avewright/proofwiki-math huggingface dataset:
Theorem: One Plus Reciprocal to the Nth

Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = \paren {1 + \dfrac 1 n}^n$.
Then $\sequence {x_n}$ converges to a limit as $n$ increases without bound.

Proof:
First we show that $\sequence {x_n}$ is increasing.
Let $a_1 = a_2 = \cdots = a_{n - 1} = 1 + \dfrac 1 {n - 1}$.
Let $a_n = 1$.
Let:
:$A_n$ be the arithmetic mean of $a_1 \ldots a_n$
:$G_n$ be the geometric mean of $a_1 \ldots a_n$
Thus:
:$A_n = \dfrac {\paren {n - 1} \paren {1 + \dfrac 1 {n - 1} } + 1} n = \dfrac {n + 1} n = 1 + \dfrac 1 n$
:$G_n = \paren {1 + \dfrac 1 {n - 1} }^{\dfrac {n - 1} n}$
By Cauchy's Mean Theorem:
: $G_n \le A_n$
Thus:
:$\paren {1 + \dfrac 1 {n - 1} }^{\frac {n - 1} n} \le 1 + \dfrac 1 n$
and so:
:$x_{n - 1} = \paren {1 + \dfrac 1 {n - 1} }^{n - 1} \le \paren {1 + \dfrac 1 n}^n = x_n$
Hence $\sequence {x_n}$ is increasing.
Next, we show that $\sequence {x_n}$ is bounded above.
Using the Binomial Theorem:
{{begin-eqn}}
{{eqn | l = \paren {1 + \frac 1 n}^n
      | r = 1 + n \paren {\frac 1 n} + \frac {n \paren {n - 1} } 2 \paren {\frac 1 n}^2 + \cdots + \paren  {\frac 1 n}^n
}}
{{eqn | r = 1 + 1 + \paren {1 - \frac 1 n} \frac 1 {2!} + \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \frac 1 {3!} + \cdots + \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {n - 1} n} \frac 1 {n!}
}}
{{eqn | o = \le
      | r = 1 + 1 + \frac 1 {2!} + \frac 1 {3!} + \cdots + \frac 1 {n!}
}}
{{eqn | o = \le
      | r = 1 + 1 + \frac 1 2 + \frac 1 {2^2} + \cdots + \frac 1 {2^n}
      | c = (because $2^{n - 1} \le n!$)
}}
{{eqn | r = 1 + \frac {1 - \paren {\frac 1 2}^n} {1 - \frac 1 2}
}}
{{eqn | r = 1 + 2 \paren {1 - \paren {\frac 1 2}^n}
}}
{{eqn | o = <
      | r = 3
}}
{{end-eqn}}
So $\sequence {x_n}$ is bounded above by $3$.
From the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {x_n}$ converges to a limit.
Theorem: Between two Real Numbers exists Rational Number

Let $a, b \in \R$ be real numbers such that $a < b$.
Then:
:$\exists r \in \Q: a < r < b$

Proof:
Suppose that $a \ge 0$.
As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.
Thus:
:$\dfrac 1 {b - a} \in \R$
By the Axiom of Archimedes:
:$\exists n \in \N: n > \dfrac 1 {b - a}$
Let $M := \set {x \in \N: \dfrac x n > a}$.
By the Well-Ordering Principle, there exists $m \in \N$ such that $m$ is the smallest element of $M$.
That is:
:$m > a n$
and, by definition of smallest element:
:$m - 1 \le a n$
As $n > \dfrac 1 {b - a}$, it follows from Ordering of Reciprocals that:
:$\dfrac 1 n < b - a$
Thus:
{{begin-eqn}}
{{eqn | l = m - 1
      | o = \le
      | r = a n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = m
      | o = \le
      | r = a n + 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac m n
      | o = \le
      | r = a + \frac 1 n
      | c = 
}}
{{eqn | o = <
      | r = a + \paren {b - a}
      | c = 
}}
{{eqn | r = b
      | c = 
}}
{{end-eqn}}
Thus we have shown that $a < \dfrac m n < b$.
That is:
: $\exists r \in \Q: a < r < b$
such that $r = \dfrac m n$.
Now suppose $a < 0$.
If $b > 0$ then $0 = r$ is a rational number such that $a < r < b$.
Otherwise we have $a < b \le 0$.
Then $0 \le -b < -a$ and there exists $r \in \Q$ such that:
:$-b < r < -a$
where $r$ can be found as above.
That is:
:$a < -r < b$
All cases have been covered, and the result follows.
{{qed}}
Theorem: Between two Real Numbers exists Rational Number

