hcju/proofwikiqa huggingface dataset:
Let $\sequence {x_n}$ be convergent in the norm $\norm {\,\cdot\,}$ to the limit $l$, then: :$\forall \epsilon \in \R_{\gt 0}: \exists N \in \N : \forall n \ge N: \norm {x_n - l} < \epsilon$ Let $n_1$ satisfy: :$\forall n \ge n_1: \norm {x_n - l} < 1$ Then $\forall n \ge n_1$: {{begin-eqn}} {{eqn | l = \norm {x_n}       | r = \norm {x_n - l + l} }} {{eqn | o = \le       | r = \norm {x_n - l} + \norm l       | c = Norm axiom (N3) (Triangle Inequality) }} {{eqn | o = \le       | r = 1 + \norm l       | c = as $n \ge n_1$ }} {{end-eqn}} Let $K = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_{n_1 - 1} }, 1 + \norm l}$. Then: :$\forall n < n_1: \norm {x_n} \le K$ :$\forall n \ge n_1: \norm {x_n} \le 1 + \norm l \le K$ Thus, by definition, $\sequence {x_n}$ is bounded. {{qed}}
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$. Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$. By the definition of a convergent sequence in a normed division ring, $\sequence {x_n} $ is convergent to the limit $l$ in $\struct {R, d}$. By Convergent Sequence in Metric Space is Bounded, $\sequence {x_n} $ is a bounded sequence in $\struct {R, d}$. By Sequence is Bounded in Norm iff Bounded in Metric, $\sequence {x_n} $ is a bounded sequence in $\struct {R, \norm {\,\cdot\,} }$. {{qed}}
Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,} }$. By modulus of limit in normed division ring, $\sequence {\norm {x_n} }$ is a convergent sequence in $\R$. By Convergent Real Sequence is Bounded, $\sequence {\norm {x_n} }$ is bounded. That is: :$\exists M \in \R_{> 0}: \forall n, \norm {x_n} = \size {\norm {x_n} } \le M$ Thus, by definition, $\sequence {x_n}$ is bounded. {{qed}}
Let $\sequence {x_n}$ be convergent to the limit $l$ in $\struct {R, \norm {\,\cdot\,}}$. By Convergent Sequence is Cauchy Sequence in Normed Division Ring, $\sequence {x_n}$ is a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$. By Cauchy Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is a bounded sequence in $\struct {R, \norm {\,\cdot\,}}$. {{qed}}
=== Proof of $(\text M 1)$ and $(\text M 4)$ === Let $x, y \in R$.  Then $\map d {x, y} = \norm {x - y} \ge 0$, and furthermore: {{begin-eqn}} {{eqn | l = \map d {x, y}       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = \norm {x - y}       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = x - y       | r = \mathbf 0_R       | c = {{NormAxiom|1}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = y }} {{end-eqn}} {{qed|lemma}} === Proof of $(\text M 2)$ === Let $x, y, z \in R$. Then: {{begin-eqn}} {{eqn | l = \map d {x, z}       | r = \norm {x - z} }} {{eqn | r = \norm {x - y + y - z}       | c =  }} {{eqn | o = \le       | r = \norm {x - y} + \norm {y - z}       | c = {{NormAxiom|3}} }} {{eqn | r = \map d {x, y} + \map d {y, z} }} {{end-eqn}} {{qed|lemma}} === Proof of $(\text M 3)$ === Let $x, y \in R$. Then:  {{begin-eqn}} {{eqn | l = \map d {x, y}       | r = \norm {x - y} }} {{eqn | r = \norm {-1 \paren {y - x} } }} {{eqn | r = \norm {-1} \times \norm {y - x}       | c = {{NormAxiom|2}} }} {{eqn | r = \norm {y - x} }} {{eqn | r = \map d {y, x} }} {{end-eqn}} {{qed|lemma}} As $d$ satisfies the metric space axioms, it is a metric. {{qed}}
