hcju/proofwikiqa huggingface dataset:
From Product of r Choose k with r Minus Half Choose k: :$\dbinom r k \dbinom {r - \frac 1 2} k = \dfrac {\dbinom {2 r} k \dbinom {2 r - k} k} {4^k}$ Setting $r = -\dfrac 1 2$: {{begin-eqn}} {{eqn | l = \dbinom {-\frac 1 2} k \dbinom {-1} k       | r = \frac 1 {4^k} \dbinom {-1} k \dbinom {-1 - k} k       | c =  }} {{eqn | ll= \leadsto       | l = \dbinom {-\frac 1 2} k       | r = \frac 1 {4^k} \dbinom {-1 - k} k       | c =  }} {{eqn | r = \frac {\paren {-1}^k} {4^k} \dbinom {k - \paren {-1 - k} - 1} k       | c = Negated Upper Index of Binomial Coefficient }} {{eqn | r = \frac {\paren {-1}^k} {4^k} \dbinom {2 k} k       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = r^{\underline {a + b} }       | r = \prod_{j \mathop = 0}^{a + b - 1} \left({r - j}\right)       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \left({\prod_{j \mathop = 0}^{a - 1} \left({r - j}\right)}\right) \left({\prod_{j \mathop = a}^{a + b - 1} \left({r - j}\right)}\right)       | c =  }} {{eqn | r = \left({\prod_{j \mathop = 0}^{a - 1} \left({r - j}\right)}\right) \left({\prod_{j \mathop = 0}^{b - 1} \left({r - a - j}\right)}\right)       | c = Translation of Index Variable of Product }} {{eqn | r = r^{\underline a} \left({r - a}\right)^{\underline b}       | c = {{Defof|Falling Factorial}} }} {{end-eqn}} {{qed}} Category:Falling Factorials omcl2pl3k5146o1fidt0yuhc3w8g0es
{{ProofWanted|Resembles Binet's Formula for Logarithm of Gamma Function}}
{{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^{\infty} \frac 1 {n^8}        | r = \map \zeta 8        | c = {{Defof|Riemann Zeta Function}} }} {{eqn | r = \paren {-1}^5 \frac {B_8 2^7 \pi^8} {8!}        | c = Riemann Zeta Function at Even Integers }} {{eqn | r = \frac 1 {30} \cdot \frac {2^7 \pi^8} {8!}       | c = {{Defof|Sequence of Bernoulli Numbers}} }} {{eqn | r = \frac {128 \pi^8} {30 \cdot 40 \, 320}        | c = {{Defof|Factorial}} }} {{eqn | r = \frac {\pi^8} {9450} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n H_k       | r = \sum_{k \mathop = 1}^n \paren {\sum_{j \mathop = 1}^k \frac 1 j}       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^n \paren {\sum_{k \mathop = j}^n \frac 1 j}       | c = Summation of i from 1 to n of Summation of j from 1 to i }} {{eqn | r = \sum_{j \mathop = 1}^n \frac {n + 1 - j} j       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^n \frac {n + 1} j - \sum_{j \mathop = 1}^n \frac j j       | c =  }} {{eqn | r = \paren {n + 1} H_n - n       | c =  }} {{end-eqn}} {{qed}}
