avewright/proofwiki-math huggingface dataset:
Theorem: Polynomial Factor Theorem

Let $\map P x$ be a polynomial in $x$ over a field $K$ of degree $n$.
Then:
:$\xi \in K: \map P \xi = 0 \iff \map P x = \paren {x - \xi} \map Q x$
where $Q$ is a polynomial of degree $n - 1$.
Hence, if $\xi_1, \xi_2, \ldots, \xi_n \in K$ such that all are different, and $\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$, then:
:$\ds \map P x = k \prod_{j \mathop = 1}^n \paren {x - \xi_j}$
where $k \in K$.

Proof:
Let $P = \paren {x - \xi} Q$.
Then:
:$\map P \xi = \map Q \xi \cdot 0 = 0$
Conversely, let $\map P \xi = 0$.
By the Division Theorem for Polynomial Forms over Field, there exist polynomials $Q$ and $R$ such that:
:$P = \paren {x - \xi} Q + R$
and:
:$\map \deg R < \map \deg {x - \xi} = 1$
Evaluating at $\xi$ we have:
:$0 = \map P \xi = \map R \xi$
But:
:$\deg R = 0$
so:
:$R \in K$
In particular:
:$R = 0$
Thus:
:$P = \paren {x - \xi} Q$
as required.
The fact that $\map \deg Q = n - 1$ follows from:
:Ring of Polynomial Forms is Integral Domain
and:
:Degree of Product of Polynomials over Integral Domain.
We can then apply this result to:
:$\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$
We can progressively work through:
:$\map P x = \paren {x - \xi_1} \map {Q_{n - 1} } x$ 
where $\map {Q_{n - 1} } x$ is a polynomial of order $n - 1$.
Then, substituting $\xi_2$ for $x$:
:$0 = \map P {\xi_2} = \paren {\xi_2 - \xi_1} \map {Q_{n - 1} } x$
Since $\xi_2 \ne \xi_1$:
:$\map {Q_{n - 1} } {\xi_2} = 0$
and we can apply the above result again:
:$\map {Q_{n - 1} } x = \paren {x - \xi_2} \map {Q_{n - 2} } x$
Thus:
:$\map P x = \paren {x - \xi_1} \paren {x - \xi_2} \map {Q_{n - 2} } x$
and we then move on to consider $\xi_3$.
Eventually we reach:
:$\map P x = \paren {x - \xi_1} \paren {x - \xi_2} \dotsm \paren {x - \xi_n} \map {Q_0} x$
$\map {Q_0} x$ is a polynomial of zero degree, that is a constant polynomial.
The result follows.
{{qed}}
Theorem: Telescoping Series/Example 1

Let $\sequence {b_n}$ be a sequence in $\R$.
Let $\sequence {a_n}$ be a sequence whose terms are defined as:
:$a_k = b_k - b_{k + 1}$
Then:
:$\ds \sum_{k \mathop = 1}^n a_k = b_1 - b_{n + 1}$

Proof:
{{begin-eqn}}
{{eqn | l = \ds \sum_{k \mathop = 1}^n a_k
      | r = \sum_{k \mathop = 1}^n \paren {b_k - b_{k + 1} }
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 1}^n b_{k + 1}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 2}^{n + 1} b_k
      | c = Translation of Index Variable of Summation
}}
{{eqn | r = b_1 + \sum_{k \mathop = 2}^n b_k - \sum_{k \mathop = 2}^n b_k - b_{n + 1}
      | c = 
}}
{{eqn | r = b_1 - b_{n + 1}
      | c = 
}}
{{end-eqn}}
If $\sequence {b_k}$ converges to zero, then $b_{n + 1} \to 0$ as $n \to \infty$.
Thus:
:$\ds \lim_{n \mathop \to \infty} a_n = b_1 - 0 = b_1$
So:
:$\ds \sum_{k \mathop = 1}^\infty a_k = b_1$
{{Qed}}
Theorem: Terms in Convergent Series Converge to Zero

