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Demuestre Identidad

PaulRS Ver Perfil del Miembro Ver los mensajes de este miembro	Apr 20 2008, 12:54 PM Publicado: #1
Doctor en Matemáticas Grupo: Usuario FMAT Mensajes: 156 Registrado: 11-June 07 Miembro Nº: 6.615 Nacionalidad: Sexo:	$TEX: $<br />\begin{gathered}<br /> {\text{Demuestre que:}}{\text{ }}\gamma = \sum\limits_{k = 2}^\infty {\tfrac{{\left( { - 1} \right)^k \cdot \zeta \left( k \right)}}<br />{k}} \hfill \\<br /> {\text{Donde }}\gamma = \mathop {\lim }\limits_{n \to + \infty } \left[ {\sum\limits_{k = 1}^n {\left( {\tfrac{1}<br />{k}} \right)} - \ln \left( {n + 1} \right)} \right]{\text{ y }}\zeta \left( s \right) = \sum\limits_{k = 1}^\infty {\left( {\tfrac{1}<br />{{k^s }}} \right)} \hfill \\ <br />\end{gathered} <br />$$ Saludos -------------------- $TEX: $\sqrt[3]{\displaystyle\sum_{i=1}^n{i^{3\cdot{\sqrt[]{3}}-1}}}\approx{\displaystyle\sum_{i=1}^n{i^{\sqrt[]{3}-1}}}$$

Jean Renard Gran... Jean Renard Granier Ver Perfil del Miembro Ver los mensajes de este miembro	Nov 14 2008, 04:11 PM Publicado: #2
Dios Matemático Supremo Grupo: Usuario FMAT Mensajes: 1.920 Registrado: 19-August 06 Desde: DIM, DCC Beauchef Miembro Nº: 1.989 Nacionalidad: Universidad: Sexo:	$TEX: $$PDQ$$$ $TEX: $$\ln \left( {\Gamma \left( {x + 1} \right)} \right) = - \gamma x + \sum\limits_{j = 2}^\infty {\frac{{\left( { - 1} \right)^j \zeta \left( j \right)}}{j}} x^j ,\left\| x \right\| < 1$$$ Aquí se piensa "hacer tender $TEX: $$x$$$ a $TEX: $$1$$$ ". $TEX: $$\Gamma \left( {1 + x} \right) = \int_0^\infty {\phi ^x e^{ - \phi } } d\phi$$$ $TEX: $$\int_0^\infty {\phi ^x e^{ - \phi } } d\phi = \mathop {\lim}\limits_{n \to \infty } \int_0^n {\phi ^x } \left( {1 - \frac{\phi }{n}} \right)^n d\phi$$$ Gracias Coneno y Paul $TEX: $$\mathop {\lim }\limits_{n \to \infty } \int_0^n {\phi ^x } \left( {1 - \frac{\phi }{n}} \right)^n d\phi$$$ $TEX: $$\mathop {\lim }\limits_{n \to \infty } n^{x + 1} \int_0^1 {\phi ^x } \left( {1 - \phi } \right)^n d\phi ,\int_0^1 {\phi ^x } \left( {1 - \phi } \right)^n d\phi = \frac{{n!}}{{\prod\limits_{k = 1}^{n + 1} {\left( {x + k} \right)} }}$$$ $TEX: $$\mathop {\lim }\limits_{n \to \infty } \frac{{n^{x + 1} n!}}{{\prod\limits_{k = 1}^{n + 1} {\left( {x + k} \right)} }}/\ln \left( {} \right)$$$ $TEX: $$\ln \left( {\Gamma \left( {1 + x} \right)} \right) = \mathop {\lim }\limits_{n \to \infty } \left[ {\left( {1 + x} \right)\ln \left( n \right) + \ln \left( {n!} \right) - \sum\limits_{k = 1}^{n + 1} {\ln \left( {x + k} \right)} } \right]$$$ $TEX: $$\mathop {\lim }\limits_{n \to \infty } \left[ {\left( {1 + x} \right)\ln \left( n \right) + \ln \left( {n!