From Sum over k to n of k Choose m by kth Harmonic Number: :$\displaystyle \sum_{k \mathop = 1}^n \binom k m H_k = \binom {n + 1} {m + 1} \left({H_{n + 1} - \frac 1 {m + 1} }\right)$ Setting $m = 0$: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \binom k 0 H_k        | r = \binom {n + 1} {0 + 1} \left({H_{n + 1} - \frac 1 {0 + 1} }\right)       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{j \mathop = 1}^n H_k        | r = \left({n + 1}\right) \left({H_{n + 1} - 1}\right)       | c = Binomial Coefficient with $0$, Binomial Coefficient with $1$ }} {{eqn | r = \left({n + 1}\right) \left({H_n + \frac 1 {n + 1} - 1}\right)       | c = {{Defof|Harmonic Number}} }} {{eqn | r = \left({n + 1}\right) H_n + \left({n + 1}\right) \frac 1 {n + 1} - \left({n + 1}\right) 1       | c =  }} {{eqn | r = \left({n + 1}\right) H_n + 1 - \left({n + 1}\right)       | c =  }} {{eqn | r = \left({n + 1}\right) H_n - n       | c =  }} {{end-eqn}} {{qed}}
Let $\left \langle {a_n}\right \rangle$ be the sequence defined as: : $\forall n \in \N_{> 0}: a_n = H_n$ where $H_n$ denotes the $n$th harmonic number. Let $G \left({z}\right)$ be the generating function for $\left \langle {a_n}\right \rangle$. From Generating Function for Sequence of Harmonic Numbers: :$G \left({z}\right) = \dfrac 1 {1 - z} \ln \left({\dfrac 1 {1 - z} }\right)$ Then: {{begin-eqn}} {{eqn | l = G' \left({z}\right)       | r = \dfrac 1 {\left({1 - z}\right)^2} \ln \left({\dfrac 1 {1 - z} }\right) + \dfrac 1 {\left({1 - z}\right)^2}       | c = Derivative of Generating Function for Sequence of Harmonic Numbers }} {{eqn | r = \dfrac 1 {1 - z} G \left({z}\right) + \dfrac 1 {\left({1 - z}\right)^2}       | c =  }} {{end-eqn}} From Generating Function for Sequence of Partial Sums of Series, $\dfrac 1 {1 - z} G \left({z}\right)$ is the generating function for $\left \langle {b_n}\right \rangle$ where: :$b_n = \displaystyle \sum_{k \mathop = 0}^n H_k$ and so: :$\dfrac 1 {\left({1 - z}\right)^2} \ln \left({\dfrac 1 {1 - z} }\right) = \displaystyle \sum_{n \mathop \ge 0} \left({\sum_{k \mathop = 0}^n H_k}\right) z^n$ From Generating Function for Natural Numbers: :$\dfrac 1 {\left({1 - z}\right)^2} = \displaystyle \sum_{n \mathop \ge 0} \left({n + 1}\right) z^n$ That is: :$G' \left({z}\right) = \displaystyle \sum_{n \mathop \ge 0} \left({\sum_{k \mathop = 0}^n H_k + \left({n + 1}\right)}\right) z^n$ Now we have: {{begin-eqn}} {{eqn | l = G' \left({z}\right)       | r = \dfrac \d {\d z} \left({\sum_{n \mathop \ge 0} H_n z^n}\right)       | c = Derivative of Generating Function for Sequence of Harmonic Numbers }} {{eqn | r = \sum_{n \mathop \ge 0} n H_n z^{n - 1}       | c = Derivative of Power }} {{end-eqn}} Equating coefficients of $z^n$ in these two expressions for $G' \left({z}\right)$: :$\displaystyle \sum_{k \mathop = 0}^n H_k + \left({n + 1}\right) = \left({n + 1}\right) H_{n + 1}$ {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^n H_k + \left({n + 1}\right)       | r = \left({n + 1}\right) H_{n + 1}       | c =  }} {{eqn | r = \left({n + 1}\right) H_n + \left({n + 1}\right) \left({\dfrac 1 {n + 1} }\right)       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 1}^n H_k + n       | r = \left({n + 1}\right) H_n       | c = $H_k = 0$ when $k = 0$, and simplification }} {{end-eqn}} The result follows. {{qed}}