Let $\sequence {a_n}$ be a sequence in any of the standard number fields $\Q$, $\R$, or $\C$.
Suppose that the series $\ds \sum_{n \mathop = 1}^\infty a_n$ converges in any of the standard number fields $\Q$, $\R$, or $\C$.
Then:
:$\ds \lim_{n \mathop \to \infty} a_n = 0$

Proof:
Let $\ds s = \sum_{n \mathop = 1}^\infty a_n$.
Then $\ds s_N = \sum_{n \mathop = 1}^N a_n \to s$ as $N \to \infty$.
Also, $s_{N - 1} \to s$ as $N \to \infty$.
Thus:
{{begin-eqn}}
{{eqn | l = a_N
      | r = \paren {a_1 + a_2 + \cdots + a_{N - 1} + a_N} - \paren {a_1 + a_2 + \cdots + a_{N - 1} }
      | c = 
}}
{{eqn | r = s_N - s_{N - 1}
      | c = 
}}
{{eqn | o = \to
      | r = s - s = 0 \text{ as } N \to \infty
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Linear Combination of Convergent Series

Let $\sequence {a_n}_{n \mathop \ge 1}$ and $\sequence {b_n}_{n \mathop \ge 1}$ be sequences of real numbers.
Let the two series $\ds\sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ converge to $\alpha$ and $\beta$ respectively.
Let $\lambda, \mu \in \R$ be real numbers.
Then the series $\ds \sum_{n \mathop = 1}^\infty \paren {\lambda a_n + \mu b_n}$ converges to $\lambda \alpha + \mu \beta$.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 1}^N \paren {\lambda a_n + \mu b_n}
      | r = \lambda \sum_{n \mathop = 1}^N a_n + \mu \sum_{n \mathop = 1}^N b_n
      | c = Linear Combination of Indexed Summations
}}
{{eqn | o = \to
      | r = \lambda \alpha + \mu \beta \text{ as } N \to \infty
      | c = Combination Theorem for Sequences
}}
{{end-eqn}}
{{qed}}
Theorem: Abel's Limit Theorem

Let $\ds \sum_{k \mathop = 0}^\infty a_k$ be a convergent series in $\R$.
Then:
:$\ds \lim_{x \mathop \to 1^-} \paren {\sum_{k \mathop = 0}^\infty a_k x^k} = \sum_{k \mathop = 0}^\infty a_k$
where $\ds \lim_{x \mathop \to 1^-}$ denotes the limit from the left.

Proof:
Let $\epsilon > 0$.
Let $\ds \sum_{k \mathop = 0}^\infty a_k$ converge to $s$.
Then its sequence of partial sums $\sequence {s_N}$, where $\ds s_N = \sum_{n \mathop = 1}^N a_n$, is a Cauchy sequence.
So:
:$\ds \exists N: \forall k, m: k \ge m \ge N: \size {\sum_{l \mathop = m}^k a_l} < \frac \epsilon 3$
From Abel's Lemma: Formulation 2, we have:
:$\ds \sum_{k \mathop = m}^n u_k v_k = \sum_{k \mathop = m}^{n - 1} \paren {\paren {\sum_{l \mathop = m}^k u_l} \paren {v_k - v_{k + 1} } } + v_n \sum_{k \mathop = m}^n u_k$
Let $u_k = a_k$ and $v_k = x^k$.
We have:
:$\ds \sum_{k \mathop = m}^n a_k x^k = \sum_{k \mathop = m}^{n - 1} \paren {\paren {\sum_{l \mathop = m}^k a_l} \paren {x^k - x^{k + 1} } } + x^n \sum_{k \mathop = m}^n a_k$
So $\forall n \in \N: n \ge m \ge N$ and $\forall x \in \R: 0 < x < 1$, we have:
{{begin-eqn}}
{{eqn | l = \size {\sum_{k \mathop = m}^n a_k x^k}
      | o = <
      | r = \paren {1 - x} \sum_{k \mathop = m}^{n-1} \frac \epsilon 3 x^k + \frac \epsilon 3 x^n
      | c = replacing instances of $\ds \sum_{l \mathop = m}^k a_l$ with $\dfrac \epsilon 3$
}}
{{eqn | o = <
      | r = \frac \epsilon 3 \paren {1 - x} \frac {1 - x^n} {1 - x} + \frac \epsilon 3 x^n
      | c = Sum of Geometric Progression
}}
{{eqn | r = \frac \epsilon 3
      | c = 
}}
{{end-eqn}}
We conclude that:
:$\ds \size {\sum_{k \mathop = N}^\infty a_k x^k} \le \frac \epsilon 3$
From the above, we have $\forall x \in \R: 0 < x < 1$:
:$\ds \size {\sum_{k \mathop = 0}^\infty a_k x^k - \sum_{k \mathop = 0}^\infty a_k} \le \sum_{k \mathop = 0}^{N - 1} \size {a_n} \paren {1 - x^n} + \frac \epsilon 3 + \frac \epsilon 3$
But for finite $n$, we have:
:$1 - x^n \to 0$ as $x \to 1^-$.
Thus:
:$\ds \sum_{k \mathop = 0}^{N-1} \size {a_n} \paren {1 - x^n} \to 0$ as $x \to 1^-$
So:
:$\ds \exists \delta > 0: \forall x: 1 - \delta < x < 1: \sum_{k \mathop = 0}^{N - 1} \size {a_n} \paren {1 - x^n} < \frac \epsilon 3$
So, for any given $\epsilon > 0$, there exists $\delta > 0$, such that:
:$\forall x \in \R: 1 - \delta < x < 1 \implies \ds \size {\sum_{k \mathop = 0}^\infty a_k x^k - \sum_{k \mathop = 0}^\infty a_k} < \epsilon$
The result follows from the definition for the limit from the left.
{{qed}}
Theorem: Abel's Limit Theorem

