hcju/proofwikiqa huggingface dataset:
We know that Terms in Convergent Series Converge to Zero. This is the contrapositive statement of this theorem. Thus, the theorem holds by Rule of Transposition. {{qed}}
We will borrow some ideas and techniques from the proof of Gauss's Lemma. Let $S = \set {a, 2 a, 3 a, \ldots, \dfrac {p - 1} 2 a}$. Let us create $S'$ from $S$ by replacing each element of $S$ by its least positive residue modulo $p$. Arranging $S'$ into increasing order, we get: :$S' = \set {b_1, b_2, \ldots, b_m, c_1, c_2, \ldots, c_n}$ where $b_m < \dfrac p 2 < c_1$ and $m + n = \dfrac {p - 1} 2$. According to the Division Theorem, we can divide any $k a \in S$ by $p$ and get: :$k a = p \times \floor {\dfrac {k a} p} + r$ where $r \in S'$. Let $k$ run from $1$ to $\dfrac {p - 1} 2$ and add up the resulting equations. We get: :$(1): \quad \displaystyle \sum_{k \mathop = 1}^{\frac {p - 1} 2} k a = \sum_{k \mathop = 1}^{\frac {p - 1} 2} p \times \floor {\frac {k a} p} + \sum_{i \mathop = 1}^{m} b_i + \sum_{j \mathop = 1}^n c_j$ As in the proof of Gauss's Lemma, all the remainders are in fact distinct elements of $S'$. From the proof of Gauss's Lemma again, we have that: :$S'' = \set {b_1, b_2, \ldots, b_m, p - c_1, p - c_2, \ldots, p - c_n} = \set {1, 2, 3, \ldots, \dfrac {p - 1} 2}$ Adding up all the elements of $S''$ we have: :$\displaystyle \sum_{k \mathop = 1}^{\frac {p - 1} 2} k = \sum_{i \mathop = 1}^m b_i + \sum_{j \mathop = 1}^n \paren {p - c_j} = \sum_{i \mathop = 1}^m b_i + p n - \sum_{j \mathop = 1}^n c_j$ Subtracting this equation from $(1)$ above gives us: :$\displaystyle \sum_{k \mathop = 1}^{\frac {p - 1} 2} k a - \sum_{k \mathop = 1}^{\frac {p - 1} 2} k = \sum_{k \mathop = 1}^{\frac {p - 1} 2} p \times \floor {\frac {k a} p} + 2 \sum_{j \mathop = 1}^n c_j - p n$ That is: :$\displaystyle \paren {a - 1} \sum_{k \mathop = 1}^{\frac {p - 1} 2} k = p \times \map \alpha {a, p} + 2 \sum_{j \mathop = 1}^n c_j - p n$ Now we are going to write this equation modulo $2$. Note that: :$\displaystyle \paren {a - 1} \sum_{k \mathop = 1}^{\frac {p - 1} 2} k$ is even, as $a$ is odd :$\displaystyle 2 \sum_{j \mathop = 1}^n c_j$ is also of course even. So: :$p \map \alpha {a, p} - p n \equiv 0 \pmod 2$ Since $p$ is odd, this amounts to saying: :$\map \alpha {a, p} \equiv n \pmod 2$ From the conclusion of Gauss's Lemma: :$\paren {\dfrac a p} = \paren {-1}^n$ Hence the result. {{qed}} {{proofread}} {{Namedfor|Ferdinand Gotthold Max Eisenstein|cat = Eisenstein}} Category:Number Theory jwto24xawat1vl6b6gobyyp7n1lqgu6
Let $p$ and $q$ be distinct odd primes. Consider the rectangle in the $x y$ plane with vertices at $\tuple {0, 0}, \tuple {\dfrac p 2, 0}, \tuple {\dfrac p 2, \dfrac q 2}, \tuple {0, \dfrac q 2}$. The number of lattice points inside this rectangle is $\dfrac {p - 1} 2 \times \dfrac {q - 1} 2$. :700px Consider the diagonal from $\tuple {0, 0}$ to $\tuple {\dfrac p 2, \dfrac q 2}$. This has the equation: :$y = \dfrac q p x$ {{AimForCont}} there were a lattice point $\tuple {a, b}$ on the diagonal. Then: :$b = \dfrac q p a$ But as $p$ and $q$ are distinct primes, this would mean: :$p$ divides $a$ and: :$q$ divides $b$ This means that $\tuple {a, b}$ has to be outside the rectangle. So there are no lattice points on the diagonal inside the rectangle. Let $A$ and $B$ be the triangular regions inside the rectangle lying respectively above and below the diagonal. Let $k \in \Z: 0 < k < \dfrac p 2$. The number of lattice points in $B$ which lie directly above the point $\tuple {k, 0}$ is: :$\floor {\dfrac q p k}$ where $\floor {\dfrac q p k}$ is the floor of $\dfrac q p k$. So the total number of lattice points in $B$ is given by: :$\displaystyle N_B = \sum_{k \mathop = 1}^{\frac{p - 1} 2} \floor {\dfrac q p k}$ Let $\map \alpha {q, p}$ be defined as: :$\displaystyle \map \alpha {q, p} = \sum_{k \mathop = 1}^{\frac {p - 1} 2} \floor {\dfrac q p k}$ In the same way, by counting the lattice points to the right of $\tuple {0, k}$, the total number of lattice points in $A$ is given by: :$N_A = \map \alpha {p, q}$ Now the total number of lattice points in the rectangle is $N_A + N_B$. But this is also equal to: :$\dfrac {p - 1} 2 \times \dfrac {q - 1} 2 = \dfrac {\paren {p - 1} \paren {q - 1} } 4$ So we have that: :$\map \alpha {p, q} + \map \alpha {q, p} = \dfrac {\paren {p - 1} \paren {q - 1} } 4$ The fact that these counts are equal relies upon the fact that there are no lattice points on the diagonal, as demonstrated above. Now we invoke Eisenstein's Lemma: :$\paren {\dfrac a p} = \paren {-1}^{\map \alpha {a, p} }$ So this gives us: {{begin-eqn}}      {{eqn | l = \paren {\frac p q} \paren {\frac q p}       | r = \paren {-1}^{\map \alpha {p, q} } \times \paren {-1}^{\map \alpha {q, p} }       | c =  }} {{eqn | r = \paren {-1}^{\map \alpha {p, q} + \map \alpha {q, p} }       | c =  }} {{eqn | r = \paren {-1}^{\frac {\paren {p - 1} \paren {q - 1} } 4}       | c =  }} {{end-eqn}} which is LQR. {{qed}} {{proofread}}
Follows from * Finite Simple Continued Fraction has Rational Value * the proof below === Proof that Every Rational Number can be expressed as a SFCF === Let $\dfrac a b$ be a rational number expressed in canonical form. That is $b > 0$ and $a \perp b = 1$. By the Euclidean Algorithm, we have: {{begin-eqn}} {{eqn | l=a       | r=q_1 b + r_1,       | rr=0 < r_1 < b       | c=or $\dfrac a b = q_1 + \dfrac {r_1} b$ }} {{eqn | l=b       | r=q_2 r_1 + r_2,       | rr=0 < r_2 < r_1       | c=or $\dfrac b {r_1} = q_2 + \dfrac {r_2} {r_1}$ }} {{eqn | l=r_1       | r=q_3 r_2 + r_3,       | rr=0 < r_3 < r_2       | c=or $\dfrac {r_2} {r_1} = q_3 + \dfrac {r_3} {r_2}$ }} {{eqn | o=\vdots }} {{eqn | l=r_{n-3}       | r=q_{n-1} r_{n-2} + r_{n-1},       | rr=0 < r_{n-1} < r_{n-2}       | c=or $\dfrac {r_{n-3} } {r_{n-2} } = q_{n-1} + \dfrac {r_{n-1} } {r_{n-2} }$ }} {{eqn | l=r_{n-2}       | r=q_n r_{n-1} + 0,       | c=or $\dfrac {r_{n-2} } {r_{n-1} } = q_n$ }} {{end-eqn}} Thus from the system of eqns at the right hand side, we get: {{begin-eqn}} {{eqn | l=\frac a b       | r=q_1 + \cfrac 1 {\frac b {r_1} }       | c= }} {{eqn | r=q_1 + \cfrac 1 {q_2 + \cfrac 1 {\frac {r_1} {r_2} } }       | c= }} {{eqn | r=q_1 + \cfrac 1 {q_2 + \cfrac 1 {q_3 + \cfrac 1 {\frac {r_2} {r_3} } } }       | c= }} {{eqn | o=\cdots       | c= }} {{eqn | r=q_1 + \cfrac 1 {q_2 + \cfrac 1 {q_3 + \cfrac 1 {\ddots \cfrac {} {q_{n-1} + \cfrac 1 {q_n} } } } }       | c= }} {{end-eqn}} This shows that $\dfrac a b$ has the SFCF $\left[{q_1, q_2, q_3, \ldots, q_n}\right]$. {{qed}}