Let $z \in B_R \paren \xi$. By definition of the open $R$-ball of $\xi$: :$\cmod {z - \xi} < R$ where $\cmod z$ denotes the complex modulus of $z$. By definition of radius of convergence, it follows that $S \paren z$ converges. Suppose $R$ is finite. Let $\epsilon = R - \cmod {z - \xi} > 0$. Now, let $w \in B_R \paren \xi$ be a complex number such that $R - \cmod {w - \xi} = \dfrac \epsilon 2$. Such a $w$ exists because, if at a loss, we can always let $w = \xi + R - \dfrac \epsilon 2$. If $R = +\infty$, then let $w \in C$ be any complex number such that $\cmod {z - \xi} < \cmod {w - \xi}$. Then: {{begin-eqn}} {{eqn | l = \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n}       | o = <       | r = \limsup_{n \mathop \to \infty} \cmod {a_n \paren {w - \xi}^n}^{1/n}       | c = as $\cmod {z - \xi} < \cmod {w - \xi}$, we have $\cmod {z - \xi}^n < \cmod {w - \xi}^n$ }} {{eqn | o = \le       | r = 1       | c = $n$th Root Test: $S \paren w$ is not divergent }} {{end-eqn}} From the $n$th Root Test, it follows that $S \paren z$ converges absolutely. {{qed}}
Let $z \notin {B_R}^- \paren \xi$. Then by definition of the closed $R$-ball of $\xi$: : $\cmod {z - \xi} > R$ where $\cmod z$ denotes the complex modulus of $z$. By definition of radius of convergence, there exists $w \in \C$ such that: : $\cmod {w - \xi} < \cmod {z - \xi}$ and $S \paren w$ is divergent. Then: {{begin-eqn}} {{eqn | l = \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n }^{1/n}       | o = >       | r = \limsup_{n \mathop \to \infty} \cmod {a_n \paren {w - \xi}^n }^{1/n} }} {{eqn | o = \ge       | r = 1       | c = $n$th Root Test: $S \paren w$ is divergent }} {{end-eqn}} From the $n$th Root Test, it follows that $S \paren z$ is divergent. {{qed}}
{{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \dfrac {\cmod {z^n} } {\cmod {z^{n - 1} } }       | r = \lim_{n \mathop \to \infty} \cmod {\dfrac {z^n} {z^{n - 1} } }       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \cmod z       | c =  }} {{eqn | r = \cmod z       | c =  }} {{end-eqn}} By the Ratio Test, it follows that: :$S$ is convergent for $\cmod z < 1$ :$S$ is divergent for $\cmod z > 1$. Hence the result by definition of radius of convergence. {{Qed}}
Applying Radius of Convergence from Limit of Sequence: Complex Case, we find that: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }       | r = \lim_{n \mathop \to \infty} \cmod {\dfrac {\frac {\paren {-1}^{n + 1} } {\paren {2 \paren {n + 1} }!} } {\frac {\paren {-1}^n} {\paren {2 n}!} } }       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \cmod {\dfrac {\paren {2 n}!} {\paren {2 \paren {n + 1} }!} }       | c = as $\cmod {\paren {-1}^n} = 1$ for all $n$ }} {{eqn | r = \lim_{n \mathop \to \infty} \cmod {\dfrac 1 {\paren {2 n + 1} \paren {2 n + 2} } }       | c =  }} {{eqn | r = 0       | c = Sequence of Powers of Reciprocals is Null Sequence }} {{end-eqn}} Hence the result. {{qed}}
Applying Radius of Convergence from Limit of Sequence: Complex Case, we find that: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }       | r = \lim_{n \mathop \to \infty} \cmod {\dfrac {\frac {\paren {-1}^{n + 1} } {\paren {2 \paren {n + 1} + 1}!} } {\frac {\paren {-1}^n} {\paren {2 n + 1}!} } }       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \cmod {\dfrac {\paren {2 n + 1}!} {\paren {2 \paren {n + 1} + 1}!