First we note that by Pascal's Rule: :$\dbinom k m = \dbinom {k + 1} {m + 1} - \dbinom k {m + 1}$ Thus: {{begin-eqn}} {{eqn | l = \dbinom k m H_k       | r = \dbinom k {m + 1} \left({H_{k + 1} - \dfrac 1 {k + 1} }\right) - \dfrac k {m + 1} H_k       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 1}^n \binom k m H_k       | r = \left({\binom 2 {m + 1} H_2 - \binom 1 {m + 1} H_1}\right)       | c =  }} {{eqn | o =        | ro= +       | r = \cdots       | c =  }} {{eqn | o =        | ro= +       | r = \left({\binom {n + 1} {m + 1} H_{n + 1} - \binom n {m + 1} H_1}\right)       | c =  }} {{eqn | o =        | ro= -       | r = \sum_{k \mathop = 1}^n \binom {k + 1} {m + 1} \frac 1 {k + 1}       | c =  }} {{eqn | r = \binom {n + 1} {m + 1} H_{n + 1} - \binom n {m + 1} H_1 - \frac 1 {m + 1} \sum_{k \mathop = 0}^n \binom k n + \frac 1 {k + 1} \binom 0 m       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 1}^n \binom k m H_k       | r = \binom {n + 1} {m + 1} \left({H_{n + 1} - \frac 1 {m + 1} }\right)       | c = Sum of Binomial Coefficients over Upper Index }} {{end-eqn}} {{qed}}
Let $S_n := \displaystyle \sum_{k \mathop \in \Z} \binom n k x^k H_k$. Then: {{begin-eqn}} {{eqn | l = S_{n + 1}       | r = \sum_{k \mathop \in \Z} \binom {n + 1} k x^k H_k       | c =  }} {{eqn | r = \sum_{k \mathop \in \Z} \left({\binom n k + \binom n {k - 1} }\right) x^k H_k       | c = Pascal's Rule }} {{eqn | r = \sum_{k \mathop \in \Z} \binom n k x^k H_k + \sum_{k \mathop \in \Z} \binom n {k - 1} x^k H_k       | c =  }} {{eqn | r = S_n + x \sum_{k \mathop \ge 1} \binom n {k - 1} x^{k - 1} \left({H_{k - 1} + \frac 1 k}\right)       | c = Definition of $S_n$ }} {{eqn | r = S_n + x \sum_{k \mathop \ge 1} \binom n {k - 1} x^{k - 1} H_{k - 1} + x \sum_{k \mathop \ge 1} \binom n {k - 1} x^{k - 1} \frac 1 k       | c =  }} {{eqn | r = S_n + x \sum_{k \mathop \ge 0} \binom n k x^k H_k + \sum_{k \mathop \ge 1} \binom n {k - 1} x^k \frac 1 k       | c = Translation of Index Variable of Summation }} {{eqn | r = S_n + x S_n + \sum_{k \mathop \ge 1} \frac k {n + 1} \binom {n + 1} k x^k \frac 1 k       | c = Definition of $S_n$, Factors of Binomial Coefficient }} {{eqn | r = \left({x + 1}\right) S_n + \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k x^k       | c = simplifying }} {{eqn | r = \left({x + 1}\right) S_n + \frac 1 {n + 1} \left({\sum_{k \mathop \ge 0} \binom {n + 1} k x^k - \binom {n + 1} 0 x^0}\right)       | c = forcing the summation to include the $k = 0$ term }} {{eqn | r = \left({x + 1}\right) S_n + \frac {\left({x + 1}\right)^{n + 1} - 1} {n + 1}       | c = Binomial Theorem and Binomial Coefficient with $0$ }} {{eqn | ll= \leadsto       | l = \frac {S_{n + 1} } {\left({x + 1}\right)^{n + 1} }       | r = \frac {S_n} {\left({x + 1}\right)^n} + \frac 1 {n + 1} - \frac 1 {\left({n + 1}\right) \left({x + 1}\right)^{n + 1} }       | c = dividing through by $\left({x + 1}\right)^{n + 1}$ }} {{end-eqn}} As $S_1 = x$, we have that: :$\dfrac {S_n} {\left({x + 1}\right)^n} = H_n - \displaystyle \sum_{k \mathop = 1}^n \frac 1 {k \left({x + 1}\right)^k}$ {{explain|Not