There is a homomorphism $\phi: U \to \R^n$, since $X^n$ is a manifold. Let $y_i$ be coordinates in $\R^n$, so that any point $z \in \R^n$ has a unique representation $y_1, \dots, y_n$. Then define $x_i: X^n \to \R$ as $x_i = y_i \circ \phi$. {{qed}} Category:Topology 26dvvieee8v6pkl2rb4t1eln4t9pzmq
Let $k \ge 1$. From Denominators of Simple Continued Fraction are Strictly Positive, $q_k q_{k - 1}>0$. From Difference between Adjacent Convergents of Simple Continued Fraction: :$C_k - C_{k - 1} = \dfrac {\paren {-1}^{k + 1} } {q_k q_{k - 1} }$ Let $k$ be even. Then: :$\exists s \in \N: k = 2 s$ and: :$C_{2 s} < C_{2 s - 1}$ Let $k$ be odd. Then: :$\exists t \in \N: k = 2 t + 1$ and: :$C_{2 t + 1 } > C_{2 t}$ Now, consider any even convergent $C_{2 s}$ and any odd convergent $C_{2 t + 1}$, with $s, t \ge 0$. If $2 s \le 2 t + 1$, then $2 s \le 2 t$, so: {{begin-eqn}} {{eqn | l = C_{2 s}       | o = \le       | r = C_{2 t}       | c = Even Convergents of Simple Continued Fraction are Strictly Increasing }} {{eqn | o = <       | r = C_{2 t + 1}       | c = by the above }} {{end-eqn}} If $2 s > 2 t + 1$, then $2 s - 1 \ge 2 t + 1$, so: {{begin-eqn}} {{eqn | l = C_{2 s}       | o = <       | r = C_{2 s - 1}       | c = by the above }} {{eqn | o = \le       | r = C_{2 t + 1}       | c = Odd Convergents of Simple Continued Fraction are Strictly Decreasing }} {{end-eqn}} In either case: :$C_{2 s} < C_{2 t + 1}$ Hence the result. {{qed}}
By definition of simple continued fraction, all partial quotients of $\sqbrk {a_0, a_1, a_2, \ldots}$ are strictly positive integers, with the possible exception of $a_0$. So: :$q_1 = a_1 \ge 1 = q_0$ :$q_2 = a_2 a_1 + 1 \ge a_1 + 1 > a_1 = q_1$ hence: :$1 = q_0 \le q_1 < q_2$. Suppose $q_k > q_{k-1} \ge 1$ for some $k \ge 2$. Then $q_{k+1} = a_{k+1} q_k + q_{k-1} \ge q_k + q_{k-1} \ge q_{k} + 1 > q_k$. So, by induction, $\sequence {q_n}$ is strictly increasing except when possibly $q_0 = q_1 = 1$. {{qed}}
Let $\left[{a_0, a_1, a_2, \ldots}\right]$ be a simple infinite continued fraction. Note that by Simple Infinite Continued Fraction Converges, a simple infinite continued fraction is indeed convergent, say to $x \in \R$. Let $p_0, p_1, \ldots$ and $q_0, q_1, \ldots$ be its numerators and denominators. Let $C_0, C_1, \ldots$ be its convergents, so that $C_n = p_n/q_n$ for $n \geq 0$. For all $n \geq 0$, from Accuracy of Convergents of Convergent Simple Infinite Continued Fraction: :$\left\vert{x - \dfrac {p_n} {q_n} }\right\vert < \dfrac 1 {q_n q_{n + 1} }$ Suppose $x$ is rational. That is, let $x = \dfrac r s$ where $r, s \in \Z$ such that $s > 0$. Then: :$0 < \left|{\dfrac r s - \dfrac {p_n} {q_n}}\right| = \dfrac {\left|{r q_n - s p_n}\right|} {s q_n} < \dfrac 1 {q_n q_{n+1}}$ (Note that $\dfrac r s \ne \dfrac {p_n} {q_n}$ or otherwise the continued fraction would be finite.) So: :$0 < \left|{r q_n - s p_n}\right| < \dfrac s {q_{n+1}}$ But Denominators of Simple Continued Fraction are Strictly Increasing. That means we can choose $n$ so that $q_{n+1} > s$. But then $\left|{r q_n - s p_n}\right|$ would be an integer lying strictly between $0$ and $1$, which cannot happen. So no such integers $r, s$ exist. Thus $x$ must be irrational. {{qed}} Category:Continued Fractions oxdnw5ini7xc2guygtd3uc242oel5f0