} }       | c = as $\cmod {\paren {-1}^n} = 1$ for all $n$ }} {{eqn | r = \lim_{n \mathop \to \infty} \cmod {\dfrac 1 {\paren {2 n + 2} \paren {2 n + 3} } }       | c =  }} {{eqn | r = 0       | c = Sequence of Powers of Reciprocals is Null Sequence }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac 1 {\paren {1 - z}^2}       | r = \paren {\dfrac 1 {1 - z} }^2       | c =  }} {{eqn | r = \paren {\sum_{n \mathop = 0}^\infty z^n}^2       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\displaystyle \sum_{k \mathop = 0}^n 1 \times 1} z^n       | c = Product of Absolutely Convergent Series }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {n + 1} z^n       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{n \mathop = 0}^\infty z^n       | r = \frac 1 {1 - z}       | c = Sum of Infinite Geometric Sequence }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d z} } {\sum_{n \mathop = 1}^\infty z^n}       | r = \dfrac \d {\d z} \frac 1 {1 - z}       | c =  }} {{eqn | r = \dfrac 1 {\paren {1 - z}^2}       | c = Power Rule for Derivatives and the Chain Rule for Derivatives }} {{end-eqn}} Now we have: {{begin-eqn}} {{eqn | o =        | r = \map {\frac \d {\d z} } {\sum_{n \mathop = 0}^\infty z^n}       | c =  }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\frac \d {\d z} z^n}       | c = Derivative of Absolutely Convergent Series }} {{eqn | r = \sum_{n \mathop = 0}^\infty n z^{n - 1}       | c = Power Rule for Derivatives }} {{eqn | r = \sum_{n \mathop = 0}^\infty n z^{n - 1}       | c = as the zeroth term vanishes when $n = 0$ }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {n + 1} z^n       | c = Translation of Index Variable of Summation }} {{end-eqn}} Hence the result. {{qed}}
Let $0 < \theta < \dfrac \pi 2$. Then $0 < \cos \theta < 1$. {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^\infty r^k \cos k \theta       | r = \dfrac {1 - r \cos \theta} {1 - 2 r \cos \theta + r^2}       | c = Sum of Infinite Series of Product of Power and Cosine: $\size r < 1$ }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 0}^\infty \cos^k \theta \cos k \theta       | r = \dfrac {1 - \cos^2 \theta } {1 - 2 \cos^2 \theta + \cos^2 \theta}       | c = setting $r = \cos \theta$ }} {{eqn | ll= \leadsto       | l = \cos^0 \theta \cos 0 \theta + \sum_{k \mathop = 1}^\infty \cos^k \theta \cos k \theta       | r = \dfrac {1 - \cos^2 \theta } {1 - \cos^2 \theta}       | c = simplifying }} {{eqn | ll= \leadsto       | l = 1 + \sum_{k \mathop = 1}^\infty \cos^k \theta \cos k \theta       | r = \dfrac {1 - \cos^2 \theta } {1 - \cos^2 \theta}       | c = simplifying }} {{eqn | ll= \leadsto       | l = 1 + \sum_{k \mathop = 1}^\infty \cos^k \theta \cos k \theta       | r = 1       | c = simplifying }} {{eqn | ll= \leadsto       | l = \cos \theta \sum_{k \mathop = 1}^\infty \cos^{k - 1} \theta \, \map \cos {n + 1} \theta       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 1}^\infty \cos^{k - 1} \theta \cos k \theta       | r = 0       | c = dividing by $\cos \theta$ }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 0}^\infty \cos^k \theta \cos k \theta       | r = 0       | c = Translation of Index Variable of Summation }} {{end-eqn}} {{qed}}
