hcju/proofwikiqa huggingface dataset:
By the Inverse Rule for Normed Division Ring: :$\exists K \in \N : \forall n > K : y_n \ne 0$. and the sequence: :$\sequence {\paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ is well-defined and a Cauchy sequence. By Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence, $\sequence {x_{K + n} }_{n \mathop \in \N}$ is a Cauchy sequence. By Product Rule for Normed Division Ring Sequences: :the sequences $\sequence { {x_{K + n} } \paren {y_{K + n} }^{-1} }_{n \mathop \in \N}$ and $\sequence {\paren {y_{K + n} }^{-1} {x_{K + n} } }_{n \mathop \in \N}$ are Cauchy sequences. {{qed}}
From Factorisation of $z^n + 1$: :$(1): \quad z^{2 n} + 1 = \displaystyle \prod_{k \mathop = 0}^{2 n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n} }$ From Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs, the roots of $(1)$ occur in conjugate pairs. Hence we can express $(1)$ as: {{begin-eqn}} {{eqn | l = z^{2 n} + 1       | r = \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n} } \prod_{k \mathop = n}^{2 n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n} }       | c =  }} {{eqn | r = \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n} } \paren {z - \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n} }       | c =  }} {{eqn | r = \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} {2 n} + \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n} } + \exp \dfrac {\paren {2 k + 1} i \pi} {2 n} \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n} }       | c =  }} {{eqn | r = \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} {2 n} + \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n} } + 1}       | c = simplifying }} {{eqn | r = \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\cos \dfrac {\paren {2 k + 1} \pi} {2 n} + i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} + \cos \dfrac {\paren {2 k + 1} \pi} {2 n} - i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} } + 1}       | c = Euler's Formula }} {{eqn | r = \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - 2 z \cos \dfrac {\paren {2 k + 1} \pi} {2 n} + 1}       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Algebra Category:Complex Roots k97ripjv8byy07nvr1l7d8zr255jrlx
From Factorisation of $z^n + 1$: :$(1): \quad z^{2 n + 1} + 1 = \displaystyle \prod_{k \mathop = 0}^{2 n} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} }$ From Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs, the roots of $(1)$ occur in conjugate pairs. Hence we can express $(1)$ as: {{begin-eqn}} {{eqn | l = z^{2 n + 1} + 1       | r = \prod_{k \mathop = 0}^n \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} } \prod_{k \mathop = n + 1}^{2 n} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} }       | c =  }} {{eqn | r =  \paren {z - \exp \dfrac {\paren {2 n + 1} i \pi} {2 n + 1} } \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} } \prod_{k \mathop = n + 1}^{2 n} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} }       | c =  }} {{eqn | r = \paren {z - e^{i \pi} } \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} } \prod_{k \mathop = n + 1}^{2 n} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} }       | c =  }} {{eqn | r = \paren {z + 1} \prod_{k \mathop = 0}^{n - 1} \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} } \paren {z - \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n + 1} }       | c =  }} {{eqn | r = \paren {z + 1} \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} + \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n + 1} } + \exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n + 1} }       | c =  }} {{eqn | r = \paren {z + 1} \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} {2 n + 1} + \exp \dfrac {-\paren {2 k + 1} i \pi} {2 n + 1} } + 1}       | c = simplifying }} {{eqn | r = \paren {z + 1} \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - z \paren {\cos \dfrac {\paren {2 k + 1} \pi} {2 n + 1} + i \sin \dfrac {\paren {2 k + 1} \pi} {2 n + 1} + \cos \dfrac {\paren {2 k + 1} \pi} {2 n + 1} - i \sin \dfrac {\paren {2 k + 1} \pi} {2 n + 1} } + 1}       | c = Euler's Formula }} {{eqn | r = \paren {z + 1} \prod_{k \mathop = 0}^{n - 1} \paren {z^2 - 2 z \cos \dfrac {\paren {2 k + 1} \pi} {2 n + 1} + 1}       | c = simplifying }} {{end-eqn}} {{qed}} Category:Algebra Category:Complex Roots tnptjbzisstmzrfmv91sc8vsts8rwjx