Let $\map g z = \map {F_1} z - \map {F_2} z$. Then: :$\forall z \in U: \map g z = 0$ Because Zeroes of Analytic Function are Isolated, and the zeroes of $g$ are not isolated, $g$ must be constant everywhere in its domain. Since $\map g z = 0$ for some $z$, it follows that $\map g z = 0$ for all $z$. Hence: :$\forall z \in V: \map {F_1} z - \map {F_2} z = 0$ and so $F_1 = F_2$. {{qed}} Category:Complex Analysis 0pyi30lw0smgutmj81bz1efsam1hghh
Suppose $f$ has no zeroes in $U$. Then the set described in the theorem is the empty set, and we're done. So we suppose $\exists z_0 \in U$ such that $\map f {z_0} = 0$. Since $f$ is analytic, there is a Taylor series for $f$ at $z_0$ which converges for $\cmod {z - z_0} < R$. Now, since $\map f {z_0} = 0,$ we know $a_0 =0$. Other $a_j$ may be $0$ as well. So let $k$ be the least number such that $a_j = 0$ for $0 \le j < k$, and $a_k \ne 0$. Then we can write the Taylor series for $f$ about $z_0$ as: :$\displaystyle \sum_{n \mathop = k}^\infty a_n \paren {z - z_0}^n = \paren {z - z_0}^k \sum_{n \mathop = 0}^\infty a_{n + k} \paren {z - z_0}^n$  where $a_k \ne 0$ (otherwise, we'd just start at $k + 1$). Now we define a new function $\map g z$, as the sum on the right hand side, which is clearly analytic in $\cmod {z - z_0} < R$.   Since it is analytic here, it is also continuous here. Since $\map g {z_0} = a_k \ne 0, \exists \epsilon >0$ so that $\forall z$ such that $\cmod {z - z_0} < \epsilon, \cmod {\map g z - a_k} < \dfrac {\cmod {a_k} } 2$. But then $\map g z $ cannot possibly be $0$ in that disk. Hence the result. {{qed}} Category:Complex Analysis jf8s5xpcz1mubifqc80p9wynul3b3f5
Follows from: *Continued Fraction Expansion of Irrational Number Converges to Number Itself *Simple Infinite Continued Fraction is Uniquely Determined by Limit. {{Qed}}
The (yet to be confirmed) meromorphic continuation of the Riemann zeta function to the half-plane $\left\{{s: \Re \left({s}\right) > 0}\right\}$ is given by Integral Representation of Riemann Zeta Function in terms of Fractional Part: :$\displaystyle (1): \quad \zeta \left({s}\right) = \frac s {s - 1} - s \int_1^\infty \left\{ {x}\right\} x^{-s - 1} \ \mathrm d x$ where $\left\{{x}\right\}$ is the fractional part of $x$. Let $\Re \left({s}\right) \le 0$. Then the value of $\zeta \left({s}\right)$ can be computed from the relation: :$\Gamma \left({\dfrac s 2}\right) \pi^{-s/2} \zeta \left({s}\right) = \Gamma \left({\dfrac{s - 1} 2}\right) \pi^{\frac {1 - s} 2} \zeta\left({1 - s}\right)$ That is: : $\xi \left({s}\right) = \xi \left({1 - s}\right)$ where $\xi$ is the completed zeta function. First we show that $(1)$ is analytic for $\Re \left({s}\right) > 0$. For $n \ge 1$, let: {{begin-eqn}} {{eqn | l = a_n       | r = s \int_n^{n + 1} \left\{ {x}\right\} x^{-s - 1}\ \mathrm d x       | c =  }} {{eqn | o = \ll       | r = s \int_n^{n + 1} x^{-s-1}\ \mathrm d x       | c =  }} {{eqn | r = -x^{-s} \Big \vert_n^{n+1}       | c =  }} {{eqn | r = \frac 1 {n^s} - \frac 1 {\left({n + 1}\right)^s}       | c = }} {{eqn | r = \frac {\left({n + 1}\right)^s - n^s} {n^s \left({n + 1}\right)^s}       | c = }} {{end-eqn}} Here $\ll$ is the order notation. By the Mean Value Theorem, for some $n \le \theta \le n+1$: :$\left({n + 1}\right)^s - n^s = s \theta^{s-1} \le s \left({n + 1}\right)^{s-1}$ Thus if $s = \sigma + i t$: :$\left\vert{a_n}\right\vert \le \left\vert{\dfrac s {n^{s+1} } } \right\vert = \dfrac {\sigma^2 + t^2} {n^{\sigma + 1} }$ Since: :$\displaystyle \zeta \left({s}\right) = \frac s {s-1} - \sum_{n \mathop \ge 1} a_n$ it follows that this representation converges absolutely uniformly on $\Re \left({s}\right) > 0$. Thus by Uniform Limit of Analytic Functions is Analytic $\zeta \left({s}\right)$ is analytic for $\Re \left({s}\right) > 0$ and $s \ne 1$. For all $s$ with $\Re \left({s}\right) < \dfrac 1 2$, $\zeta \left({s}\right)$ is simply the reflection of $\zeta$ in the upper half plane. Therefore, $\zeta$ is also analytic for all $s$ with $\Re \left({s}\right) < 0$. {{qed}} Category:Riemann Zeta Function 7h65czo7r8xjtv4s4n9ro8a4tp6u3i2
The Continued Fraction Algorithm gives the following system of equations: {{begin-eqn}} {{eqn | l = x       | r = \left[{a_0, x_2}\right]       | c =  }} {{eqn | r = \left[{a_0, a_1, x_3}\right]       | c =  }} {{eqn | r = \ldots       | c =  }} {{eqn | r = \left[{a_0, a_1, \ldots, a_n, x_{n + 1} }\right]       | c =  }} {{eqn | r = \ldots       | c =  }} {{end-eqn}} and {{begin-eqn}} {{eqn | l = \left\vert{x - \frac {p_n} {q_n} }\right\vert       | r = \left\vert{\left[{a_0, a_1, \ldots, a_n, x_{n + 1} }\right] - \frac {p_n} {q_n} }\right\vert       | c =  }} {{eqn | r = \left\vert{\frac {x_{n + 1} p_n + p_{n - 1} } {x_{n + 1} q_n + q_{n - 1} } - \frac {p_n} {q_n} }\right\vert       | c = Value of Finite Continued Fraction equals Numerator Divided by Denominator }} {{eqn | r = \left\vert{\frac {p_{n - 1} q_n - p_n q_{n - 1} } {q_n \left({x_{n + 1} q_n + q_{n - 1} }\right)} }\right\vert       | c =  }} {{eqn | r = \frac 1 {q_n \left({x_{n + 1} q_n + q_{n - 1} }\right)}       | c = Difference between Adjacent Convergents of Simple Continued Fraction }} {{end-eqn}} Now from the Continued Fraction Algorithm: :$x_{n + 1} = \left[{a_{n + 1}, a_{n + 2}, a_{n + 3}, \ldots}\right]$ So: : $a_{n + 1} < x_{n + 1} < a_{n + 1} + 1$ Therefore: :$\left\vert{x - \dfrac {p_n} {q_n} }\right\vert < \dfrac 1 {q_n \left({a_{n + 1} q_n + q_{n - 1} }\right)} = \dfrac 1 {q_n q_{n + 1} }$ This gives the {{LHS}} of the inequality when $n = k + 1$. We also have: :$\left\vert{x - \dfrac {p_n} {q_n} }\right\vert > \dfrac 1 {q_n \left({\left({a_{n + 1} + 1}\right) q_n + q_{n - 1} }\right)}$ {{begin-eqn}} {{eqn | l = \left\vert{x - \frac {p_n} {q_n} }\right\vert       | o = >       | r = \frac 1 {q_n \left({\left({a_{n + 1} + 1}\right) q_n + q_{n - 1} }\right)}       | c =  }} {{eqn | r = \frac 1 {q_n \left({a_{n + 1} q_n + q_{n - 1} }\right) + q_n^2}       | c =  }} {{eqn | r = \frac 1 {q_n \left({q_{n + 1} + q_n}\right)}       | c =  }} {{eqn | o = >       | r = \frac 1 {q_n \left({q_{n + 1} + q_{n + 1} }\right)}       | c =  }} {{eqn | r = \frac 1 {2 q_n q_{n + 1} }       | c =  }} {{end-eqn}} This gives the {{RHS}} of the inequality when $n = k$. For the middle inequality, note that: :$q_{k + 2} = a_{k + 2} q_{k + 1} + q_k > q_k + q_k = 2 q_k$ So: :$\dfrac 1 {q_{k + 1} q_{k + 2} } \le \dfrac 1 {2 q_k q_{k + 1} }$ {{qed}}