Let $a, b \in \R$ be real numbers such that $a < b$.
Then:
:$\exists r \in \Q: a < r < b$

Proof:
As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.
Thus:
:$\dfrac 1 {b - a} \in \R$
By the Axiom of Archimedes:
:$\exists n \in \N: n > \dfrac 1 {b - a}$
Let $M := \set {x \in \Z: x > a n}$.
By Set of Integers Bounded Below has Smallest Element, there exists $m \in \Z$ such that $m$ is the smallest element of $M$.
That is:
:$m > a n$
and, by definition of smallest element:
:$m - 1 \le a n$
As $n > \dfrac 1 {b - a}$, it follows from Ordering of Reciprocals that:
:$\dfrac 1 n < b - a$
Thus:
{{begin-eqn}}
{{eqn | l = m - 1
      | o = \le
      | r = a n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = m
      | o = \le
      | r = a n + 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac m n
      | o = \le
      | r = a + \frac 1 n
      | c = 
}}
{{eqn | o = <
      | r = a + \paren {b - a}
      | c = 
}}
{{eqn | r = b
      | c = 
}}
{{end-eqn}}
Thus we have shown that $a < \dfrac m n < b$.
That is:
:$\exists r \in \Q: a < r < b$
such that $r = \dfrac m n$.
{{qed}}
Theorem: Power over Factorial

Let $x \in \R: x > 0$ be a positive real number.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = \dfrac {x^n} {n!}$.
Then $\sequence {x_n}$ converges to zero.

Proof:
We need to show that $x_n \to 0$ as $n \to \infty$.
Let $N \in \N$ be the smallest natural number which satisfies $N > x$.
(From the Axiom of Archimedes, such an $N$ always exists.)
First we show that:
: $\forall n > N: \dfrac {x^n} {n!} \le \dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{n - N + 1}$
Note that as $N > x$, $\dfrac x N < 1$.
Also:
:$m > n \implies \dfrac x m < \dfrac x n$
Thus:
{{begin-eqn}}
{{eqn | l = \frac {x^n} {n!}
      | r = \frac x 1 \frac x 2 \cdots \frac x {N - 1} \frac x N \frac x {N + 1} \cdots \frac x n
      | c = 
}}
{{eqn | r = \frac {x^{N - 1} } {\paren {N - 1}!} \frac x N \frac x {N + 1} \cdots \frac x n
      | c = 
}}
{{eqn | o = \le
      | r = \frac {x^{N - 1} } {\paren {N - 1}!} \paren {\frac x N}^{n - N + 1}
      | c = as $\dfrac x {N + 1}, \dfrac x {N + 2}, \ldots, \dfrac x n < \dfrac x N$
}}
{{eqn | r = \frac {x^{N - 1} } {\paren {N - 1}!} \paren {\frac x N}^{1 - N} \paren {\frac x N}^n
      | c = 
}}
{{end-eqn}}
As $\dfrac x N < 1$, it follows from Sequence of Powers of Number less than One that $\paren {\dfrac x N}^n \to 0$ as $n \to \infty$.
For a given $x$ and $N$, $\dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N}$ is constant.
Thus by the Multiple Rule for Real Sequences:
:$\dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N} \paren {\dfrac x N}^n \to 0$ as $n \to \infty$
As (from above):
:$\dfrac {x^n} {n!} \le \dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N} \paren {\dfrac x N}^n$
the result follows from the Squeeze Theorem for Real Sequences.
{{qed}}
Theorem: Sequence of Powers of Reciprocals is Null Sequence

Let $r \in \Q_{>0}$ be a strictly positive rational number.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as:
:$x_n = \dfrac 1 {n^r}$
Then $\sequence {x_n}$ is a null sequence.

