hcju/proofwikiqa huggingface dataset:
=== Definition 1 implies Definition 2 === Let $f$ be a mapping which fulfils the condition: :$\forall y \in T: \exists x \in \Dom f: f \paren x = y$ From Image is Subset of Codomain: : $\Img f \subseteq T$ It remains to be proved that: : $T \subseteq \Img f$ Thus: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = T }} {{eqn | ll= \leadsto       | l = \exists x \in \Dom f: f \paren x       | r = y       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \Img f       | c ={{Defof|Image of Mapping}} }} {{eqn | ll= \leadsto       | l = T       | o = \subseteq       | r = \Img f       | c = {{Defof|Subset}} }} {{end-eqn}} Thus by definition of set equality: : $\Img f = T$ and by definition of image of mapping: : $f \sqbrk S = T$ Hence $f$ is a surjection by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $f$ be a mapping which fulfils the condition: : $f \sqbrk S = T$ that is: : $\Img f = T$ Then by definition of set equality: : $T \subseteq \Img f$ Hence: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = T }} {{eqn | ll= \implies       | l = y       | o = \in       | r = \Img f }} {{end-eqn}} So by the definition of the image of $f$: : $\exists x \in \Dom f: f \paren x = y$ Hence $f$ is a surjection by definition 1. {{Qed}}
Let $y \in \mathcal R \left [{\operatorname{Dom} \left({\mathcal R}\right)}\right]$. {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \mathcal R \left [{\operatorname{Dom} \left({\mathcal R}\right)}\right]       | c =  }} {{eqn | lll= \implies       | ll= \exists x \in \operatorname{Dom} \left({\mathcal R}\right):       | l = \left({x, y}\right)       | o = \in       | r = \mathcal R       | c = Definition of Image of Subset under Relation }} {{eqn | lll= \implies       | l = y       | o = \in       | r = \operatorname{Im} \left ({\mathcal R}\right)       | c = Definition of Image of Relation }} {{eqn | lll= \implies       | l = \mathcal R \left [{\operatorname{Dom} \left({\mathcal R}\right)}\right]       | o = \subseteq       | r = \operatorname{Im} \left ({\mathcal R}\right)       | c = Definition of Subset }} {{end-eqn}} Let $y \in \operatorname{Im} \left ({\mathcal R}\right)$. {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \operatorname{Im} \left ({\mathcal R}\right)       | c =  }} {{eqn | lll= \implies       | ll= \exists x \in S:       | l = \left({x, y}\right)       | o = \in       | r = \mathcal R       | c = Definition of Image of Relation }} {{eqn | lll= \implies       | l = y       | o = \in       | r = \mathcal R \left [{\operatorname{Dom} \left({\mathcal R}\right)}\right]       | c = Definition of Domain of Relation }} {{eqn | lll= \implies       | l = \operatorname{Im} \left ({\mathcal R}\right)       | o = \subseteq       | r = \mathcal R \left [{\operatorname{Dom} \left({\mathcal R}\right)}\right]       | c = Definition of Subset }} {{end-eqn}} The result follows by definition of set equality. {{qed}} Category:Relation Theory 5y8jw6ej980i5w3sw6mn13q53vrki50
=== Necessary Condition === Suppose $\neg \paren {\exists x, y, z \in S: x \mathrel {\RR} y \land y \mathrel {\RR} z}$. Then $\RR$ is both transitive and antitransitive vacuously. {{qed|lemma}} === Sufficient Condition === Suppose $\RR$ is both transitive and antitransitive. {{AimForCont}} it is not the case that $\neg \paren {\exists x, y, z \in S: x \mathrel {\RR} y \land y \mathrel {\RR} z}$. Then $\exists x, y, z \in S: x \mathrel {\RR} y \land y \mathrel {\RR} z$. By transitivity: :$x \mathrel {\RR} z$ By antitransitivity: :$\neg \paren {x \mathrel {\RR} z}$ This is a contradiction. Hence we must have $\neg \paren {\exists x, y, z \in S: x \mathrel {\RR} y \land y \mathrel {\RR} z}$. {{qed}}