{{finish|First it needs to be established that those points are in fact zeroes. This follows directly and trivially from Riemann Zeta Function at Non-Positive Integers, but needs to be made explicit here.}} First we note that by Zeroes of Gamma Function, $\Gamma$ has no zeroes on $\C$. Therefore, the completed Riemann zeta function: :$\map \xi s = \dfrac 1 2 s \paren {s - 1} \pi^{-s/2} \map \Gamma {\dfrac s 2} \, \map \zeta s$ has the same zeroes as $\zeta$. Additionally by Functional Equation for Riemann Zeta Function, we have: :$\map \xi s = \map \xi {1 - s}$ for all $s \in \C$. Therefore if: :$\map \zeta s \ne 0$ for all $s$ with $\map \Re s > 1$, then also: :$\map \zeta s \ne 0$ for all $s$ with $\map \Re s < 0$. Let us consider $\map \Re s > 1$. We have: :$\displaystyle \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$ where here and in the following $p$ ranges over the primes. Therefore, we have: :$\displaystyle \map \zeta s \prod_p \paren {1 - p^{-s} } = 1$ All of the factors of this infinite product can be found in the product: :$\displaystyle \prod_{n \mathop = 2}^\infty \paren {1 - n^{-s} }$ which converges absolutely. This follows because from Convergence of P-Series: :$\displaystyle \sum_{k \mathop = 1}^\infty k^{-s}$ converges absolutely. Hence: :$\displaystyle \prod_p \paren {1 - p^{-s} }$ converges absolutely. So by the fact that: :$\displaystyle \map \zeta s \prod_p \paren {1 - p^{-s} } = 1$ we know $\map \zeta s$ cannot possibly be zero for any point in the region in question. {{qed}}
First we examine the location of the poles. We examine the Weierstrass form of the Gamma function: :$\dfrac 1 {\map \Gamma z} = z e^{\gamma z} \displaystyle \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }$ The factors of the product clearly do not tend to zero. So the product is zero only if one of the factors is zero. This occurs when $1 + \dfrac z n = 0$. This occurs at $z \in \set {-1, -2, \dotsc}$. We also have the expression outside the product to consider. From Exponential Tends to Zero and Infinity, the exponential function is never $0$. So this expression is zero whenever $z = 0$. Furthermore, if some factor in the product is zero, this excludes the possibility that any other factor is zero. So the zeros are simple. Hence: :$\dfrac 1 {\map \Gamma z} = 0 \iff z \in \set {0, -1, -2, -3, \ldots}$ Therefore $\Gamma$ has simple poles at these points. Next we show that $\Gamma$ is analytic on $\map \Re z > 0$.  By Gamma Difference Equation this proves that $\Gamma$ is analytic on $\C \setminus \set {0, -1, -2, -3, \dotsc}$. Let the functions $\Gamma_n$ be defined by :$\displaystyle \map {\Gamma_n} z = \int_0^n t^{z - 1} e^{-t} \rd t$ Clearly each $\Gamma_n$ is analytic. So by Uniform Limit of Analytic Functions is Analytic it is sufficient to show that $\Gamma_n \to \Gamma$ locally uniformly. By Modulus of Complex Integral: :$\displaystyle \cmod {\map \Gamma z - \map {\Gamma_n} Z } \le \int_n^\infty t^{x - 1} e^{-t} \rd t$ where $x = \map \Re z$.  