sure why}} From Power Series Expansion for Logarithm of 1 over 1-z we have: :$\ln \dfrac 1 {1 - z} = \displaystyle \sum_{k \mathop \ge 1} \frac {z^k} k$ Thus: {{begin-eqn}} {{eqn | l = \ln \left({1 + \frac 1 x}\right)       | r = \ln \dfrac 1 {1 - \dfrac 1 {x + 1} }       | c =  }} {{eqn | r = \sum_{k \mathop \ge 1} \frac 1 {k \left({x + 1}\right)^k}       | c =  }} {{end-eqn}} So as $n \to \infty$, $\displaystyle \sum_{k \mathop \ge 1} \frac 1 {k \left({x + 1}\right)^k}$ converges, and so: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \frac 1 {k \left({x + 1}\right)^k}       | o = <       | r = \frac 1 {\left({n + 1}\right) \left({x + 1}\right)^{n + 1} } \sum_{k \mathop \ge 0} \frac 1 {\left({x + 1}\right)^k}       | c =  }} {{eqn | r = \frac 1 {\left({n + 1}\right) \left({x + 1}\right)^n x}       | c =  }} {{end-eqn}} The result follows. {{qed}}
:$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 n = \underbrace 1_{s_0} + \underbrace {\frac 1 2 + \frac 1 3}_{s_1} + \underbrace {\frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7}_{s_2} + \cdots$  where $\displaystyle s_k = \sum_{i \mathop = 2^k}^{2^{k + 1} \mathop - 1} \frac 1 i$ From Ordering of Reciprocals: : $\forall m, n \in \N_{>0}: m > n: \dfrac 1 m < \dfrac 1 n$ so each of the summands in a given $s_k$ is less than $\dfrac 1 {2^k}$. The number of summands in a given $s_k$ is $2^{k + 1} - 2^k = 2 \times 2^k - 2^k = 2^k$, and so: :$s_k < \dfrac {2^k} {2^k} = 1$ Hence the harmonic sum $H_{2^m}$ satisfies the following inequality: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^{2^m} \frac 1 n       | r = \sum_{k \mathop = 0}^m \left({s_k}\right)       | c =  }} {{eqn | o = <       | r = \sum_{a \mathop = 0}^m 1       | c =  }} {{eqn | r = 1 + m       | c =  }} {{end-eqn}} Hence the result. {{qed}}
If $r$ is rational this has already been established. {{MissingLinks|Find where.}} Let $d$ denote the difference between $x^{\psi^1}$ and $x^{\psi^2}$: {{begin-eqn}} {{eqn | l = d       | r = x^{\psi^2} - x^{\psi^1}       | c =  }} {{eqn | r = x^{\psi^1} \left({x^{\frac 1 {10^k} } - 1}\right)       | c =  }} {{eqn | r = x^{\psi^1} \left({x^{\frac 1 {10^k} } - 1}\right)       | c =  }} {{end-eqn}} It follows from Nth Root of 1 plus x not greater than 1 plus x over n that: :$d < \dfrac {x^{n + 1} \left({x - 1}\right)} {10^k}$ Thus as $k \to \infty$, $d \to 0$. The result follows from the Squeeze Theorem. {{qed}}
From Change of Base of Logarithm: :$\log_a x = \log_a b \ \log_b x$ Substituting $a = 10$ and $b = 2$ gives: :$\log_{10} x = \left({\lg x}\right) \left({\log_{10} 2}\right)$ The common logarithm of $2$: :$\log_{10} 2 = 0 \cdotp 30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$ can be calculated or looked up. {{qed}}