Let $\left({x, y}\right) \in \mathcal R$ for some $x, y \in S$. Then as $\mathcal R$ is antitransitive: :$\left({x, x}\right) \notin \mathcal R$ and so as $\mathcal R$ is transitive and $\left({x, x}\right) \notin \mathcal R$: :$\left({y, x}\right) \notin \mathcal R$ That is, $\mathcal R$ is asymmetric. {{qed}}
Consider Figure $\text I$: {{:Definition:Figure of Categorical Syllogism/I}} Let the major premise of $Q$ be denoted $\text{Maj}$. Let the minor premise of $Q$ be denoted $\text{Min}$. Let the conclusion of $Q$ be denoted $\text{C}$. $M$ is: : the subject of $\text{Maj}$ : the predicate of $\text{Min}$. So, in order for $M$ to be distributed, either: : $(1): \quad$ From Universal Categorical Statement Distributes its Subject: $\text{Maj}$ must be universal or: : $(2): \quad$ From Negative Categorical Statement Distributes its Predicate: $\text{Min}$ must be negative. Suppose $\text{Min}$ is a negative categorical statement. Then by Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative: : $\text{C}$ is a negative categorical statement. From $(2)$: : $P$ is distributed in $\text{C}$. From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise: : $P$ is distributed in $\text{Maj}$. From Negative Categorical Statement Distributes its Predicate: : $\text{Maj}$ is a negative categorical statement. Thus both: : $\text{Min}$ is a negative categorical statement : $\text{Maj}$ is a negative categorical statement. But from No Valid Categorical Syllogism contains two Negative Premises, this means $Q$ is invalid. Thus $\text{Min}$ is not a negative categorical statement in Figure $\text I$. As $\text{Min}$ needs to be an affirmative categorical statement, $M$ is not distributed in $\text{Min}$. From Middle Term of Valid Categorical Syllogism is Distributed at least Once, this means $M$ must be distributed in $\text{Maj}$. As $M$ is the subject of $\text{Maj}$ in Figure $\text I$, it follows from $(1)$ that: : $\text{Maj}$ is a universal categorical statement. Hence, in order for $Q$ to be valid: : $\text{Maj}$ must be a universal categorical statement : $\text{Min}$ must be an affirmative categorical statement. {{qed}}
Consider Figure $\text{II}$: {{:Definition:Figure of Categorical Syllogism/II}} Let the major premise of $Q$ be denoted $\text{Maj}$. Let the minor premise of $Q$ be denoted $\text{Min}$. Let the conclusion of $Q$ be denoted $\text{C}$. $M$ is: : the predicate of $\text{Maj}$ : the predicate of $\text{Min}$. So, in order for $M$ to be distributed, either: : $(1): \quad$ From Negative Categorical Statement Distributes its Predicate: $\text{Maj}$ must be negative or: : $(2): \quad$ From Negative Categorical Statement Distributes its Predicate: $\text{Min}$ must be negative. Note that from No Valid Categorical Syllogism contains two Negative Premises, it is not possible for both $\text{Maj}$ and $\text{Min}$ to be negative. From Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative: : $\text{C}$ is a negative categorical statement. From Negative Categorical Statement Distributes its Predicate: : $P$ is distributed in $\text{C}$. From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise: : $P$ is distributed in $\text{Maj}$. From Universal Categorical Statement Distributes its Subject: : $\text{Maj}$ is a universal categorical statement. Hence, in order for $Q$ to be valid: : $\text{Maj}$ must be a universal categorical statement : Either $\text{Maj}$ or $\text{Min}$, and therefore $\text{C}$, must be a negative categorical statement. {{qed}}