As Sine Function is Absolutely Convergent and Cosine Function is Absolutely Convergent, we have: {{begin-eqn}} {{eqn | l = \cos \theta + i \sin \theta       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n} } {\paren {2 n}!} + i \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {\theta^{2 n + 1} } {\paren {2 n + 1}!}       | c = {{Defof|Complex Cosine Function}} and {{Defof|Complex Sine Function}} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\paren {-1}^n \dfrac {\theta^{2 n} } {\paren {2 n}!} + i \paren {-1}^n \dfrac {\theta^{2 n + 1} } {\paren {2 n + 1}!} }       | c = Sum of Absolutely Convergent Series }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\dfrac {\paren {i \theta}^{2 n} } {\paren {2 n}!} + \dfrac {\paren {i \theta}^{2 n + 1} } {\paren {2 n + 1}!} }       | c = {{Defof|Imaginary Unit}} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \dfrac {\paren {i \theta}^n} {n!} }} {{eqn | r = e^{i \theta}       | c = {{Defof|Exponential Function/Complex|subdef = Sum of Series|Complex Exponential Function}} }} {{end-eqn}} {{qed}}
Recall the definition of the real cosine function: {{begin-eqn}} {{eqn | l = \cos x       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n!} }       | c =  }} {{eqn | r = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \frac {x^6} {6!} + \cdots + \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} + \cdots       | c =  }} {{end-eqn}} Recall the definition of the exponential as a power series: {{begin-eqn}} {{eqn | l = e^x       | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}       | c =  }} {{eqn | r = 1 + x + \frac {x^2} 2 + \frac {x^3} 6 + \cdots + \frac {x^n} {n!} + \cdots       | c =  }} {{end-eqn}} Then, starting from the {{RHS}}: {{begin-eqn}} {{eqn | l = \frac {e^{i x} + e^{-i x} } 2       | r = \frac 1 2 \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i x}^n} {n!} + \sum_{n \mathop = 0}^\infty \frac {\paren {-i x}^n} {n!} } }} {{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^n + \paren {-i x}^n} {n!} }       | c = Cosine Function is Absolutely Convergent }} {{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^{2 n} + \paren {-i x}^{2 n} } {\paren {2 n}!} + \frac {\paren {i x}^{2 n + 1} + \paren {-i x}^{2 n + 1} } {\paren {2 n + 1}!} }       | c = split into even and odd $n$ }} {{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n} + \paren {-i x}^{2 n} } {\paren {2 n}!}       | c = $\paren {-i x}^{2 n + 1} = -\paren {i x}^{2 n + 1}$ }} {{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {2 \paren {i x}^{2 n} } {\paren {2 n}!}       | c = $\left({ -1 }\right)^{2n} = 1$ }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n} } {\paren {2 n}!}       | c = cancel $2$ }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}       | c = $i^{2 n} = \paren {-1}^n$ }} {{eqn | r = \cos x }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | n = 1       | l = e^{i x}       | r = \cos x + i \sin x       | c = Euler's Formula }} {{eqn | n = 2       | l = e^{-i x}       | r = \cos x - i \sin x       | c = Euler's Formula: Corollary }} {{eqn | ll= \leadsto       | l = e^{i x} + e^{-i x}       | r = \paren {\cos x + i \sin x} + \paren {\cos x - i \sin x}       | c = $(1) + (2)$ }} {{eqn | r = 2 \cos x       | c = simplifying }} {{eqn | ll= \leadsto       | l = \frac {e^{i x} + e^{-i x} } 2       | r = \cos x       | c =  }} {{end-eqn}} {{qed}}
Let $0$ be the zero of $\struct {R, \norm {\,\cdot\,} }$. Let $d$ denote the metric induced by $\norm {\, \cdot \,}$, that is: :$\map d {x, y} = \norm {x - y}$ From Metric Induced by Norm on Normed Division Ring is Metric we have that $d$ is indeed a metric. Then, from the Reverse Triangle Inequality as applied to metric spaces: :$\forall x, y, z \in R: \bigsize {\norm {x - z} - \norm {y - z} } \le \norm {x - y}$ Then: :$\forall x, y \in R: \bigsize {\norm x - \norm y} = \bigsize {\norm{x - 0} - \norm{y - 0} } \le \norm {x - y}$ {{qed}} Category:Triangle Inequality Category:Normed Division Rings s4owpzx2q00sisb93nn4yho7dlsvgj2