Proof:
Let $\epsilon \in \R_{>0}$.
We need to show that:
:$\exists N \in \N: n > N \implies \size {\dfrac 1 {n^r} } < \epsilon$
That is, that $n^r > 1 / \epsilon$.
Let us choose $N = \ceiling {\paren {1 / \epsilon}^{1/r} }$.
By Reciprocal of Strictly Positive Real Number is Strictly Positive and power of positive real number is positive, it follows that:
:$\paren {\dfrac 1 \epsilon}^{1/r} \gt 0$
Then by Positive Power Function on Non-negative Reals is Strictly Increasing:
:$\forall n > N: n^r > N^r \ge 1 / \epsilon$
{{qed}}
Theorem: Euler's Number: Limit of Sequence implies Limit of Series

Let Euler's number $e$ be defined as:
:$\ds e := \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n$
Then:
:$\ds e = \sum_{k \mathop = 0}^\infty \frac 1 {k!}$
That is:
:$e = \dfrac 1 {0!} + \dfrac 1 {1!} + \dfrac 1 {2!} + \dfrac 1 {3!} + \dfrac 1 {4!} \cdots$

Proof:
We expand $\paren {1 + \dfrac 1 n}^n$ by the Binomial Theorem, using that $\dfrac {n - k} n = 1 - \dfrac k n$:
{{begin-eqn}}
{{eqn | l = \paren {1 + \frac 1 n}^n
      | r = 1 + n \paren {\frac 1 n} + \frac {n \paren {n - 1} } 2 \paren {\frac 1 n}^2 + \cdots + \paren {\frac 1 n}^n
      | c =
}}
{{eqn | r = \frac 1 {0!} + \frac 1 {1!} + \paren {1 - \frac 1 n} \frac 1 {2!} + \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \frac 1 {3!} + \cdots + \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {n - 1} n} \frac 1 {n!}
      | c =
}}
{{end-eqn}}
Take one of the terms in the above:
:$x = \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \cdots \paren {1 - \dfrac {k - 1} n} \dfrac 1 {k!}$
From Sequence of Powers of Reciprocals is Null Sequence, $\dfrac 1 n \to 0$ as $n \to \infty$.
From the Combination Theorem for Sequences:
:$\forall \lambda \in \R: \dfrac \lambda n \to 0$ as $n \to \infty$
:$\forall \lambda \in \R: 1 - \dfrac \lambda n \to 1$ as $n \to \infty$
:$x = \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \cdots \paren {1 - \dfrac {k - 1} n} \dfrac 1 {k!} \to \dfrac 1 {k!}$ as $n \to \infty$
Hence:
:$\ds \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n = \frac 1 {0!} + \frac 1 {1!} + \frac 1 {2!} + \frac 1 {3!} + \cdots = \sum_{k \mathop = 0}^\infty \frac 1 {k!}$
{{qed}}
Theorem: Euler's Number: Limit of Sequence implies Limit of Series

Let Euler's number $e$ be defined as:
:$\ds e := \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n$
Then:
:$\ds e = \sum_{k \mathop = 0}^\infty \frac 1 {k!}$
That is:
:$e = \dfrac 1 {0!} + \dfrac 1 {1!} + \dfrac 1 {2!} + \dfrac 1 {3!} + \dfrac 1 {4!} \cdots$

Proof:
It will be shown that:
:$\ds \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n = \sum_{k \mathop = 0}^\infty \frac 1 {k!}$
Let $t_n := \paren {1 + \dfrac 1 n}^n$
Then:
:$t_n = \dfrac 1 {0!} + \dfrac 1 {1!} + \paren {1 - \dfrac 1 n} \dfrac 1 {2!} + \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \dfrac 1 {3!} + \cdots + \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \cdots \paren {1 - \dfrac {n-1} n} \dfrac 1 {n!}$
Now let:
:$\ds s_m := \sum_{k \mathop = 0}^m \frac 1 {k!}$
We have that:
:$\forall n: t_n \le s_n$
Hence:
:$\limsup \paren {t_n} \le e$
Now, for all $m$, for $n \ge m$:
:$t_n \ge \dfrac 1 {0!} + \dfrac 1 {1!} + \paren {1 - \dfrac 1 n} \dfrac 1 {2!} + \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \dfrac 1 {3!} + \cdots + \paren {1 - \dfrac 1 n} \paren {1 - \dfrac 2 n} \cdots \paren {1 - \dfrac {m - 1} n} \dfrac 1 {m!}$
Hence, for all $m$, we have the right side as being a sequence in $n$, and then:
:$\ds \liminf \paren {t_n} \ge \sum_{k \mathop = 0}^m \frac 1 {m!}$
Since this is true for all $m$:
:$\liminf \paren {t_n} \ge e$
So $\ds \lim \paren {t_n}$ exists and is equal to $e$.
{{qed}}
Theorem: Degree Equation