Consider Figure $\text {III}$: {{:Definition:Figure of Categorical Syllogism/III}} Let the major premise of $Q$ be denoted $\text{Maj}$. Let the minor premise of $Q$ be denoted $\text{Min}$. Let the conclusion of $Q$ be denoted $\text{C}$. $M$ is: : the subject of $\text{Maj}$ : the subject of $\text{Min}$. So, in order for $M$ to be distributed, either: : $(1): \quad$ From Universal Categorical Statement Distributes its Subject: $\text{Maj}$ must be universal or: : $(2): \quad$ From Universal Categorical Statement Distributes its Subject: $\text{Min}$ must be universal. Suppose $\text{Min}$ to be a negative categorical statement. Then by No Valid Categorical Syllogism contains two Negative Premises: : $\text{Maj}$ is an affirmative categorical statement. From Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative: : $\text{C}$ is a negative categorical statement. From Negative Categorical Statement Distributes its Predicate: : $P$ is distributed in $\text{C}$. From Negative Categorical Statement Distributes its Predicate: : $P$ is undistributed in $\text{Maj}$. From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise: : $P$ is distributed in $\text{Maj}$. That is, $P$ is both distributed and undistributed in $\text{Maj}$. From this Proof by Contradiction it follows that $\text{Min}$ is an affirmative categorical statement. Thus from Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative: : if $\text{C}$ is a negative categorical statement, then so is $\text{Maj}$ {{qed|lemma}} We have that $\text{Min}$ is an affirmative categorical statement. Hence from Negative Categorical Statement Distributes its Predicate: : $S$ is undistributed in $\text{Min}$. From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise: : $S$ is undistributed in $\text{C}$. So from Universal Categorical Statement Distributes its Subject: : $\text{C}$ is a particular categorical statement. {{qed|lemma}} Hence, in order for $Q$ to be valid: : $\text{C}$ must be a particular categorical statement : if $\text{C}$ is a negative categorical statement, then so is $\text{Maj}$. {{qed}}
Consider Figure $\text {IV}$: {{:Definition:Figure of Categorical Syllogism/IV}} Let the major premise of $Q$ be denoted $\text{Maj}$. Let the minor premise of $Q$ be denoted $\text{Min}$. Let the conclusion of $Q$ be denoted $\text{C}$. $M$ is: : the predicate of $\text{Maj}$ : the subject of $\text{Min}$. We have: :Middle Term of Valid Categorical Syllogism is Distributed at least Once. So, in order for $M$ to be distributed, either: : From Negative Categorical Statement Distributes its Predicate: $\text{Maj}$ must be negative or: : From Universal Categorical Statement Distributes its Subject: $\text{Min}$ must be universal. Both may be the case. Thus $(1)$ is seen to hold. {{qed|lemma}} Let $\text{C}$ be a negative categorical statement. From Negative Categorical Statement Distributes its Predicate: : $P$ is distributed in $\text{C}$. From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise: : $P$ is distributed in $\text{Maj}$. So from Universal Categorical Statement Distributes its Subject: : $\text{Maj}$ is a universal categorical statement. Thus $(2)$ is seen to hold. {{qed|lemma}} Let $\text{C}$ be a universal categorical statement. From Universal Categorical Statement Distributes its Subject: : $S$ is distributed in $\text{C}$. From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise: : $S$ is distributed in $\text{Min}$. From Negative Categorical Statement Distributes its Predicate: : $S$ is a negative categorical statement. Thus $(3)$ is seen to hold. {{qed}}
From Elimination of all but 48 Categorical Syllogisms as Invalid there are $12$ possible patterns of categorical syllogism per figure: :$\begin{array}{cccccc} AAA & AAI & AEE & AEO & AII & AOO \\ EAE & EAO & EIO & IAI & IEO & OAO \\ \end{array}$ === Figure $\text I$ === Consider Figure $\text I$: {{:Definition:Figure of Categorical Syllogism/I}} From Valid Syllogism in Figure I needs Affirmative Minor Premise and Universal Major Premise, the patterns: :$AEE$, $AEO$, $AOO$ and $IEO$ can be eliminated as they all have a negative minor premise, and: :$IAI$ and $OAO$ can be eliminated as they all have a particular major premise. Thus the only patterns in Figure $\text I$ that may be valid are: :$\begin{array}{rl} \text{I} & AAA \\ \text{I} & AII \\ \text{I} & EAE \\ \text{I} & EIO \\ \text{I} & AAI \\ \text{I} & EAO \\ \end{array}$ {{qed|lemma}} === Figure $\text {II}$ === Consider Figure $\text {II}$: {{:Definition:Figure of Categorical Syllogism/II}} From Valid Syllogism in Figure II needs Negative Conclusion and Universal Major Premise, the patterns: :$AAA$, $AAI$, $AII$ and $IAI$ can be eliminated as they all have an affirmative conclusion, and: :$IEO$ and $OAO$ can be eliminated as they all have a particular major premise. Thus the only patterns in Figure $\text {II}$ that may be valid are: :$\begin{array}{rl} \text{II} & EAE \\ \text{II} & AEE \\ \text{II} & AOO \\ \text{II} & EIO \\ \text{II} & EAO \\ \text{II} & AEO \\ \end{array}$ {{qed|lemma}} === Figure $\text {III}$ === Consider Figure $\text {III}$: {{:Definition:Figure of Categorical Syllogism/III}} From Valid Syllogism in Figure III needs Particular Conclusion and if Negative then Negative Major Premise, the patterns: :$AAA$, $AEE$ and $EAE$ can be eliminated as they all have a universal conclusion, and: :$AEO$, $AOO$ and $IEO$ can be eliminated as they all have a negative minor premise. Thus the only patterns in Figure $\text {III}$ that may be valid are: :$\begin{array}{rl} \text{III} & AAI \\ \text{III} & AII \\ \text{III} & IAI \\ \text{III} & EAO \\ \text{III} & EIO \\ \text{III} & OAO \\ \end{array}$ {{qed|lemma}} === Figure $\text {IV}$ === Consider Figure $\text {IV}$: {{:Definition:Figure of Categorical Syllogism/IV}} From Valid Syllogisms in Figure IV, the patterns: :$AII$ and $EOO$ can be eliminated as they have an affirmative major premise and a particular minor premise. :$IEO$ and $OAO$ can be eliminated as they have a negative conclusion and a particular major premise. :$AAA$ and $EAE$ can be eliminated as they have a universal conclusion and a negative minor premise. Thus the only patterns in Figure $\text {IV}$ that may be valid are: :$\begin{array}{rl} \text{IV} & AAI \\ \text{IV} & AEE \\ \text{IV} & EAO \\ \text{IV} & EIO \\ \text{IV} & IAI \\ \text{IV} & AEO \\ \end{array}$ {{qed|lemma}} It remains to be established whether these $24$ patterns actually do represent valid categorical syllogisms. {{qed}}