Since $l \ne 0$, by norm axiom (N1):  $\norm l > 0$. Let us choose $N$ such that: :$\forall n > N: \norm {x_n - l} < \dfrac {\norm l} 2$ Then: {{begin-eqn}} {{eqn | l = \norm {x_n - l}       | o = <       | r = \frac {\norm l} 2       | c =  }} {{eqn | ll= \leadsto       | l = \norm l - \norm {x_n}       | o = \le       | r = \norm {x_n - l}       | c = Reverse Triangle Inequality }} {{eqn | o = <       | r = \frac {\norm l} 2       | c =  }} {{eqn | ll= \leadsto       | l = \norm {x_n}       | o = >       | r = \norm l - \frac {\norm l} 2       | c =  }} {{eqn | r = \frac {\norm l} 2       | c =  }} {{end-eqn}} {{qed}} Category:Sequences Category:Limits of Sequences Category:Normed Division Rings t6eaiqi7jbcka4w01d7fpn9fkegn4oz
Recall the definition of the sine function: {{begin-eqn}} {{eqn | l = \sin x       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}       | c =  }} {{eqn | r = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7} {7!} + \cdots + \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} + \cdots       | c =  }} {{end-eqn}} Recall the definition of the exponential as a power series: {{begin-eqn}} {{eqn | l = \exp x       | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}       | c =  }} {{eqn | r = 1 + \frac x {1!} + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots + \frac {x^n} {n!} + \cdots       | c =  }} {{end-eqn}} Then, starting from the {{RHS}}: {{begin-eqn}} {{eqn | l = \frac {e^{i x} - e^{-i x} } {2 i}       | r = \frac 1 {2 i} \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i x}^n} {n!} - \sum_{n \mathop = 0}^\infty \frac {\paren {-i x}^n} {n!} }       | c =  }} {{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^n - \paren {-i x}^n} {n!} }       | c =  }} {{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^{2 n} - \paren {-i x}^{2 n} } {\paren {2 n}!} + \frac {\paren {i x}^{2 n + 1} - \paren {-i x}^{2 n + 1} } {\paren {2 n + 1}!} }       | c = split into even and odd $n$ }} {{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n + 1} - \paren {-i x}^{2 n + 1} } {\paren {2 n + 1}!}       | c = as $\paren {-i x}^{2 n} = \paren {i x}^{2 n}$ }} {{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {2 \paren {i x}^{2 n + 1} } {\paren {2 n + 1}!}       | c = as $\paren {-1}^{2 n + 1} = -1$ }} {{eqn | r = \frac 1 i \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n + 1} } {\paren {2 n + 1}!}       | c = cancel $2$ }} {{eqn | r = \frac 1 i \sum_{n \mathop = 0}^\infty \frac {i \paren {-1}^n x^{2 n + 1} } {\paren {2 n + 1}!}       | c = as $i^{2 n + 1} = i \paren {-1})^n $ }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1!} }       | c = cancel $i$ }} {{eqn | r = \sin x }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | n = 1       | l = e^{i x}       | r = \cos x + i \sin x       | c = Euler's Formula }} {{eqn | n = 2       | l = e^{-i x}       | r = \cos x - i \sin x       | c = Euler's Formula: Corollary }} {{eqn | ll= \leadsto       | l = e^{i x} - e^{-i x}       | r = \paren {\cos x + i \sin x} - \paren {\cos x - i \sin x}       | c = $(1) - (2)$ }} {{eqn | r = 2 i \sin x       | c = simplifying }} {{eqn | ll= \leadsto       | l = \frac {e^{i x} - e^{-i x} } {2 i}       | r = \sin x       | c =  }} {{end-eqn}} {{qed}}