From Factorisation of $z^n + 1$: :$z^4 + 1 = \displaystyle \prod_{k \mathop = 0}^4 \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} 5}$ Thus: :$z = \set {\exp \dfrac {\paren {2 k + 1} i \pi} 5}$ {{begin-eqn}} {{eqn | n = k = 0       | l = z       | r = \cos \dfrac \pi 5 + i \sin \dfrac \pi 5       | c =  }} {{eqn | n = k = 1       | l = z       | r = \cos \dfrac {3 \pi} 5 + i \sin \dfrac {3 \pi} 5       | c =  }} {{eqn | n = k = 2       | l = z       | r = \cos \dfrac {5 \pi} 5 + i \sin \dfrac {5 \pi} 5       | c =  }} {{eqn | r = \cos \pi + i \sin \pi       | c =  }} {{eqn | r = -1       | c = Euler's Identity }} {{eqn | n = k = 3       | l = z       | r = \cos \dfrac {7 \pi} 5 + i \sin \dfrac {7 \pi} 5       | c =  }} {{eqn | r = \cos \dfrac {-3 \pi} 5 + i \sin \dfrac {-3 \pi} 5       | c =  }} {{eqn | r = \cos \dfrac {3 \pi} 5 - i \sin \dfrac {3 \pi} 5       | c =  }} {{eqn | n = k = 4       | l = z       | r = \cos \dfrac {9 \pi} 5 + i \sin \dfrac {9 \pi} 5       | c =  }} {{eqn | r = \cos \dfrac {-\pi} 5 + i \sin \dfrac {-\pi} 5       | c =  }} {{eqn | r = \cos \dfrac \pi 5 - i \sin \dfrac \pi 5       | c =  }} {{end-eqn}} {{qed}}
=== $(1)$ implies $(2)$ === Let $\sequence {z_n}$ be a convergent complex sequence by definition 1. Then by definition: :$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \cmod {z_n - c} < \epsilon$ Let $z_n = x_n + i y_n$. Let $c = a + i b$. Let $\epsilon \in \R_{>0}$. Let $N \in \R: n > N \implies \cmod {z_n - c} < \epsilon$. Then: {{begin-eqn}} {{eqn | l = \sqrt {\paren {x_n - a}^2 + \paren {y_n - b}^2}       | o = <       | r = \epsilon       | c = {{Defof|Complex Modulus}} }} {{eqn | ll= \leadsto       | l = \paren {x_n - a}^2 + \paren {y_n - b}^2       | o = <       | r = \epsilon^2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x_n - a}^2       | o = <       | r = \epsilon^2       | c =  }} {{eqn | l = \paren {y_n - b}^2       | o = <       | r = \epsilon^2       | c =  }} {{eqn | ll= \leadsto       | l = \size {x_n - a}       | o = <       | r = \epsilon       | c =  }} {{eqn | l = \size {y_n - b}       | o = <       | r = \epsilon       | c =  }} {{end-eqn}} Thus $\sequence {z_n}$ is a convergent complex sequence by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $\sequence {z_n} = \sequence {x_n + y_n}$ be a convergent complex sequence by definition 2. Then by definition: :$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {x_n - a} < \epsilon \text { and } \size {y_n - b} < \epsilon$ where $a + i b = c$. Let $\epsilon \in \R_{>0}$. Let $N \in \R: n > N \implies \size {x_n - a} < \dfrac \epsilon 2 \text { and } \size {y_n - b} < \dfrac \epsilon 2$. Then: {{begin-eqn}} {{eqn | l = \cmod {z_n − c}       | r = \cmod {x_n + i y_n − \paren {a + i b} }       | c =  }} {{eqn | r = \cmod {x_n − a + i \paren {y_n − b} }       | c =  }} {{eqn | o = \le       | r = \cmod {x_n − a} + \cmod {i \paren {y_n − b} }       | c = Triangle Inequality for Complex Numbers }} {{eqn | r = \cmod {x_n − a} + \cmod {y_n − b}       | c = {{Defof|Complex Modulus}} }} {{eqn | o = <       | r = \frac \epsilon 2 + \cmod {y_n − b}       | c = as $\cmod {x_n − a} < \dfrac \epsilon 2$ }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2       | c = as $\cmod {y_n − b} < \dfrac \epsilon 2$ }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Thus $\sequence {z_n}$ is a convergent complex sequence by definition 1. {{qed}}