Since $\Gamma$ is contradictory, there is some WFF $\mathbf A$ such that both $\mathbf A$ and $\neg \mathbf A$ occur along $\Gamma$. Since $\Gamma'$ is an extension of $\Gamma$, $\mathbf A$ and $\neg \mathbf A$ also occur along $\Gamma'$. Hence $\Gamma'$ is contradictory. {{qed}} Category:Propositional Tableaus qma3zjj5lx66bs29ndupq530a06tdmc
Suppose $T$ has no finished branch. Then since $T$ is finished, every branch of $T$ is contradictory. Hence $T$ is a tableau confutation. {{qed}}
Note first that: :$\displaystyle \sum_{n \mathop = 1}^\infty \size {\frac {\paren {-1}^\paren {n - 1} } n} = \sum_{n \mathop = 1}^\infty \frac 1 n$ which is divergent by Harmonic Series is Divergent. Next note that $\size {\dfrac 1 n}$ is a basic null sequence: :$\displaystyle \lim_{n \mathop \to \infty} \dfrac 1 n = 0$ and that: : $\forall n \in \N_{>0}: \dfrac 1 n > 0$ From Reciprocal Sequence is Strictly Decreasing: :$\dfrac 1 n > \dfrac 1 {n + 1}$ The result follows from the Alternating Series Test. {{qed}}
Let $\epsilon > 0$. From Tail of Convergent Series tends to Zero, it follows that there exists $N \in \N$ such that: :$\displaystyle \sum_{n \mathop = N}^\infty \size {a_n} < \epsilon$ By definition, a permutation is bijective. Hence we can find $M \in \N$ such that: :$\set {1, \ldots, N - 1} \subseteq \set {\map \pi 1, \ldots, \map \pi M}$ Let $m \in \N$, and put $B = \set {n \in N: \map {\pi^{-1} } n > m}$. For all $m \ge M$, it follows that: {{begin-eqn}} {{eqn | l = \size {\sum_{n \mathop = 1}^\infty a_n - \sum_{k \mathop = 1}^m a_{\map \pi k} }       | r = \size {\sum_{n \mathop = 1}^\infty a_n \map {\chi_B} n}       | c = where $\chi_B$ is the characteristic function of $B$ }} {{eqn | o = \le       | r = \sum_{n \mathop = 1}^\infty \size {a_n} \map {\chi_B} n       | c = by Triangle Inequality }} {{eqn | o = \le       | r = \sum_{n \mathop = N}^\infty \size {a_n}       | c = as $\chi_B = 0$ for all $n < N$ }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} By definition of convergent series, it follows that: :$\displaystyle \sum_{n \mathop = 1}^\infty a_n = \lim_{m \mathop \to \infty} \sum_{k \mathop = 1}^m a_{\map \pi k} = \sum_{k \mathop = 1}^\infty a_{\map \pi k}$ {{qed}}
For all $N \in \N$, we have: :$\displaystyle \sum_{n \mathop = 1}^N \size {a_n \map {\chi_A} n} \le \sum_{n \mathop = 1}^N \size {a_n} \le \sum_{n \mathop = 1}^\infty \size {a_n}$ It follows that: :$\displaystyle \sum_{n \mathop = 1}^\infty \size {a_n \map {\chi_A} n} \le \sum_{n \mathop = 1}^\infty \size {a_n}$ Then $\displaystyle \sum_{n \mathop = 1}^\infty a_n \map {\chi_A} n$ is absolutely convergent. It follows from Manipulation of Absolutely Convergent Series: Permutation that the order of the terms in the series does not matter. As $\map {\chi_A} n = 0$ for all $n \notin A$, we have: :$\displaystyle \sum_{n \mathop = 1}^\infty a_n \map {\chi_A} n = \sum_{n \mathop \in A} a_n$ {{qed}} Category:Series Category:Convergence 099d9r1iq961cebm8qrb9gzp2qxkbxd
{{begin-eqn}} {{eqn | l = c \sum_{n \mathop = 1}^\infty a_n       | r = c \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N a_n }} {{eqn | r = \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N c a_n       | c = Multiple Rule for Sequences }} {{eqn | r = \sum_{n \mathop = 1}^\infty c a_n }} {{end-eqn}} {{qed}} Category:Series Category:Convergence nvhddmods7jgv3h5uh9umhpimeim6t8