{{AimForCont}} $x = \dfrac p q$ is the smallest strictly positive rational number. By definition of strictly positive: :$0 < \dfrac p q$ Let us calculate the mediant of $0$ and $\dfrac p q$: :$\dfrac 0 1 < \dfrac {0 + p} {1 + q} < \dfrac p q$ The inequalities follow from Mediant is Between. Thus $\dfrac p {1 + q}$ is a strictly positive rational number which is smaller than $\dfrac p q$. Thus $\dfrac p q$ cannot be the smallest strictly positive rational number. The result follows by Proof by Contradiction. {{qed}}
From Change of Base of Logarithm: :$\log_a x = \log_a b \ \log_b x$ Substituting $a = e$ and $b = 10$ gives: :$\ln x = \paren {\ln 10} \paren {\log_{10} x}$ The Natural Logarithm of 10: :$\ln 10 = 2 \cdotp 30258 \, 50929 \, 94 \ldots$ can be calculated or looked up. {{qed}}
From Change of Base of Logarithm: :$\log_a x = \dfrac {\log_b x} {\log_b a}$ Substituting $a = e$ and $b = 10$ gives: :$\ln x = \dfrac {\log_{10} x} {\log_{10} e}$ The Common Logarithm of e: :$\log_{10} e = 0 \cdotp 43429 \, 44819 \, 03 \ldots$ can then be calculated or looked up. {{qed}} Category:Examples of Change of Base of Logarithm 9pk1pf7ln90gee1arnrnnu7hei93er2
From Change of Base of Logarithm: :$\log_a x = \log_a b \, \log_b x$ Substituting $a = 10$ and $b = e$ gives: :$\log_{10} x = \paren {\log_{10} e} \paren {\ln x}$ The Common Logarithm of e: :$\log_{10} e = 0 \cdotp 43429 \, 44819 \, 03 \ldots$ can be calculated or looked up. {{qed}}
From Change of Base of Logarithm: :$\log_a x = \dfrac {\log_b x} {\log_b a}$ Substituting $a = 10$ and $b = e$ gives: :$\log_{10} x = \dfrac {\ln x} {\ln 10}$ as by definition of $\ln x$: :$\ln x := \log_e x$ The Natural Logarithm of 10: :$\ln 10 = 2 \cdotp 30258 \, 50929 \, 94 \ldots$ can then be calculated or looked up. {{qed}}
From Change of Base of Logarithm: :$\log_a x = \dfrac {\log_b x} {\log_b a}$ Substituting $a = e$ and $b = 10$ gives: :$\log_e x = \dfrac {\log_{10} x} {\log_{10} e}$ The Common Logarithm of 2: :$\log_{10} 2 = 0 \cdotp 30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$ can then be calculated or looked up. {{qed}} Category:Examples of Common Logarithms Category:Examples of Binary Logarithms Category:Examples of Change of Base of Logarithm etifvsevdac1nxxs8fz4g3nk62chban
{{begin-eqn}} {{eqn | l = \sqrt [n] {\dfrac a b}       | r = \paren {\dfrac a b}^{1 / n}       | c = {{Defof|Root (Analysis)|$n$th Root}} }} {{eqn | r = \paren {a \times \dfrac 1 b}^{1 / n}       | c =  }} {{eqn | r = a^{1 / n} \times \paren {\dfrac 1 b}^{1 / n}       | c = Exponent Combination Laws/Power of Product }} {{eqn | r = a^{1 / n} \times b^{-1 / n}       | c = Exponent Combination Laws/Negative Power }} {{eqn | r = a^{1 / n} \times \dfrac 1 {b^{1 / n} }       | c = Exponent Combination Laws/Negative Power }} {{eqn | r = \sqrt [n] a \times \dfrac 1 {\sqrt [n] b}       | c = {{Defof|Root (Analysis)|$n$th Root}} }} {{eqn | r = \dfrac {\sqrt [n] a} {\sqrt [n] b}       | c =  }} {{end-eqn}} {{qed}}
By definition of logarithm: {{begin-eqn}} {{eqn | l = y       | r = \log_b b       | c =  }} {{eqn | ll= \leadstoandfrom       | l = b^y       | r = b       | c = {{Defof|Real General Logarithm}} }} {{eqn | ll= \leadstoandfrom       | l = y       | r = 1       | c = {{Defof|Power to Real Number}} }} {{end-eqn}} {{qed}} Category:Logarithms nzbjueziu6rt7577su6ztppps8eck02