By definition of convergent complex sequence: :$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \cmod {z_n - c} < \epsilon$ From the Reverse Triangle Inequality: :$\size {\cmod x - \cmod y} \le \cmod {x - y}$ and the result follows. {{qed}}
From Sum Rule for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$ From Multiple Rule for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {-y_n} = -m$ Hence: :$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + \paren {-y_n} } = l + \paren {-m}$ The result follows. {{qed}}
Let $\epsilon > 0$ be given. Then $\dfrac \epsilon 2 > 0$. We are given that: : $\displaystyle \lim_{n \mathop \to \infty} z_n = c$ : $\displaystyle \lim_{n \mathop \to \infty} w_n = d$ By definition of the limit of a complex sequence, we can find $N_1$ such that: :$\forall n > N_1: \cmod {z_n - c} < \dfrac \epsilon 2$ where $\cmod {z_n - c}$ denotes the complex modulus of $z_n - c$. Similarly we can find $N_2$ such that: :$\forall n > N_2: \cmod {w_n - d} < \dfrac \epsilon 2$ Let $N = \max \set {N_1, N_2}$. Then if $n > N$, both the above inequalities will be true: :$n > N_1$ :$n > N_2$ Thus $\forall n > N$: {{begin-eqn}} {{eqn | l = \cmod {\paren {z_n + w_n} - \paren {c + d} }       | r = \cmod {\paren {z_n - c} + \paren {w_n - d} }       | c =  }} {{eqn | o = \le       | r = \cmod {z_n - l} + \cmod {w_n - m}       | c = Triangle Inequality for Complex Numbers }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Hence the result: : $\displaystyle \lim_{n \mathop \to \infty} \paren {z_n + w_n} = c + d$ {{qed}}
Let $\epsilon > 0$ be given. Then $\dfrac \epsilon 2 > 0$. We are given that: : $\displaystyle \lim_{n \mathop \to \infty} z_n = c$ : $\displaystyle \lim_{n \mathop \to \infty} w_n = d$ Let: :$z_n = x_n + i y_n$ :$w_n = r_n + i s_n$ :$c = a + i b$ :$d = l + i m$ where each of $x_n, y_n, r_n, s_n, a, b, l, m \in \R$ are real. By definition: $\sequence {z_n}$ '''converges to the limit $c = a + i b$''' {{iff}} both: :$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {x_n - a} < \epsilon \text { and } \size {y_n - b} < \epsilon$ $\sequence {w_n}$ '''converges to the limit $d = l + i m$''' {{iff}} both: :$\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {r_n - l} < \epsilon \text { and } \size {s_n - m} < \epsilon$ where $\size x$ denotes the absolute value of $x \in \R$. Then: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \paren {z_n + w_n}       | r = \lim_{n \mathop \to \infty} \paren {x_n + i y_n + r_n + i s_n}       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n + r_n + i \paren {y_n + s_n} }       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n + r_n} + i \lim_{n \mathop \to \infty} \paren {y_n + s_n}       | c = {{Defof|Convergent Sequence/Complex Numbers|Convergent Complex Sequence|index = 1}} }} {{eqn | r = \paren {a + l} + i \paren {b + m}       | c =  }} {{eqn | r = \paren {a + i b} + \paren {l + i m}       | c =  }} {{eqn | r = c + d       | c =  }} {{end-eqn}} {{qed}}
From Sum Rule for Complex Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {z_n + w_n} = c + d$ From Multiple Rule for Complex Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {-w_n} = -d$ Hence: :$\displaystyle \lim_{n \mathop \to \infty} \paren {z_n + \paren {-w_n} } = c + \paren {-d}$ The result follows. {{qed}}