Let $a \in \C$ with $\map \Re a > 0$. Let $D$ be an open disk of radius $r$ about $a$. By definition of exponential function, we have the expansion: :$e^z = 1 + z + \dfrac {z^2} 2 + \dotsb$ from which we see that: :$\forall \alpha \in \R_{>0}: z^\alpha = \map o {e^z}$ where $o$ denores little-o notation. In particular: :$\exists c \in \R_{>0}: \forall t \ge 1: e^{-t} \le c t^{-\map \Re a - r - 1}$ Then: {{begin-eqn}} {{eqn | l = \cmod {\map \Gamma z - \map {\Gamma_n} z}       | o = \le       | r = \int_n^\infty t^{x - 1} c t^{-\map \Re a -r - 1} \rd t }} {{eqn | o = \le       | r = c\int_n^\infty t^{-2} \rd t }} {{eqn | o = \le       | r = \frac c n }} {{end-eqn}} So $\Gamma_n \to \Gamma$ uniformly in $D$, and the proof is complete. {{qed}}
Suppose $\exists z$ such that $\Gamma \left({z}\right) = 0$. We examine the Euler form of the gamma function, which is defined for $\C - \left\{{0, -1, -2, \ldots}\right\}$. The Euler form, equated with zero, yields :$\displaystyle 0 = \frac 1 z \prod_{n \mathop = 1}^\infty \left({\left({1 + \frac 1 n}\right)^z \left({1 + \frac z n}\right)^{-1} }\right)$ It is clear that $\dfrac 1 z \ne 0$, so we may divide this out for $z$ in the area of definition. Now it is clear that as $n \to \infty$, each of the two halves of the term in the product will tend to $1$ for any $z$, and there is no $z$ which yields zero for any $n$ in either of the product terms. Hence this product will not equal $0$ anywhere. This leaves only the question of the behavior on $\left\{{0,-1,-2, \ldots}\right\}$, which is discussed at Poles of Gamma Function. {{qed}} Category:Gamma Function j6ltofli311h7s3qpuw7qxxu20t5wbb
Let $z \in \C$, with $\map \Re z > 0$. Then: {{begin-eqn}} {{eqn | l = \map \Gamma {z + 1}       | r = \int_0^\infty t^z e^{-t} \rd t       | c = }} {{eqn | r = \bigintlimits {-t^z e^{-t} } 0 \infty + z \int_0^\infty t^{z - 1} e^{-t} \rd t       | c = Integration by Parts }} {{eqn | r = z \int_0^\infty t^{z - 1} e^{-t} \rd t       | c = }} {{eqn | r = z \, \map \Gamma z       | c = }} {{end-eqn}} If $z \in \C \setminus \set {0, -1, -2, \ldots}$ such that $\map \Re z \le 0$, then the statement holds by the definition of $\Gamma$ in this region. {{qed}}
By Euler's form of the Gamma function: {{begin-eqn}} {{eqn | l = \frac {\map \Gamma {z + 1} } {\map \Gamma z}       | r = \paren {\frac 1 {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \frac 1 n}^{z + 1} } {1 + \frac {z + 1} n} } \div \paren {\frac 1 z \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \frac {\paren {1 + \frac 1 n}^z} {1 + \frac z n} }       | c =  }} {{eqn | r = \frac z {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \paren {\frac {\paren {1 + \frac 1 n}^{z + 1} \paren {1 + \frac z n} } {\paren {1 + \frac 1 n}^z \paren {1 + \frac {z + 1} n} } }       | c =  }} {{eqn | r = \frac z {z + 1} \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \paren {\frac {\paren {1 + \frac 1 n} \paren {z + n} } {z + n + 1} }       | c =  }} {{eqn | r = z \lim_{m \mathop \to \infty} \frac {m + 1} {z + m + 1} = z       | c =  }} {{end-eqn}} {{qed}}