By definition of logarithm: {{begin-eqn}} {{eqn | l = \log_b 1       | r = \frac {\log_e 1} {\log_e b}       | c = Change of Base of Logarithm }} {{eqn | r = \frac 0 {\log_e b}       | c = Natural Logarithm of 1 is 0 }} {{eqn | r = 0       | c = whatever $\log_e b$ happens to be }} {{end-eqn}} {{qed}}
{{AimForCont}} $\log_b \left({-1}\right) = y \in \R$. Then: :$b^y = -1 < 0$ But from Power of Positive Real Number is Positive: :$b^y > 0$ The result follows by Proof by Contradiction. {{qed}}
From Change of Base of Logarithm: :$\log_a x = \dfrac {\log_b x} {\log_b a}$ Substituting $a = 8$ and $b = 2$ gives: :$\log_8 x = \dfrac {\log_2 x} {\log_2 8}$ We have that: {{begin-eqn}} {{eqn | l = 2^3       | r = 8       | c =  }} {{eqn | ll= \leadsto       | l = \lg 8       | r = \log_2 2^3       | c =  }} {{eqn | r = 3 \log_2 2       | c =  }} {{eqn | r = 3       | c = Logarithm to Own Base equals 1 }} {{end-eqn}} The result follows. {{qed}}
Let $n$ have $d$ digits. Then: {{begin-eqn}} {{eqn | l = b^{d - 1}       | o = \le       | m = n       | mo= <       | r = b^d       | c = Basis Representation Theorem }} {{eqn | ll= \leadsto       | l = b^{d - 1}       | o = <       | m = n + 1       | mo= \le       | r = b^d       | c =  }} {{eqn | ll= \leadsto       | l = d - 1       | o = <       | m = \map {\log_b} {n + 1}       | mo= \le       | r = d       | c =  }} {{eqn | ll= \leadsto       | o =        | m = \ceiling {\map {\log_b} {n + 1} }       | mo= =       | r = d       | c = Integer equals Ceiling iff Number between Integer and One Less }} {{end-eqn}} {{qed}} Category:Number Theory 7uk6ird8jwcsa8ian5bgrtxks94kyxo
Let $d$ be the number of bits that may be needed to represent $n$. Let $n$ have $m$ digits. Then: :$n \le 10^m - 1$ and so: {{begin-eqn}} {{eqn | l = d       | r = \ceiling {\map {\log_2} {\paren {10^m - 1} + 1} }       | c = Number of Digits to Represent Integer in Given Number Base }} {{eqn | r = \ceiling {\map {\log_2} {10^m} }       | c =  }} {{eqn | r = \ceiling {m \log_2 10}       | c =  }} {{eqn | r = \ceiling {\dfrac m {\log_{10} 2} }       | c = Reciprocal of Logarithm }} {{end-eqn}} {{qed}}
For $n = 1$ :$\map \arg {z^1} = 1 \cdot \map \arg z$ Assuming the result is true for $n = k$, we have: {{begin-eqn}} {{eqn | l = \map \arg {z^{k + 1} }       | r = \map \arg {z z^k} }} {{eqn | r = \map \arg z + \map \arg {z^k}       | c = Argument of Product equals Sum of Arguments }} {{eqn | r = \map \arg z + k \map \arg z       | c = by our induction hypothesis }} {{eqn | r = \paren {k + 1} \map \arg z }} {{end-eqn}} {{qed}} q7sayp8wd0d8htaxqjuls43da5tek8v
Proof by Counterexample: {{proof wanted|No idea at the moment. Have to find a series of $a_{i j}$ which is conditionally convergent for a start.}}