Let $\epsilon > 0$. We need to find $N$ such that: : $\forall n > N: \cmod {\lambda z_n - \lambda c} < \epsilon$ If $\lambda = 0$ the result is trivial. So, assume $\lambda \ne 0$. Then $\cmod \lambda > 0$ from the definition of the modulus of $\lambda$. Hence $\dfrac \epsilon {\cmod \lambda} > 0$. We have that $z_n \to c$ as $n \to \infty$. Thus it follows that: : $\exists N: \forall n > N: \cmod {z_n - c} < \dfrac \epsilon {\cmod \lambda}$ That is: : $\forall n > N: \cmod \lambda \cmod {z_n - c} < \epsilon$ But we have: {{begin-eqn}} {{eqn | l = \cmod \lambda \cmod {z_n - c}       | r = \cmod {\lambda \paren {z_n - c} }       | c = Complex Modulus of Product of Complex Numbers }} {{eqn | r = \cmod {\lambda x_n - \lambda l}       | c =  }} {{end-eqn}} Hence: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda c$ {{qed}} Category:Combination Theorems for Sequences tfbqx2zl15xj1mnq5sky0k6294531dj
From the Multiple Rule for Complex Sequences, we have: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda z_n} = \lambda c$ :$\displaystyle \lim_{n \mathop \to \infty} \paren {\mu w_n} = \mu d$ The result now follows directly from the Sum Rule for Complex Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda z_n + \mu w_n} = \lambda c + \mu d$ {{qed}} Category:Combination Theorems for Sequences 6k3brs9fz1m9fx3wsb3uqxqtribvl1f
Let $\epsilon > 0$ be given. Then $\dfrac \epsilon 2 > 0$. Since $\sequence {x_n}$ is convergent to $l$, we can find $N_1$ such that: :$\forall n > N_1: \norm {x_n - l} < \dfrac \epsilon 2$ Similarly, for $\sequence {y_n}$ we can find $N_2$ such that: :$\forall n > N_2: \norm {y_n - m} < \dfrac \epsilon 2$ Now let $N = \max \set {N_1, N_2}$. Then if $n > N$, both the above inequalities will be true. Thus $\forall n > N$: {{begin-eqn}} {{eqn | l = \norm {\paren {x_n + y_n} - \paren {l + m} }       | r = \norm {\paren {x_n - l} + \paren {y_n - m} }       | c =  }} {{eqn | o = \le       | r = \norm {x_n - l} + \norm {y_n - m}       | c = {{NormAxiom|3}} }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2 = \epsilon       | c =  }} {{end-eqn}} Hence: :$\sequence {x_n + y_n}$ is convergent to $l + m$. {{qed}} Category:Combination Theorem for Sequences in Normed Division Rings msbv7wuxm011galapa42gjw9m7l1zhn
As $z_n \to c$ as $n \to \infty$, it follows from Modulus of Limit that $\size {w_n} \to \size d$ as $n \to \infty$. As $d \ne 0$, it follows from the definition of the modulus of $d$ that $\size d > 0$. From Sequence Converges to Within Half Limit, we have $\exists N: \forall n > N: \size {w_n} > \dfrac {\size d} 2$. Now, for $n > N$, consider: {{begin-eqn}} {{eqn | l = \size {\frac {z_n} {w_n} - \frac c d}       | r = \size {\frac {d z_n - w_n c} {d w_n} }       | c =  }} {{eqn | o = <       | r = \frac 2 {\size m^2} \size {d z_n - w_n c}       | c =  }} {{end-eqn}} By the above, $d z_n - w_n c \to d c - d c = 0$ as $n \to \infty$. The result follows by the Squeeze Theorem for Sequences of Complex Numbers (which applies as well to real as to complex sequences). {{qed}}
Follows directly from Product Rule for Normed Division Ring Sequences, setting: :$\sequence {y_n} := \sequence {x_n}$ and: :$\sequence {x_n} := \tuple {\lambda, \lambda, \lambda, \ldots}$ {{qed}}
From the Multiple Rule for Sequences in Normed Division Ring, we have: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$ :$\displaystyle \lim_{n \mathop \to \infty} \paren {\mu y_n} = \mu m$ The result now follows directly from the Sum Rule for Sequences in Normed Division Ring: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x_n + \mu y_n} = \lambda l + \mu m$ {{qed}}
From Sum Rule for Sequences in Normed Division Ring: :$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n} = l + m$ From Multiple Rule for Sequences in Normed Division Ring: :$\displaystyle \lim_{n \mathop \to \infty} \paren {-y_n} = -m$ Hence: :$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + \paren {-y_n} } = l + \paren {-m}$ The result follows. {{qed}}