Let $E$, $K$ and $F$ be fields.
Let $E / K$ and $K / F$ be finite field extensions.
Then:
:$E / F$ is a finite field extension
:$\index E F = \index E K \index K F$
where $\index E F$ denotes the degree of $E / F$

Proof:
First, note that $E / F$ is a field extension as $F \subseteq K \subseteq E$.
Suppose that $\index E K = m$ and $\index K F = n$.
Let $\alpha = \set {a_1, \ldots, a_m}$ be a basis of $E / K$, and $\beta = \set {b_1, \ldots, b_n}$ be a basis of $K / F$.
We wish to prove that the set:
:$\gamma = \set {a_i b_j: 1 \le i \le m, 1 \le j \le n}$
is a basis of $E / F$.
As $\alpha$ is a basis of $E / K$, we have, for all $c \in E$:
:$\ds c = \sum_{i \mathop = 1}^m c_i a_i$, for some $c_i \in K$.
Define $\ds b := \sum_{j \mathop = 1}^n b_j$ and $d_i := \dfrac {c_i} b$.
Note $b \ne 0$ since $\beta$ is linearly independent over $F$, and $d_i \in K$ since $b, c_i \in K$.
Now we have:
:$\ds c = \sum_{i \mathop = 1}^m \frac {c_i} b \cdot b \cdot a_i = \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n d_i a_i b_j$
Thus $\gamma$ is seen to be a spanning set of $E / F$.
To show $\gamma$ is linearly independent, suppose that for some $c_{i j} \in F$:
:$\ds \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n c_{i j} a_i b_j = 0$
Then we have (as fields are commutative rings):
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n c_{i j} a_i b_j
      | r = 0
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall i \in \set {1, \ldots, m}
      | l = \sum_{j \mathop = 1}^n c_{i j} b_j
      | r = 0
      | c = {{Defof|Linearly Independent Set}}: $\alpha$ over $K$
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall i \in \set {1, \ldots, m}, j \in \set {1, \ldots, n}
      | l = c_{i j}
      | r = 0
      | c = {{Defof|Linearly Independent Set}}: $\beta$ over $F$
}}
{{end-eqn}}
Hence $\gamma$ is a linearly independent spanning set; thus it is a basis.
Recalling the definition of $\gamma$ as $\set {a_i b_j: 1 \le i \le m, 1 \le j \le n}$, we have:
:$\size {\gamma} = m n = \index E K \index K F$
as was to be shown.
{{qed}}
Theorem: Convergent Sequence in Metric Space has Unique Limit

Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $M$.
Then $\sequence {x_n}$ can have at most one limit in $M$.

Proof:
Suppose $\ds \lim_{n \mathop \to \infty} x_n = l$ and $\ds \lim_{n \mathop \to \infty} x_n = m$.
Let $\epsilon > 0$.
Then, provided $n$ is sufficiently large:
{{begin-eqn}}
{{eqn | l = \map d {l, m}
      | o = \le
      | r = \map d {l, x_n} + \map d {x_n, m}
      | c = Triangle Inequality
}}
{{eqn | o = <
      | r = \epsilon + \epsilon
      | c = {{Defof|Limit of Sequence (Metric Space)}}
}}
{{eqn | r = 2 \epsilon
}}
{{end-eqn}}
So $0 \le \dfrac {\map d {l, m} } 2 < \epsilon$.
This holds for ''any'' value of $\epsilon > 0$.
Thus from Real Plus Epsilon it follows that $\dfrac {\map d {l, m} } 2 = 0$, that is, that $l = m$.
{{qed}}
Theorem: Convergent Sequence in Metric Space has Unique Limit

Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $M$.
Then $\sequence {x_n}$ can have at most one limit in $M$.

Proof:
We have that a Metric Space is Hausdorff.
The result then follows from Convergent Sequence in Hausdorff Space has Unique Limit.
{{qed}}
Theorem: Lower and Upper Bounds for Sequences

Let $\sequence {x_n}$ be a sequence in $\R$.
Let $x_n \to l$ as $n \to \infty$.
Then:
:$(1): \quad \forall n \in \N: x_n \ge a \implies l \ge a$
:$(2): \quad \forall n \in \N: x_n \le b \implies l \le b$