Let $\ds \sum_{k \mathop = 0}^\infty a_k$ be a convergent series in $\R$.
Then:
:$\ds \lim_{x \mathop \to 1^-} \paren {\sum_{k \mathop = 0}^\infty a_k x^k} = \sum_{k \mathop = 0}^\infty a_k$
where $\ds \lim_{x \mathop \to 1^-}$ denotes the limit from the left.

Proof:
{{tidy|Please attempt to understand the house style. Seriously, I haven't got the patience to tidy up.}}
{{MissingLinks}}
Since $\ds \sum_{k \mathop = 0}^\infty a_k$ converges and $\cmod{x^k}\le1$ and $\sequence{x^k}$ is decreasing,
by Abel's Test for Uniform Convergence $\ds \sum_{k \mathop = 0}^\infty a_kx^k$ converges uniformly on $[0,1]$ and hence to a continuous function by Uniform Limit Theorem.
Theorem: Alternating Series Test

Let $\sequence {a_n}_{N \mathop \ge 0}$ be a decreasing sequence of positive terms in $\R$ which converges with a limit of zero.
That is, let $\forall n \in \N: a_n \ge 0, a_{n + 1} \le a_n, a_n \to 0$ as $n \to \infty$
Then the series:
:$\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} a_n = a_1 - a_2 + a_3 - a_4 + \dotsb$ 
converges.

Proof:
First we show that for each $n > m$, we have $0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$.
Theorem: Euler's Identity

:$e^{i \pi} + 1 = 0$

Proof:
Follows directly from Euler's Formula $e^{i z} = \cos z + i \sin z$, by plugging in $z = \pi$:
:$e^{i \pi} + 1 = \cos \pi + i \sin \pi + 1 = -1 + i \times 0 + 1 = 0$
{{qed}}
Theorem: Intermediate Value Theorem

Let $I \subseteq \R$ be a real interval.
Let $f: I \to \R$ be continuous on $I$.
Then $f$ is a Darboux function.
That is:
Let $a, b \in I$.
Let $k \in \R$ lie between $\map f a$ and $\map f b$.
That is, either:
:$\map f a < k < \map f b$
or:
:$\map f b < k < \map f a$
Then $\exists c \in \openint a b$ such that $\map f c = k$.