Let $d$ denote the metric induced by $\norm {\, \cdot \,}$, that is, :$d \tuple {x, y} = \norm {x - y}$ By definition of convergence in a normed division ring: :$\sequence {x_n}$ converges to $l$ in $\struct {R, \norm {\, \cdot \,} }$ {{iff}} $\sequence {x_n}$ converges to $l$ in the metric space $\struct {R, d}$. We can then apply the result of Limit of Subsequence equals Limit of Sequence for metric spaces to get the result for normed division rings. {{qed}} Category:Sequences Category:Normed Division Rings nja981douhnr2aoylkfzgebvaz9xo90
By the Inverse Rule for Normed Division Ring: :$\exists k \in \N : \forall n \in \N : y_{k + n} \ne 0$ and the sequence: :$\sequence { {y_{k + n} }^{-1} }$ is well-defined and convergent with: :$\displaystyle \lim_{n \mathop \to \infty} {y_{k + n} }^{-1} = m^{-1}$ By Limit of Subsequence equals Limit of Sequence, $\sequence {x_{k + n} }$ is convergent with: :$\displaystyle \lim_{n \mathop \to \infty} x_{k + n} = l$ By Product Rule for Normed Division Ring Sequences: :$\displaystyle \lim_{n \mathop \to \infty} x_{k + n} \ {y_{k + n} }^{-1} = l m^{-1}$ :$\displaystyle \lim_{n \mathop \to \infty} {y_{k + n} }^{-1} \ x_{k + n} = m^{-1} l$ {{qed}} Category:Combination Theorem for Sequences in Normed Division Rings cfkhm7xuqj4mr0odhepsxdcjkooqcwm
{{begin-eqn}} {{eqn | o =        | r = 2 \sin \alpha \cos \beta }} {{eqn | r = 2 \paren {\dfrac {\exp \paren {i \alpha} - \exp \paren {-i \alpha} } {2 i} } \paren {\dfrac {\exp \paren {i \beta} + \exp \paren {-i \beta} } 2}       | c = Sine Exponential Formulation and Cosine Exponential Formulation }} {{eqn | r = \frac 1 {2 i} \paren {\exp \paren {i \alpha} - \exp \paren {-i \alpha} } \paren {\exp \paren {i \beta} + \exp \paren {-i \beta} }  }} {{eqn | r = \frac 1 {2 i} \paren {\exp \paren {i \paren {\alpha + \beta} } - \exp \paren {-i \paren {\alpha + \beta} } + \exp \paren {i \paren {\alpha - \beta} } - \exp \paren {-i \paren {\alpha - \beta} } }  }} {{eqn | r = \frac {\exp \paren {i \paren {\alpha + \beta} } - \exp \paren {-i \paren {\alpha + \beta} } } {2 i} + \frac {\exp \paren {i \paren {\alpha - \beta} } - \exp \paren {-i \paren {\alpha - \beta} } } {2 i} }} {{eqn | r = \sin \paren {\alpha + \beta} + \sin \paren {\alpha - \beta} }} {{end-eqn}} {{qed}}
From the algebraic definition of the real sine function: :$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$ it follows that $\sin x$ is a real function. Thus $\sin^2 x \ge 0$. From Sum of Squares of Sine and Cosine, we have that $\cos^2 x + \sin^2 x = 1$. Thus it follows that: :$\sin^2 x = 1 - \cos^2 x \le 1$ From Ordering of Squares in Reals and the definition of absolute value, we have that: :$x^2 \le 1 \iff \size x \le 1$ The result follows. {{qed}}
Let $K \in \R_{>0}$ be an arbitrary real number. Let $p = \ln {2 K}$. Let $z = i p$, where $i$ denotes the imaginary unit. Then: {{begin-eqn}} {{eqn | l = \cos z       | r = \dfrac {\exp \paren {i \paren {i p} } + \exp \paren {-i \paren {i p} } } 2       | c = Cosine Exponential Formulation }} {{eqn | r = \dfrac {\exp p + \exp \paren {-p} } 2       | c = simplifying }} {{eqn | o = >       | r = \dfrac {\exp p} 2       | c = as $\exp \paren {-p} > 0$ }} {{eqn | r = \dfrac {\exp \paren {\ln {2 K} } } 2       | c =  }} {{eqn | r = K       | c =  }} {{eqn | ll= \leadsto       | l = \cmod {\cos \paren {i p} }       | o = >       | r = K       | c =  }} {{end-eqn}} Thus for any $K$ we can find $z \in \C$ such that $\cmod {\cos z} > K$. Hence the result by definition of unbounded. {{qed}}