Proof:
$(1): \quad \forall n \in \N: x_n \ge a \implies l \ge a$:
Let $\epsilon > 0$.
Then:
:$\exists N \in \N: n > N \implies \size {x_n - l} < \epsilon$
So from Negative of Absolute Value:
:$l - \epsilon < x_n < l + \epsilon$
But $x_n \ge a$, so:
:$a \le x_n < l + \epsilon$
Thus, for ''any'' $\epsilon > 0$:
:$a < l + \epsilon$
From Real Plus Epsilon it follows that $a \le l$.
{{qed|lemma}}
$(2): \quad \forall n \in \N: x_n \le b \implies l \le b$:
If $x_n \le b$ it follows that $-x_n \ge -b$ and the above result can be used.
{{qed}}
Theorem: Divergent Sequence may be Bounded

While every Convergent Sequence is Bounded, it does not follow that every bounded sequence is convergent.
That is, there exist bounded sequences which are divergent.

Proof:
Let $\sequence {x_n}$ be the sequence in $\R$ which forms the basis of Grandi's series, defined as:
:$x_n = \paren {-1}^n$
It is clear that $\sequence {x_n}$ is bounded: above by $1$ and below by $-1$.
{{AimForCont}} $x_n \to l$ as $n \to \infty$.
Let $\epsilon > 0$.
Then $\exists N \in \R: \forall n > N: \size {\paren {-1}^n - l} < \epsilon$.
But there are values of $n > N$ for which $\paren {-1}^n = \pm 1$.
It follows that $\size {1 - l} < \epsilon$ and $\size {-1 - l} < \epsilon$.
From the Triangle Inequality for Real Numbers, we have:
{{begin-eqn}}
{{eqn | l = 2
      | r = \size {1 - \paren {-1} }
      | c = 
}}
{{eqn | o = \le
      | r = \size {1 - l} + \size {l - \paren {-1} }
      | c = 
}}
{{eqn | o = <
      | r = 2 \epsilon
      | c = 
}}
{{end-eqn}}
This is a contradiction whenever $\epsilon < 1$.
Thus $\sequence {x_n}$ has no limit and, while definitely bounded, is unmistakably divergent.
{{qed}}
Theorem: Divergent Sequence may be Bounded

While every Convergent Sequence is Bounded, it does not follow that every bounded sequence is convergent.
That is, there exist bounded sequences which are divergent.

Proof:
Let $\sequence {x_n}$ be the sequence in $\R$ which forms the basis of Grandi's series,  defined as:
:$x_n = \paren {-1}^n$
It is clear that $\sequence {x_n}$ is bounded: above by $1$ and below by $-1$.
Note the following subsequences of $\sequence {x_n}$:
:$(1): \quad \sequence {x_{n_r} }$ where $\sequence {n_r}$ is the integer sequence defined as $n_r = 2 r$
:$(2): \quad \sequence  {x_{n_s} }$ where $\sequence {n_s}$ is the integer sequence defined as $n_s = 2 s + 1$.
We have that:
:$\sequence {x_{n_r} }$ is the sequence $1, 1, 1, 1, \ldots$
:$\sequence {x_{n_s} }$ is the sequence $-1, -1, -1, -1, \ldots$
So $\sequence {x_n}$ has two subsequences with different limits.
From Limit of Subsequence equals Limit of Real Sequence, that means $\sequence {x_n}$ can not be convergent.
{{qed}}
Theorem: Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism

Let $\struct {R, +_R, \circ}$ and $\struct {S, +_S, *}$ be rings whose zeros are $0_R$ and $0_S$ respectively.
Let $\phi: R \to S$ be a ring homomorphism.
If $R$ is a division ring, then either:
:$(1): \quad \phi$ is a monomorphism (that is, $\phi$ is injective)
:$(2): \quad \phi$ is the zero homomorphism (that is, $\forall a \in R: \map \phi a = 0_S$).

Proof:
We have that:
:The kernel of a homomorphism is an ideal of $R$
:the only ideals of a division ring are trivial.
So $\map \ker \phi = \set {0_R}$ or $R$.  
If $\map \ker \phi = \set {0_R}$, then $\phi$ is injective by Kernel is Trivial iff Monomorphism.
If $\map \ker \phi = R$, $\phi$ is the zero homomorphism by definition.
{{qed}}
Theorem: Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism

Let $\struct {R, +_R, \circ}$ and $\struct {S, +_S, *}$ be rings whose zeros are $0_R$ and $0_S$ respectively.
Let $\phi: R \to S$ be a ring homomorphism.
If $R$ is a division ring, then either:
:$(1): \quad \phi$ is a monomorphism (that is, $\phi$ is injective)
:$(2): \quad \phi$ is the zero homomorphism (that is, $\forall a \in R: \map \phi a = 0_S$).