Proof:
As the codomain of $f$ is $\closedint a b$, it follows that the image of $f$ is a subset of $\closedint a b$.
Thus:
:$\map f a \ge a$
and
:$\map f b \le b$
Let us define the real function $g: \closedint a b \to \R$ by:
:$\map g x = \map f x - x$
Then by the Combined Sum Rule for Continuous Real Functions, $\map g x$ is continuous on $\closedint a b$.
But:
:$\map g a \ge 0$ and $\map g b \le 0$
By the Intermediate Value Theorem:
:$\exists \xi \in \openint a b: \map g \xi = 0$
Thus:
:$\map f \xi = \xi$
{{qed}}
Theorem: Intermediate Value Theorem

Let $I \subseteq \R$ be a real interval.
Let $f: I \to \R$ be continuous on $I$.
Then $f$ is a Darboux function.
That is:
Let $a, b \in I$.
Let $k \in \R$ lie between $\map f a$ and $\map f b$.
That is, either:
:$\map f a < k < \map f b$
or:
:$\map f b < k < \map f a$
Then $\exists c \in \openint a b$ such that $\map f c = k$.

Proof:
This theorem is a restatement of Image of Real Interval under Continuous Real Function is Real Interval.
From Image of Real Interval under Continuous Real Function is Real Interval, the image of $\openint a b$ under $f$ is also a real interval (but not necessarily open).
Thus if $k$ lies between $\map f a$ and $\map f b$, it must be the case that:
:$k \in \Img {\openint a b}$
The result follows.
{{qed}}
Theorem: Tail of Convergent Series tends to Zero

Let $\sequence {a_n}_{n \mathop \ge 1}$ be a sequence of real numbers.
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a convergent series.
Let $N \in \N_{\ge 1}$ be a natural number.
Let $\ds \sum_{n \mathop = N}^\infty a_n$ be the tail of the series $\ds \sum_{n \mathop = 1}^\infty a_n$.
Then:
:$\ds \sum_{n \mathop = N}^\infty a_n$ is convergent
:$\ds \sum_{n \mathop = N}^\infty a_n \to 0$ as $N \to \infty$.
That is, the tail of a convergent series tends to zero.

Proof:
Let $\sequence {s_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$.
Let $\sequence {s'_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = N}^\infty a_n$.
It will be shown that $\sequence {s'_n}$ fulfils the Cauchy criterion.
That is:
:$\forall \epsilon \in \R_{>0}: \exists N: \forall m, n > N: \size {s'_n - s'_m} < \epsilon$
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
As $\sequence {s_n}$ is convergent, it conforms to the Cauchy criterion by Convergent Sequence is Cauchy Sequence.
Thus:
:$\exists N: \forall m, n > N: \size {s_n - s_m} < \epsilon$
Now:
{{begin-eqn}}
{{eqn | l = s_n
      | r = \sum_{k \mathop = 1}^n a_k
      | c =  
}}
{{eqn | r = \sum_{k \mathop = 1}^{N - 1} a_k + \sum_{k \mathop = N}^n a_k 
      | c = Indexed Summation over Adjacent Intervals
}}
{{eqn | r = s_{N - 1} + s'_n
}}
{{end-eqn}}
and similarly:
:$s_m = s_{N - 1} + s'_m$
Thus:
:$s'_n = s_n - s_{N - 1}$
and:
:$s'_m = s_m - s_{N - 1}$
So:
{{begin-eqn}}
{{eqn | l = \size {s_n - s_m}
      | o = <
      | r = \epsilon
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {s_n - s_{N - 1} - s_m + s_{N - 1} }
      | o = <
      | r = \epsilon
      | c =
}}
{{eqn | ll= \leadsto
      | l = \size {\paren {s_n - s_{N - 1} } - \paren {s_m - s_{N - 1} } }
      | o = <
      | r = \epsilon
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {\size {s_n - s_{N - 1} } - \size {s_m - s_{N - 1} } }
      | o = <
      | r = \epsilon
      | c = Triangle Inequality
}}
{{eqn | ll= \leadsto
      | l = \size {s'_n - s'_m}
      | o = <
      | r = \epsilon
      | c = 
}}
{{end-eqn}}
So $\ds \sum_{n \mathop = N}^\infty a_n$ fulfils the Cauchy criterion.
By Convergent Sequence is Cauchy Sequence it follows that it is convergent.
Now it is shown that $\ds \sum_{n \mathop = N}^\infty a_n \to 0$ as $N \to \infty$.
We have that $\sequence {s_n}$ is convergent,
Let its limit be $l$.
Thus we have:
:$\ds l = \sum_{n \mathop = 1}^\infty a_n = s_{N - 1} + \sum_{n \mathop = N}^\infty a_n$
So:
:$\ds \sum_{n \mathop = N}^\infty a_n = l - s_{N - 1}$
But $s_{N - 1} \to l$ as $N - 1 \to \infty$.
The result follows.
{{qed}}
Theorem: Comparison Test