We shall make use of the Weierstrass M-Test to prove this result. To begin with, for each $n \in N$, define for $x \in D$: :$\map {f_n} x = a_n \paren {x - \xi}^n$ We have: {{begin-eqn}} {{eqn | l = \size {\map {f_n} x}       | r = \size {a_n \paren {x - \xi}^n} }} {{eqn | o = \le       | r = \size {a_n \rho^n}       | c = $x \in D$ }} {{end-eqn}} Define $M_n = \size {a_n \rho^n}$. Then $M_n$ is an upper bound for $\size {\map {f_n} x}$. So, by the definition of a supremum: :$\displaystyle \sup_{x \mathop \in D} \size {\map {f_n} x} \le M_n$ Next, to apply the Weierstrass M-Test, it remains to be shown that $\displaystyle \sum_{n \mathop = 0}^\infty M_n = \sum_{n \mathop = 0}^\infty \size {a_n \rho^n}$ converges. Choose $x = \xi + \rho$ and note that: :$\size {x - \xi} = \size \rho = \rho < R$ Hence, by definition of radius of convergence, $x$ falls within the interval of convergence of $S$. Thus $\displaystyle S = \sum_{n \mathop = 0}^\infty a_n (x - \xi)^n = \sum_{n \mathop = 0}^\infty a_n \rho^n$ converges. By Existence of Interval of Convergence of Power Series, a power series always converges absolutely at all points in its interval of convergence. Hence $\displaystyle \sum_{n \mathop = 0}^\infty M_n = \sum_{n \mathop = 0}^\infty \size {a_n \rho^n}$ also converges. Finally then, having satisfied all the requirements to use Weierstrass M-Test, we can conclude that $\displaystyle \sum_{n \mathop = 0}^\infty \map {f_n} x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ converges uniformly in $D$. {{qed}}
Let $\rho \in \R$ such that $0 \le \rho < R$. From Power Series Converges Uniformly within Radius of Convergence, $\map f x$ is uniformly convergent on $\set {x: \size {x - \xi} \le \rho}$. From Real Polynomial Function is Continuous, each of $\map {f_n} x = a_n x^n$ is a continuous function of $x$. The result follows from Uniformly Convergent Series of Continuous Functions is Continuous. {{qed}}
Let $\rho \in \R$ such that $0 \le \rho < R$. From Power Series Converges Uniformly within Radius of Convergence, $\map f x$ is uniformly convergent on $\set {x: \size {x - \xi} \le \rho}$. From Real Polynomial Function is Continuous, each of $\map {f_n} x = a_n x^n$ is a continuous function of $x$. Then from Definite Integral of Uniformly Convergent Series of Continuous Functions: :$\displaystyle \int_a^b \map f x \rd x = \sum_{n \mathop = 0}^\infty \int_a^b a_n x^n \rd x$ The final result follows from Integral of Power. {{qed}}
Let $\rho \in \R$ such that $0 \le \rho < R$. From Power Series Converges Uniformly within Radius of Convergence, $\map f x$ is uniformly convergent on $\set {x: \size {x - \xi} \le \rho}$. From Real Polynomial Function is Continuous, each of $\map {f_n} x = a_n x^n$ is a continuous function of $x$. From Power Rule for Derivatives: :$\dfrac \d {\d x} x^n = n x^{n - 1}$ and again from Real Polynomial Function is Continuous, each of $\dfrac \d {\d x} \map {f_n} x = n a_n x^{n - 1}$ is a continuous function of $x$. Then from Derivative of Uniformly Convergent Series of Continuously Differentiable Functions: {{begin-eqn}} {{eqn | l = \frac \d {\d x} \map f x       | r = \sum_{n \mathop = 0}^\infty \frac \d {\d x} a_n x^n }} {{eqn | r = \sum_{n \mathop = 0}^\infty n a_n x^{n - 1}       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty n a_n x^{n - 1}       | c = as $n a_n x^{n - 1} = 0$ when $n = 0$ }} {{end-eqn}} {{qed}}