Proof:
From Surjection by Restriction of Codomain, we can restrict the codomain of $\phi$ and consider the mapping $\phi': R \to \Img R$
As $\phi'$ is now a surjective homomorphism, it is by definition an epimorphism.
Then an Epimorphism from Division Ring to Ring is either null or an isomorphism.
As an isomorphism is by definition injective, the result follows.
{{qed}}
Theorem: Reciprocal of Null Sequence

Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\forall n \in \N: x_n > 0$.
Then:
:$x_n \to 0$ as $n \to \infty$ {{iff}} $\size {\dfrac 1 {x_n} } \to \infty$ as $n \to \infty$

Proof:
Suppose $x_n \to 0$ as $n \to \infty$.
Let $H > 0$.
So $H^{-1} > 0$.
Since $x_n \to 0$ as $n \to \infty$:
:$\exists N: \forall n > N: \size {x_n} < H^{-1}$
That is:
:$\size {\dfrac 1 {x_n} } > H$.
So:
:$\exists N: \forall n > N: \size {\dfrac 1 {x_n} } > H$
and thus:
:$\sequence {\size {\dfrac 1 {x_n} } }$ diverges to $+\infty$.
{{qed|lemma}}
Suppose $\size {\dfrac 1 {x_n} } \to \infty$ as $n \to \infty$.
By reversing the argument above, we see that $x_n \to 0$ as $n \to \infty$.
{{qed}}
Theorem: Modulus of Limit

Let $X$ be one of the standard number fields $\Q, \R, \C$.
Let $\sequence {x_n}$ be a sequence in $X$.
Let $\sequence {x_n}$ be convergent to the limit $l$.
That is, let $\ds \lim_{n \mathop \to \infty} x_n = l$.
Then
:$\ds \lim_{n \mathop \to \infty} \cmod {x_n} = \cmod l$
where $\cmod {x_n}$ is the modulus of $x_n$.

Proof:
By the Triangle Inequality, we have:
:$\cmod {\cmod {x_n} - \cmod l} \le \cmod {x_n - l}$
Hence by the Squeeze Theorem and Convergent Sequence Minus Limit, $\cmod {x_n} \to \cmod l$ as $n \to \infty$.
{{Qed}}
Theorem: Strictly Increasing Sequence of Natural Numbers

Let $\N_{>0}$ be the set of natural numbers without zero:
:$\N_{>0} = \set {1, 2, 3, \ldots}$
Let $\sequence {n_r}$ be a strictly increasing sequence in $\N_{>0}$.
Then:
:$\forall r \in \N_{>0}: n_r \ge r$

Proof:
This is to be proved by induction on $r$.
For all $r \in \N_{>0}$, let $\map P r$ be the proposition $n_r \ge r$.
Theorem: Limit of Subsequence equals Limit of Sequence

Let $T = \struct {S, \tau}$ be a topological space.
Let $\sequence {x_n}$ be a sequence in $T$.
Let $l \in S$ be a limit of $\sequence {x_n}$.
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then $l$ is a limit of $\sequence {x_{n_r} }$
That is, the limit of a convergent sequence in a topological space equals the limit of any subsequence of it.

Proof:
Let $U \in \tau$ be an open set such that $l \in U$.
By definition of convergence, we have:
:$\exists N \in \N: \forall n > N: x_n \in U$.
When $r > N$, we have $n_r > n_N > N$ by Strictly Increasing Sequence of Natural Numbers.
It follows that:
:$\exists N \in \N: \forall r > N: x_{n_r} \in U$.
Therefore, as $U$ was arbitrary, we have established that $l$ is a limit of $\sequence {x_{n_r} }$, by definition of convergence.
{{qed}}
Category:Convergence
Category:Limits of Sequences
Category:Subsequences
4lblkjw63pxky44739nroi0zb78zpws
Theorem: Root of Number Greater than One

Let $x \in \R$ be a real number.
Let $n \in \N_{>0}$ be a natural number such that $n > 0$.
Then:
:$x \ge 1 \implies x^{1/n} \ge 1$
where $x^{1/n}$ is the $n$th root of $x$.