Let $\ds \sum_{n \mathop = 1}^\infty b_n$ be a convergent series of positive real numbers.
Let $\sequence {a_n}$ be a sequence $\R$ or sequence in $\C$.
Let $\forall n \in \N_{>0}: \cmod {a_n} \le b_n$.
Then the series $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.

Proof:
Let $\epsilon > 0$.
As $\ds \sum_{n \mathop = 1}^\infty b_n$ converges, its tail tends to zero.
So:
:$\ds \exists N: \forall n > N: \sum_{k \mathop = n + 1}^\infty b_k < \epsilon$
Let $\sequence {s_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$.
Then $\forall n > m > N$:
{{begin-eqn}}
{{eqn | l = \cmod {s_n - s_m}
      | r = \cmod {\paren {a_1 + a_2 + \cdots + a_n} - \paren {a_1 + a_2 + \cdots + a_m} }
      | c = 
}}
{{eqn | r = \cmod {a_{m + 1} + a_{m + 2} + \cdots + a_n}
      | c = Indexed Summation over Adjacent Intervals
}}
{{eqn | o = \le
      | r = \cmod {a_{m + 1} } + \cmod {a_{m + 2} } + \cdots + \cmod {a_n}
      | c = Triangle Inequality for Indexed Summations
}}
{{eqn | o = \le
      | r = b_{m + 1} + b_{m + 2} + \cdots + b_n
      | c = 
}}
{{eqn | o = \le
      | r = \sum_{k \mathop = n + 1}^\infty b_k
}}
{{eqn | o = <
      | r = \epsilon
}}
{{end-eqn}}
So $\sequence {s_n}$ is a Cauchy sequence.
The result follows from Real Number Line is Complete Metric Space or Complex Plane is Complete Metric Space.
{{qed}}
Theorem: Ratio Test

Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers in $\R$, or a series of complex numbers in $\C$.
Let the sequence $\sequence {a_n}$ satisfy:
:$\ds \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} } = l$
:If $l > 1 $, then $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
:If $l < 1 $, then $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.

Proof:
From the statement of the theorem, it is necessary that $\forall n: a_n \ne 0$; otherwise $\size {\dfrac {a_{n + 1} } {a_n} }$ is not defined.
Here, $\size {\dfrac {a_{n + 1} } {a_n} }$ denotes either the absolute value of $\dfrac {a_{n + 1} } {a_n}$, or the complex modulus of $\dfrac {a_{n + 1} } {a_n}$.
Theorem: Limit of Subsequence of Bounded Sequence

Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be bounded.
Let $b \in \R$ be a real number.
Suppose that $\forall N: \exists n > N: x_n \ge b$.
Then $\sequence {x_n}$ has a subsequence which converges to a limit $l \ge b$.

Proof:
Let us pick $N \in \N$.
Then $\exists n_1 > N: x_{n_1} \ge b$.
Again, $\exists n_2 > n_1: x_{n_2} \ge b$.
And so on: for each $n_k$ we find, $\exists n_{k+1} > n_k: x_{n_{k+1}} \ge b$.
In this way we can build a subsequence of $\sequence {x_n}$ each of whose terms are $b$ or bigger.
By the Bolzano-Weierstrass Theorem, this subsequence itself contains a subsequence $\sequence {x_{n_r} }$ which is convergent.
Now, suppose $x_{n_r} \to l$ as $r \to \infty$.
Since $x_{n_r} \ge b$ it follows from Lower and Upper Bounds for Sequences that $l \ge b$.
{{qed}}
Theorem: Terms of Bounded Sequence Within Bounds

Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be bounded.
Let the limit superior of $\sequence {x_n}$ be $\overline l$.
Let the limit inferior of $\sequence {x_n}$ be $\underline l$.
Then:
:$\forall \epsilon > 0: \exists N: \forall n > N: x_n < \overline l + \epsilon$
:$\forall \epsilon > 0: \exists N: \forall n > N: x_n > \underline l - \epsilon$

Proof:
=== Upper Bound ===
First we show that:
:$\forall \epsilon > 0: \exists N: \forall n > N: x_n < \overline l + \epsilon$
{{AimForCont}} this proposition were to be false.
That would mean that for some $\epsilon > 0$ it would be true that for each $N$ we would be able to find $n > N$ such that $x_n \ge \overline l + \epsilon$.
From Limit of Subsequence of Bounded Sequence it would follow that there exists a convergent subsequence whose limit was $l \ge \overline l + \epsilon$.
This would contradict the definition of $\overline l$.
{{qed|lemma}}
Theorem: Convergence of Limsup and Liminf

Let $\sequence {x_n}$ be a sequence in $\R$.
Let the limit superior of $\sequence {x_n}$ be $\overline l$.
Let the limit inferior of $\sequence {x_n}$ be $\underline l$.
Then $\sequence {x_n}$ converges to a limit $l$ {{iff}} $\overline l = \underline l = l$.
Hence a bounded real sequence converges {{iff}} all its convergent subsequences have the same limit.

Proof:
=== Sufficient Condition ===
First, suppose that $\overline l = \underline l = l$.
Let $\epsilon > 0$.
By Terms of Bounded Sequence Within Bounds:
:$\exists N_1: \forall n > N_1: x_n < l + \epsilon$
Similarly:
:$\exists N_2: \forall n > N_2: x_n > l - \epsilon$
So take $N = \max \set {N_1, N_2}$.
If $n > N$, both the above inequalities hold at the same time.
So $l - \epsilon < x_n < l + \epsilon$ and so by Negative of Absolute Value:
:$\size {x_n - l} < \epsilon$
Thus $x_n \to l$ as $n \to \infty$.
{{qed|lemma}}
Theorem: Limsup and Liminf are Limits of Bounds

Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be bounded.
Let $\ds \overline l = \limsup_{n \mathop \to \infty} x_n$ be the limit superior and $\ds \liminf_{n \mathop \to \infty} x_n$ the limit inferior of $\sequence {x_n}$.
Then:
:$\ds \overline l = \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \mathop \ge n} x_k}$
:$\ds \underline l = \liminf_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\inf_{k \mathop \ge n} x_k}$

Proof:
First we show that:
:$\ds \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \mathop \ge n} x_k}$
Let $M_n = \ds \sup_{k \mathop \ge n} x_k$.
By Supremum of Subset, the sequence $\sequence {M_n}$ decreases, for:
:$m \ge n \implies \set {k \in \N: k \ge m} \subseteq \set {k \in \N: k \ge n}$
A lower bound for $\sequence {x_n}$ is also a lower bound for $\sequence {M_n}$.
So from the Monotone Convergence Theorem (Real Analysis) it follows that $\sequence {M_n}$ converges.
Suppose $M_n \to M$ as $n \to \infty$.
From the Bolzano-Weierstrass Theorem there exists a convergent subsequence of $\sequence {x_n}$.
Let $L$ be the set of all numbers which are the limit of some subsequence of $\sequence {x_n}$.
Let $l \in L$ be arbitrary.
Let $\sequence {x_{n_r} }$ be a convergent subsequence of $\sequence {x_n}$ such that $x_{n_r} \to l$ as $r \to \infty$.
Then $\forall n_r \ge n: x_{n_r} \le M_n$.
Hence $l \le M_n$ by Lower and Upper Bounds for Sequences and hence (from the same theorem) $l \le M$.
This is true for all $l \in L$, so $\overline l = \limsup_{n \mathop \to \infty} x_n \le M$.
Now, from Terms of Bounded Sequence Within Bounds, we have that:
:$\forall \epsilon > 0: \exists n: \forall k \ge n: x_k < \overline l + \epsilon$
Thus $\overline l + \epsilon$ is an upper bound for $\set {x_k: k \ge n}$.
So $M \le M_n \le \overline l + \epsilon$.
Thus from Real Plus Epsilon, $M \le \overline l$.
Thus we conclude that $M = \overline l$ and hence:
:$\ds \overline l = \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \ge n} x_k}$
{{handwaving|"similar argument"}}
:$\ds \liminf_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop to \infty} } {\inf_{k \mathop \ge n} x_k}$
can be proved using a similar argument.
{{qed}}
Theorem: Nth Root Test

Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers $\R$ or complex numbers $\C$.
Let the sequence $\sequence {a_n}$ be such that the limit superior $\ds \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = l$.
Then:
:If $l > 1$, the series $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
:If $l < 1$, the series $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.

Proof:
=== Absolute Convergence ===
Let $l < 1$.
Then let us choose $\epsilon > 0$ such that $l + \epsilon < 1$.
Consider the real sequence $\sequence {b_n}$ defined by $\sequence {b_n} = \sequence {\size {a_n} }$.
Here, $\size {a_n}$ denotes either the absolute value of $a_n$, or the complex modulus of $a_n$.
Then:
:$\ds l = \limsup_{n \mathop \to \infty} {b_n}^{1/n}$
It follows from Terms of Bounded Sequence Within Bounds that for sufficiently large $n$,:
:$b_n < \paren {l + \epsilon}^n$
By Sum of Infinite Geometric Sequence, the series $\ds \sum_{n \mathop = 1}^\infty \paren {l + \epsilon}^n$ converges.
By the comparison test, $\ds \sum_{n \mathop = 1}^\infty b_n$ converges.
Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely by the definition of absolute convergence.
{{qed|lemma}}
Theorem: Series of Power over Factorial Converges

The series $\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges for all real values of $x$.

Proof:
If $x = 0$ the result is trivially true as:
:$\forall n \ge 1: \dfrac {0^n} {n!} = 0$
If $x \ne 0$ we have:
:$\size {\dfrac {\paren {\dfrac {x^{n + 1} } {\paren {n + 1}!} } } {\paren {\dfrac {x^n} {n!} } } } = \dfrac {\size x} {n + 1} \to 0$
as $n \to \infty$.
This follows from the results:
:Sequence of Powers of Reciprocals is Null Sequence, where $\dfrac 1 n \to 0$ as $n \to \infty$
:The Squeeze Theorem for Real Sequences, as $\dfrac 1 {n + 1} < \dfrac 1 n$
:The Multiple Rule for Real Sequences, putting $\lambda = \size x$.
Hence by the Ratio Test: $\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ converges.
{{qed}}
Alternatively, the Comparison Test could be used but this is more cumbersome in this instance.
Another alternative is to view this as an example of Radius of Convergence of Power Series over Factorial setting $\xi = 0$.
Theorem: Area of Triangle in Terms of Side and Altitude

The area of a triangle $\triangle ABC$ is given by:
:$\dfrac {c \cdot h_c} 2 = \dfrac {b \cdot h_b} 2 = \dfrac {a \cdot h_a} 2$
where:
:$a, b, c$ are the sides
:$h_a, h_b, h_c$ are the altitudes from $A$, $B$ and $C$ respectively.

Proof:
:400px
Construct a point $D$ so that $\Box ABDC$ is a parallelogram.
From Opposite Sides and Angles of Parallelogram are Equal:
:$\triangle ABC \cong \triangle DCB$
hence their areas are equal.
The Area of Parallelogram is equal to the product of one of its bases and the associated altitude.
Thus 
{{begin-eqn}}
{{eqn | l = \paren {ABCD}
      | r = c \cdot h_c
}}
{{eqn | ll= \leadsto
      | l = 2 \paren {ABC}
      | r = c \cdot h_c
      | c = because congruent surfaces have equal areas
}}
{{eqn | l = \paren {ABC}
      | r = \frac {c \cdot h_c} 2
}}
{{end-eqn}}
where $\paren {XYZ}$ is the area of the plane figure $XYZ$.
A similar argument can be used to show that the statement holds for the other sides.
{{qed}}