {50} }^3 + \paren {\dfrac 1 {50} }^4 + \cdots = \ds \sum_{k \mathop \ge 1} \paren {\dfrac 1 {50} }^k$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {50} }^k
      | r = \dfrac {1 / 50} {1 - 1 / 50}
      | c = Sum of Infinite Geometric Sequence: Corollary 1
}}
{{eqn | r = \dfrac 1 {50 - 1}
      | c = 
}}
{{eqn | r = \dfrac 1 {49}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Product of Absolutely Convergent Products is Absolutely Convergent

Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let $\ds \prod_{n \mathop = 1}^\infty a_n$ converge absolutely.
Let $\ds \prod_{n \mathop = 1}^\infty b_n$ converge absolutely.
Then $\ds \prod_{n \mathop = 1}^\infty a_nb_n$ converges absolutely.

Proof:
We have:
:$a_n b_n - 1 = \paren {a_n - 1} \paren {b_n - 1} + \paren {a_n - 1} + \paren {b_n - 1}$
By the Triangle Inequality:
:$\norm {a_n b_n - 1} \le \norm {a_n - 1} \norm {b_n - 1} + \norm {a_n - 1} + \norm {b_n -1}$
By the absolute convergence, $\ds \sum_{n \mathop = 1}^\infty \norm {a_n - 1}$ and $\ds \sum_{n \mathop = 1}^\infty \norm {b_n - 1}$ converge.
By Inner Product of Absolutely Convergent Series, $\ds \sum_{n \mathop = 1}^\infty \norm {a_n - 1} \norm{b_n -1}$ converges.
By the Comparison Test, $\ds \sum_{n \mathop = 1}^\infty \norm {a_n b_n - 1}$ converges.
{{qed}}
Theorem: Convergent Product Satisfies Cauchy Criterion

Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ be convergent.
Then it satisfies Cauchy's criterion for products.

Proof:
Let $\epsilon > 0$.
Let $n_0 \in \N$ be such that $\ds \prod_{n \mathop = n_0}^\infty a_n$ converges to some $a \in \mathbb K \setminus \set 0$.
By Convergent Sequence is Cauchy Sequence, there exists $N_0 \ge n_0$ such that:
:$\ds \norm {\prod_{n \mathop = n_0}^k a_n - \prod_{n \mathop = n_0}^l a_n} \le \epsilon$
for $k, l \ge N_0$.
By Sequence Converges to Within Half Limit, there exists $N_1 \ge n_0$ such that:
:$\ds \norm {\prod_{n \mathop = n_0}^M a_n} \ge \frac {\norm a}2$
for $M \ge N_1$.
Let $N = \max \set {N_0, N_1}$.
For $N + 1 \le k \le l$:
{{begin-eqn}}
{{eqn | l = \norm {\prod_{n \mathop = k}^l a_n - 1}
      | r = \norm {\frac {\prod_{n \mathop = n_0}^l a_n - \prod_{n \mathop = n_0}^{k - 1} a_n} {\prod_{n \mathop = n_0}^{k - 1} } }
      | c = 
}}
{{eqn | o = \le
      | r = \frac {2 \epsilon} {\norm{a} }
      | c = $l, k - 1 \ge N_0$ and $k - 1 \ge N_1$
}}
{{end-eqn}}
Hence the result.
{{qed}}