Proof:
Let $y = x^{1/n}$.
From the definition of the $n$th root of $x$, it follows that $x = y^n$.
We will show by induction that $\forall n \in \N_{>0}: y^n \ge 1 \implies y \ge 1$.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$y^n \ge 1 \implies y \ge 1$
Theorem: Limit of Root of Positive Real Number

Let $x \in \R: x > 0$ be a real number.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as:
:$x_n = x^{1 / n}$
Then $x_n \to 1$ as $n \to \infty$.

Proof:
Let us define $a_1 = a_2 = \cdots = a_{n-1} = 1$ and $a_n = x$.
Let $G_n$ be the geometric mean of $a_1, \ldots, a_n$.
Let $A_n$ be the arithmetic mean of $a_1, \ldots, a_n$.
From their definitions:
:$G_n = x^{1/n}$
and:
:$A_n = \dfrac {n - 1 + x} n = 1 + \dfrac{x - 1} n$
From Arithmetic Mean is Never Less than Geometric Mean:
:$x^{1/n} \le 1 + \dfrac{x - 1} n$
That is:
:$x^{1/n} - 1 \le \dfrac{x - 1} n$
There are two cases to consider: $x \ge 1$ and $0 < x < 1$.
Let $x \ge 1$.
From Root of Number Greater than One, it follows that:
:$x^{1/n} \ge 1$
Thus:
:$0 \le x^{1/n} - 1 \le \dfrac 1 n \paren {x - 1}$
But from Sequence of Powers of Reciprocals is Null Sequence:
:$\dfrac 1 n \to 0$ as $n \to \infty$
From the Combination Theorem for Sequences:
:$\dfrac 1 n \paren {x - 1} \to 0$ as $n \to \infty$
Thus by the Squeeze Theorem:
:$x^{1/n} - 1 \to 0$ as $n \to \infty$
Hence, again from the Combination Theorem for Sequences:
:$x^{1/n} \to 1$ as $n \to \infty$
Now let $0 < x < 1$.
Then $x = \dfrac 1 y$ where $y > 1$.
But from the above:
:$y^{1/n} \to 1$ as $n \to \infty$
Hence by the Combination Theorem for Sequences:
:$x^{1/n} = \dfrac 1 {y^{1/n} } \to \dfrac 1 1 = 1$ as $n \to \infty$
{{qed}}
Theorem: Limit of Root of Positive Real Number

Let $x \in \R: x > 0$ be a real number.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as:
:$x_n = x^{1 / n}$
Then $x_n \to 1$ as $n \to \infty$.

Proof:
We consider the case where $x \ge 1$; when $0 < x < 1$ the proof can be completed as for proof 1.
From Root of Number Greater than One:
:$x^{1/n} \ge 1$
Hence $\sequence {x^{1/n} }$ is bounded below by $1$.
Now consider $x^{1/n} / x^{1 / \paren {n + 1} }$:
{{begin-eqn}}
{{eqn | l = \frac {x^{1/n} } {x^{\frac 1 {n + 1} } }
      | r = x^{\frac 1 n - \frac 1 {n + 1} }
      | c = 
}}
{{eqn | r = x^{\frac {n + 1 - n} {n \paren {n + 1} } }
      | c = 
}}
{{eqn | r = x^{\frac 1 {n \paren {n + 1} } }
      | c = 
}}
{{eqn | o = >
      | r = 1
      | c = Root of Number Greater than One
}}
{{end-eqn}}
So:
:$x^{1/n} > x^{\frac 1 {n + 1} }$
and so $\sequence {x^{1 / n} }$ is strictly decreasing.
Hence from the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {x^{1 / n} }$ converges to a limit $l$ and that $l \ge 1$.
Now, since we know that $\sequence {x^{1 / n} }$ is convergent, we can apply Limit of Subsequence equals Limit of Real Sequence.
That is, any subsequence of $\sequence {x^{1 / n} }$ must also converge to $l$.
So we take the subsequence:
:$\sequence {x^{1 / {2 n} } }$
From what has just been shown:
:$x^{1 / {2 n} } \to l$ as $n \to \infty$
Using the Combination Theorem for Sequences, we have:
:$x^{1 / n} = x^{1 / {2 n} } \cdot x^{1 / {2 n} } \to l \cdot l = l^2$ as $n \to \infty$
But a Convergent Real Sequence has Unique Limit, so $l^2 = l$ and so $l = 0$ or $l = 1$.
But $l \ge 1$ and so $l = 1$.
Hence the result.
{{qed}}