Let $z_n = x_n + i y_n$. Let $w_n = u_n + i v_n$. Let $c = a + i b$ Let $d = e + i f$. By definition of convergent complex sequence: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} z_n       | r = c       | c =  }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} x_n + i \lim_{n \mathop \to \infty} y_n       | r = a + i b       | c = {{Defof|Convergent Complex Sequence}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} w_n       | r = d       | c =  }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} u_n + i \lim_{n \mathop \to \infty} v_n       | r = e + i f       | c = {{Defof|Convergent Complex Sequence}} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} z_n w_n       | r = \lim_{n \mathop \to \infty} \paren {\paren {x_n u_n - y_n v_n} + i \paren {y_n u_n + x_n v_n} }       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n u_n - y_n v_n} + i \lim_{n \mathop \to \infty} \paren {y_n u_n + x_n v_n}       | c = {{Defof|Convergent Complex Sequence}} }} {{eqn | r = \paren {\lim_{n \mathop \to \infty} \paren {x_n u_n} - \lim_{n \mathop \to \infty} \paren {y_n v_n} } + i \paren {\lim_{n \mathop \to \infty} \paren {y_n u_n} + \lim_{n \mathop \to \infty} \paren {x_n v_n} }       | c = Sum Rule for Real Sequences }} {{eqn | r = \paren {\lim_{n \mathop \to \infty} \paren {x_n} \lim_{n \mathop \to \infty} \paren {u_n} - \lim_{n \mathop \to \infty} \paren {y_n} \lim_{n \mathop \to \infty} \paren {v_n} } + i \paren {\lim_{n \mathop \to \infty} \paren {y_n} \lim_{n \mathop \to \infty} \paren {u_n} + \lim_{n \mathop \to \infty} \paren {x_n} \lim_{n \mathop \to \infty} \paren {v_n} }       | c = Product Rule for Real Sequences }} {{eqn | r = \paren {a e - b f} + i \paren {b e + a f}       | c =  }} {{eqn | r = \paren {a + i b} \paren {e + i f}       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = c d       | c =  }} {{end-eqn}} {{qed}}
Let $z_n = u_n + i v_n$. We have that: {{begin-eqn}} {{eqn | l = \cmod {z_n}       | r = \sqrt { {u_n}^2 + {v_n}^2}       | c =  }} {{eqn | o = >       | r = \sqrt { {u_n}^2}       | c =  }} {{eqn | o = >       | r = \size {u_n}       | c =  }} {{end-eqn}} and similarly: :$\cmod {z_n} > \size {v_n}$ From the Comparison Test, the series $\displaystyle \sum_{n \mathop = 1}^\infty u_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty v_n$ are absolutely convergent. From Absolutely Convergent Series is Convergent: Real Numbers, $\displaystyle \sum_{n \mathop = 1}^\infty u_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty v_n$ are convergent. By Convergence of Series of Complex Numbers by Real and Imaginary Part, it follows that $\displaystyle \sum_{n \mathop = 1}^\infty z_n$ is convergent. {{qed}}
We do a case analysis. === $(1): \quad x \ge 0, y \ge 0$ === {{begin-eqn}} {{eqn | l = x       | o = \ge       | r = 0       | c =  }} {{eqn | l = y       | o = \ge       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \size {x + y}       | r = x + y       | c =  }} {{eqn | r = \size x + \size y       | c =  }} {{end-eqn}} {{qed|lemma}} === $(2): \quad x \le 0, y \le 0$ === {{begin-eqn}} {{eqn | l = x       | o = \le       | r = 0       | c =  }} {{eqn | l = y       | o = \le       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \size {x + y}       | r = -x - y       | c =  }} {{eqn | r = \size x + \size y       | c =  }} {{end-eqn}} {{qed|lemma}} === $(3): \quad x \ge 0, y \le 0$ === We have that $\size x = x$ and $\size y = -y$. In this case we show: :$\size {x + y} \le \max \set {\size x, \size y}$ Let $\size x \le \size y$. Then: {{begin-eqn}} {{eqn | l = x       | o = \le       | r = -y }} {{eqn | ll= \leadsto       | l = y        | o = \le       | r = y + x       | c = as $x \ge 0$ }} {{eqn | o = \le       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \size {x + y}        | r = -\paren {x + y} }} {{eqn | o = \le       | r = -y }} {{eqn | r = \size y }} {{end-eqn}} Let $\size x \ge \size y$. Then: {{begin-eqn}} {{eqn | l = x       | r = -y       | o = \ge }} {{eqn | ll= \leadsto       | l = x       | o = \ge       | r = x + y       | c = as $y \le 0$ }} {{eqn | o = \ge       | r = 0 }} {{eqn | ll= \leadsto       | l = \size {x + y} =        | r = x + y }} {{eqn | o = \le       | r = x }} {{eqn | r = \size x }} {{end-eqn}} We have $\max \set {a, b} \le a + b$ for positive real numbers $a$ and $b$.  The result follows by taking $a = \size x$ and $b = \size y$. {{qed|lemma}} === $(4): \quad x \le 0, y \ge 0$ === Follows by symmetry from the case $(3)$. {{qed}}