From the definition of the complex exponential function: :$\exp z := e^x \paren {\cos y + i \sin y}$ The result follows by definition of the real part of a complex number. {{qed}}
From the definition of the complex exponential function: :$\exp z := e^x \paren {\cos y + i \sin y}$ The result follows by definition of the imaginary part of a complex number. {{qed}}
From Sine of Complex Number: :$\map \sin {x + i y} = \sin x \cosh y + i \cos x \sinh y$ The result follows by definition of the real part of a complex number. {{qed}}
From Sine of Complex Number: :$\sin \paren {x + i y} = \sin x \cosh y + i \cos x \sinh y$ The result follows by definition of the imaginary part of a complex number. {{qed}}
{{begin-eqn}} {{eqn | l = \sin \paren {x + i y}       | r = \sin x \cosh y + i \cos x \sinh y       | c = Sine of Complex Number }} {{eqn | ll= \leadsto       | l = \cmod {\sin z}^2       | r = \paren {\sin x \cosh y}^2 + \paren {\cos x \sinh y}^2       | c = {{Defof|Complex Modulus}} }} {{eqn | r = \sin^2 x \paren {1 + \sinh^2 y} + \cos^2 x \sinh^2 y       | c = Difference of Squares of Hyperbolic Cosine and Sine }} {{eqn | r = \sin^2 x + \paren {\sin^2 x + \cos^2 x} \sinh^2 y       | c = rearranging }} {{eqn | r = \sin^2 x + \sinh^2 y       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = \cmod {\sin z}       | r = \sqrt {\sin^2 x + \sinh^2 y}       | c =  }} {{end-eqn}} {{qed}}
{{AimForCont}} $\tan z_0 = i$. {{begin-eqn}} {{eqn | l = \dfrac {\sin z_0} {\cos z_0}       | r = i       | c = {{Defof|Tangent Function}} }} {{eqn | ll= \leadsto       | l = \sin z_0       | r = i \cos z_0       | c =  }} {{eqn | ll= \leadsto       | l = \sin^2 z_0       | r = -\cos^2 z_0       | c =  }} {{eqn | ll= \leadsto       | l = \sin^2 z_0 + \cos^2 z_0       | r = 0       | c =  }} {{end-eqn}} This contradicts the theorem Sum of Squares of Sine and Cosine: :$\sin^2 z_0 + \cos^2 z_0 = 1$ Hence the result by Proof by Contradiction. {{qed}}
Let $\epsilon \in \R_{>0}$, and let $z \in \C$. Suppose that $\cmod {z - \xi} = R - \epsilon$. By definition of radius of convergence, it follows that $S \paren z$ is absolutely convergent. From the $n$th Root Test: :$\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n} \le 1$ By Multiple Rule for Complex Sequences, this inequality can be rearranged to obtain: :$\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \le \dfrac 1 {\cmod {z - \xi} } = \dfrac 1 {R - \epsilon}$ As $\epsilon > 0$ was arbitrary, it follows that: :$\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \le \dfrac 1 R$ {{qed|lemma}} Now, suppose that $\cmod {z - \xi} = R + \epsilon$. Then $S \paren z$ is divergent, so the $n$th Root Test shows that: :$\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n} \ge 1$ which we can rearrange to obtain: :$\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \ge \dfrac 1 {\cmod {z - \xi} } = \dfrac 1 {R - \epsilon}$ As $\epsilon > 0$ was arbitrary, it follows that: :$\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \ge \dfrac 1 R$ {{qed|lemma}} Thus: :$\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = \dfrac 1 R$ Hence the result. {{qed}}