From Sum of Infinite Geometric Sequence, putting $-x$ for $x$: :$(1): \quad \displaystyle \sum_{n \mathop = 0}^\infty \paren {-x}^n = \frac 1 {1 + x}$ for $-1 < x < 1$. From Power Series Converges Uniformly within Radius of Convergence, $(1)$ is uniformly convergent on every closed interval within the interval $\openint {-1} 1$. From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term: {{begin-eqn}} {{eqn | l = \int_0^x \frac 1 {1 + t} \rd t       | r = \sum_{n \mathop = 0}^\infty \int_0^x \paren {-t}^n \rd t       | c =  }} {{eqn | ll= \leadsto       | l = \map \ln {1 + x}       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{n + 1} } {n + 1}       | c = Primitive of Reciprocal and Integral of Power }} {{eqn | ll= \leadsto       | l = \map \ln {1 + x}       | r = \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {x^n} n       | c = letting $n \to n - 1$ }} {{end-eqn}} {{qed}}
From the definition of the Newton-Mercator Series: {{begin-eqn}} {{eqn | l = \map \ln {1 + x}       | r = x - \dfrac {x^2} 2 + \dfrac {x^3} 3 - \dfrac {x^4} 4 + \cdots       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n x^n       | c =  }} {{end-eqn}} This is valid for $-1 < x \le 1$. Setting $x = 1$: {{begin-eqn}} {{eqn | l = \map \ln {1 + 1}       | r = 1 - \dfrac {1^2} 2 + \dfrac {1^3} 3 - \dfrac {1^4} 4 + \cdots       | c =  }} {{eqn | ll= \leadsto       | l = \ln 2       | r = 1 - \dfrac 1 2 + \dfrac 1 3 - \dfrac 1 4 + \dfrac 1 5 - \cdots       | c =  }} {{end-eqn}} The fact that it does indeed converge is shown in Alternating Harmonic Series is Conditionally Convergent. {{qed}}
From Derivative of Sine Function: :$\dfrac \d {\d x} \sin x = \cos x$ From Derivative of Cosine Function: :$\dfrac \d {\d x} \cos x = -\sin x$ Hence: {{begin-eqn}} {{eqn | l = \dfrac {\d^2} {\d x^2} \sin x       | r = -\sin x       | c =  }} {{eqn | l = \dfrac {\d^3} {\d x^3} \sin x       | r = -\cos x       | c =  }} {{eqn | l = \dfrac {\d^4} {\d x^4} \sin x       | r = \sin x       | c =  }} {{end-eqn}} and so for all $m \in \N$: {{begin-eqn}} {{eqn | ll= m = 4 k:       | l = \dfrac {\d^m} {\d x^m} \sin x       | r = \sin x       | c =  }} {{eqn | ll= m = 4 k + 1:       | l = \dfrac {\d^m} {\d x^m} \sin x       | r = \cos x       | c =  }} {{eqn | ll= m = 4 k + 2:       | l = \dfrac {\d^m} {\d x^m} \sin x       | r = -\sin x       | c =  }} {{eqn | ll= m = 4 k + 3:       | l = \dfrac {\d^m} {\d x^m} \sin x       | r = -\cos x       | c =  }} {{end-eqn}} where $k \in \Z$. This leads to the Maclaurin series expansion: {{begin-eqn}} {{eqn | l = \sin x       | r = \sum_{r \mathop = 0}^\infty \paren {\frac {x^{4 k} } {\paren {4 k}!} \map \sin 0 + \frac {x^{4 k + 1} } {\paren {4 k + 1}!} \map \cos 0 - \frac {x^{4 k + 2} } {\paren {4 k + 2}!} \map \sin 0 - \frac {x^{4 k + 3} } {\paren {4 k + 3}!} \map \cos 0}       | c =  }} {{eqn | r = \sum_{r \mathop = 0}^\infty \paren {\frac {x^{4 k + 1} } {\paren {4 k + 1}!} - \frac {x^{4 k + 3} } {\paren {4 k + 3}!} }       | c = Sine of Zero is Zero, Cosine of Zero is One }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}       | c = setting $n = 2 k$ }} {{end-eqn}} From Series of Power over Factorial Converges, it follows that this series is convergent for all $x$. {{qed}}