The summation on the {{LHS}} is vacuous: :$\displaystyle \sum_{j \mathop = 0}^{-1} x^j = 0$ while on the {{RHS}} we have: {{begin-eqn}} {{eqn | l = \frac {x^{\paren {-1} + 1} - 1} {x - 1}       | r = \frac {x^0 - 1} {x - 1}       | c =  }} {{eqn | r = \frac 0 {x - 1}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} as long as $x \ne 1$. However, the theorem itself is based on the assumption that $n \ge 0$, so while the result is correct, the derivation to achieve it is not. {{qed}}
The summation on the {{LHS}} is vacuous: :$\displaystyle \sum_{j \mathop = 0}^{-2} x^j = 0$ while on the {{RHS}} we have: {{begin-eqn}} {{eqn | l = \frac {x^{\paren {-2} + 1} - 1} {x - 1}       | r = \frac {x^{-1} - 1} {x - 1}       | c =  }} {{eqn | r = \frac {1 / x - 1} {x - 1}       | c =  }} {{eqn | r = \frac {\paren {1 - x} / x} {x - 1}       | c =  }} {{eqn | r = \frac {1 - x} {x \paren {x - 1} }       | c =  }} {{eqn | r = -\frac 1 x       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \displaystyle \sum_{j \mathop = 0}^n \dfrac 1 {7^j}       | r = \frac {1 - \frac 1 7^{n + 1} } {1 - \frac 1 7}       | c =  }} {{eqn | r = \frac {7 - 7 \frac 1 7^{n + 1} } {7 - 1}       | c = multiplying top and bottom by $7$ }} {{end-eqn}} Hence the result. {{qed}}
From Sum of Arithmetic-Geometric Sequence: :$\displaystyle \sum_{j \mathop = 0}^n \paren {a + j d} r^j = \frac {a \paren {1 - r^{n + 1} } } {1 - r} + \frac {r d \paren {1 - \paren {n + 1} r^n + n r^{n + 1} } } {\paren {1 - r}^2}$ Hence: {{begin-eqn}} {{eqn | l = \displaystyle \sum_{j \mathop = 0}^n j \, 2^j       | r = \frac {0 \paren {1 - 2^{n + 1} } } {1 - 2} + \frac {2 \times 1 \paren {1 - \paren {n + 1} 2^n + n 2^{n + 1} } } {\paren {1 - 2}^2}       | c = putting $a = 0, d = 1, r = 2$ }} {{eqn | r = 2 \paren {1 - \paren {n + 1} 2^n + n 2^{n + 1} }       | c = initial simplification }} {{eqn | r = 2 - \paren {n + 1} 2^{n + 1} + n 2^{n + 2}       | c = further simplification }} {{end-eqn}} Hence the result. {{qed}}
From Exchange of Order of Summation with Dependency on Both Indices: :$\map {S'} j$ denotes the propositional function: ::there exists an $i$ such that both $\map R i$ and $\map S {i, j}$ hold :$\map {R'} {i, j}$ denotes the propositional function: ::both $\map R i$ and $\map S {i, j}$ hold. The definition of $\map {R'} {i, j}$ follows immediately: :$\map R i := \paren {n = k i \text { for some } k \in \Z}$ and: :$\map S {i, j} := \paren {1 \le j < i}$ Then: {{begin-eqn}} {{eqn | l = \map {S'} j       | r = \paren {\exists i \in \Z: n = k i}       | c =  }} {{eqn | o =        | ro= \land       | r = \paren {1 \le j < i}       | c =  }} {{end-eqn}} By Absolute Value of Integer is not less than Divisors, it follows from $\map R i$ that $i \le n$. That is, for $\map {S'} j$ to hold, $i \le n$. But for all $j \in \Z$ such that $1 \le j < n$ it follows that $i = n$ fulfils the condition that $n = k i$. Hence: :$\map {S'} j := \paren {1 < j \le n}$ {{qed}}