Let: :$\displaystyle s_n = \sum_{i \mathop = 0}^n a_i$   Then $\sequence {s_n}$ is a sequence in $\R$.  From Cauchy's Convergence Criterion, $\sequence {s_n}$ is convergent {{iff}} it is a Cauchy sequence. For $m > n$ we have: :$\size {s_m - s_n} = \size {a_{n + 1} + a_{n + 2} + \cdots + a_m}$ {{qed}} gbk2q579o6ovvkb0f8s0k3cqscj6nui
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded. Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$. Let $M = \max \set {K, \norm m}$. Then: :$\norm m \le M$ and: :$\forall n: \norm{x_n} \le M$ Let $\epsilon > 0$ be given. Then $\dfrac \epsilon {2 M} > 0$. As $\sequence {x_n}$ converges to $l$, we can find $N_1$ such that: :$\forall n > N_1: \norm {x_n - l} < \dfrac \epsilon {2 M}$ Similarly, for $\sequence {y_n}$ we can find $N_2$ such that: :$\forall n > N_2: \norm {y_n - m} < \dfrac \epsilon {2 M}$ Now let $N = \max \set {N_1, N_2}$. Then if $n > N$, both the above inequalities will be true. Thus $\forall n > N$: {{begin-eqn}} {{eqn | l = \norm {x_n y_n - l m}       | r = \norm {x_n y_n - x_n m + x_n m - l m}       | c =  }} {{eqn | o = \le       | r = \norm {x_n y_n - x_n m} + \norm {x_n m - l m}       | c = Axiom (N3) of norm (Triangle Inequality). }} {{eqn | r = \norm {x_n \paren {y_n - m } } + \norm {\paren {x_n - l } m}       | c =  }} {{eqn | o = \le       | r = \norm {x_n} \cdot \norm {y_n - m} + \norm {x_n - l} \cdot \norm m       | c = Axiom (N2) of norm (Multiplicativity). }} {{eqn | o = \le       | r = M \cdot \norm {y_n - m} + \norm {x_n - l} \cdot M       | c = since $\sequence {x_n}$ is bounded by $M$ and $\norm m \le M$ }} {{eqn | o = \le       | r = M \cdot \dfrac \epsilon {2 M} + \dfrac \epsilon {2 M} \cdot M       | c =  }} {{eqn | r = \dfrac \epsilon 2 + \dfrac \epsilon 2       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} Hence: :$\sequence {x_n y_n}$ is convergent. It follows that: :$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$ {{qed}}
By Convergent Sequence in Normed Division Ring is Bounded, $\sequence {x_n}$ is bounded. Suppose $\norm {x_n} \le K$ for $n = 1, 2, 3, \ldots$. Then for $n = 1, 2, 3, \ldots$: {{begin-eqn}} {{eqn | l = \norm {x_n y_n - l m}       | r = \norm {x_n y_n - x_n m + x_n m - l m}       | c =  }} {{eqn | o = \le       | r = \norm {x_n y_n - x_n m} + \norm {x_n m - l m}       | c = Axiom (N3) of norm (Triangle Inequality) }} {{eqn | r = \norm {x_n} \norm {y_n - m} + \norm {x_n - l} \norm m        | c = Axiom (N2) of norm (Multiplicativity) }} {{eqn | o = \le       | r = K \norm {y_n - m} + \norm m \norm {x_n - l}       | c = as $\sequence {x_n}$ is bounded by $K$ }} {{eqn | o = =:       | r = z_n       | c =  }} {{end-eqn}} We note that $\sequence {z_n}$ is a real sequence. But $x_n \to l$ as $n \to \infty$. So from Definition:Convergent Sequence in Normed Division Ring: :$\norm {x_n - l} \to 0$ as $n \to \infty$ Similarly $\norm {y_n - m} \to 0$ as $n \to \infty$. From the Combined Sum Rule for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x'_n + \mu y'_n} = \lambda l' + \mu m'$, $z_n \to 0$ as $n \to \infty$ By applying the Squeeze Theorem for Sequences of Complex Numbers (which applies as well to real as to complex sequences): :$\sequence {\norm {x_n y_n - l m}}$ converges to $0$ in $\R$. By definition of a convergent sequence in a normed division ring: :$\sequence{x_n y_n}$ is convergent in $R$ It follows that: :$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n y_n} = l m$ {{qed}}