Let $\left({\mathcal P, \preceq}\right)$ be a partially ordered set. The Hausdorff Maximal Principle states that there is a maximal chain in $\left({\mathcal P, \preceq}\right)$. If this maximal chain has an upper bound in $\mathcal P$, then such an upper bound is a maximal element of $\mathcal P$ itself, by the maximality of the chain. {{qed}}
Let $X$ be a set. If $X = \varnothing$ the theorem holds vacuously, so assume $X$ is not empty. Let $\mathcal W \in \mathcal P(X)$ be the set of all well-ordered subsets of $X$. Then $\mathcal W$ is partially ordered by $\subseteq$. Furthermore, there is an upper bound for every chain in $\mathcal W$: the union of all chains in $\mathcal W$. Thus the hypotheses of Zorn's Lemma hold and we can conclude that $\mathcal W$ has a maximal element. Seeking a contradiction suppose that $\mathcal W \subsetneqq X$. Let $(E,\subseteq)$ be the maximal well ordering on $X$, as defined above, and assume $x_0 \in X \setminus E$.  Define an ordering $\preceq'$ on $X \cup \{ x_0 \}$ as follows: :$ \preceq' := \begin{cases} x \preceq y & x,y \in E \\ x_0 \preceq x & x \in E,x_0 \in X \setminus E \end{cases}$ Then $\preceq'$ properly includes $\preceq$, contradicting the maximality of $\preceq$. Thus the well ordering on $\mathcal W$ is a well ordering on all of $X$. {{qed}}
Define: :$A_n = \set {x \in X: \map f x >\dfrac 1 n, n \in \N_{\ge 1} }$ Then $\sequence {A_n}_{n \mathop \in \N} \uparrow A$ is an exhausting sequence of sets, where: :$A = \set {x \in X: \map f x > 0}$ Suppose $A$ is uncountable. From Countable Union of Countable Sets is Countable, necessarily there is some $A_{n_0}$ which is uncountable. Then: {{begin-eqn}} {{eqn | l = \displaystyle \sum_{x \mathop \in X} \map f x       | r = \sup \set {\sum_{x \mathop \in F} \map f x: F \subseteq X, F \text{ finite} } }} {{eqn | o = \ge       | r = \sup \set {\sum_{x \mathop \in F} \map f x: F \subseteq A_{n_0}, F \text{ finite} } }} {{eqn | o = \ge        | r = \sup \set {\sum_{x \mathop \in F} \frac 1 {n_0} : F \subseteq A_{n_0}, F \text{ finite} } }} {{eqn | r = \sum_{n \mathop = 1}^{\infty} \frac 1 {n_0} }} {{eqn | r = +\infty }} {{end-eqn}} Thus: :$\displaystyle \sum_{x \mathop \in X} \map f x = +\infty$ Otherwise, suppose then that $A$ is countably infinite. Then there is a bijection $g: \N \leftrightarrow A$ Define $B_N = g \sqbrk {\set {1, 2, \ldots, N - 1, N} }$ Then every finite subset $F$ of $A$ is contained in some $B_N$. This implies the inequality: :$\displaystyle \sum_{x \mathop \in F} \map f x \le \sum_{n \mathop = 1}^N \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$ for each $F$ finite. Taking the supremum of this inequality over $N$: :$\displaystyle \sum_{x \mathop \in F} \map f x \le \sum_{n \mathop = 1}^\infty \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$ Taking the supremum of this inequality over $F$: :$\displaystyle \sum_{x \mathop \in X} \map f x \le \sum_{n \mathop = 1}^\infty \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$ Thus: :$\displaystyle \sum_{x \mathop \in X} \map f x = \sum_{n \mathop = 1}^\infty \map f {\map g n}$  for some bijection $g: \N \leftrightarrow A$ {{qed}}