} \right) - \sum\limits_{k = 1}^{n + 1} {\ln \left( {x + k} \right)} } \right]/ + \left( {\ln \left( {n + 1} \right) - \ln \left( {n + 1} \right)} \right)$$$ $TEX: $$\mathop {\lim }\limits_{n \to \infty } \left[ {\left( {1 + x} \right)\ln \left( n \right) - \ln \left( {n + 1} \right) + \sum\limits_{k = 1}^{n + 1} {\ln \left( k \right)} - \sum\limits_{k = 1}^{n + 1} {\ln \left( {x + k} \right)} } \right]$$$ $TEX: $$\mathop {\lim }\limits_{n \to \infty } \left[ {\left( {1 + x} \right)\ln \left( n \right) - \ln \left( {n + 1} \right) - \sum\limits_{k = 1}^{n + 1} {\left( {\ln \left( {x + k} \right) - \ln \left( k \right)} \right)} } \right]$$$ $TEX: $$\mathop {\lim }\limits_{n \to \infty } \left[ {\left( {1 + x} \right)\ln \left( n \right) - \ln \left( {n + 1} \right) - \sum\limits_{k = 1}^{n + 1} {\ln \left( {1 + \frac{x}{k}} \right)} } \right]/ + \left( {\sum\limits_{k = 1}^{n + 1} {\frac{x}{k} - \sum\limits_{k = 1}^{n + 1} {\frac{x}{k}} } } \right)$$$ $TEX: $$\mathop {\lim }\limits_{n \to \infty } \left[ {x\left( {\ln \left( n \right) - \sum\limits_{k = 1}^{n + 1} {\frac{1}<br />{k}} } \right) - \ln \left( {1 + \frac{1}{n}} \right) - \sum\limits_{k = 1}^{n + 1} {\left( {\ln \left( {1 + \frac{x}{k}} \right) - \frac{x}{k}} \right)} } \right]$$$ $TEX: $$\mathop {\lim }\limits_{n \to \infty } \left[ {x\left( {\ln \left( n \right) - \sum\limits_{k = 1}^{n + 1} {\frac{1}<br />{k}} } \right) - \ln \left( {1 + \frac{1}{n}} \right) - \sum\limits_{k = 1}^{n + 1} {\sum\limits_{j = 2}^\infty {\frac{{\left( { - 1} \right)^{j - 1} }}{j}\frac{{x^j }}{{k^j }}} } } \right]$$$ $TEX: $$\mathop {\lim }\limits_{n \to \infty } \left[ {\underbrace {x\left( {\ln \left( n \right) - \sum\limits_{k = 1}^{n + 1} {\frac{1}{k}} } \right) - \ln \left( {1 + \frac{1}{n}} \right)}_{ - \gamma x} + \underbrace { - \sum\limits_{k = 1}^{n + 1} {\sum\limits_{j = 2}^\infty {\frac{{\left( { - 1} \right)^{j - 1} }}{j}\frac{{x^j }}{{k^j }}} } }_{\sum\limits_{j = 2}^\infty {\sum\limits_{k = 1}^\infty {\frac{{\left( { - 1} \right)^j }}{j}\frac{{x^j }}{{k^j }}} = \sum\limits_{j = 2}^\infty {\frac{{\left( { - 1} \right)^j \zeta \left( j \right)}}{j}x^j } } }} \right]$$$ $TEX: $$\therefore \ln \left( {\Gamma \left( {1 + x} \right)} \right) = - \gamma x + \sum\limits_{j = 2}^\infty {\frac{{\left( { - 1} \right)^j \zeta \left( j \right)}}{j}x^j } ,\left\| x \right\| < 1$$$ $TEX: $$QED$$$ ¿Qué pasa cuando $TEX: $$x \to 1$$$ ? Supongo que se sobreentiende la dirección $TEX: $$\ln \left( {\Gamma \left( 2 \right)} \right) = - \gamma + \sum\limits_{j = 2}^\infty {\frac{{\left( { - 1} \right)^j \zeta \left( j \right)}}{j}}$$$ $TEX: $$\ln \left( {\Gamma \left( 2 \right)} \right) = \ln \left( {1\Gamma \left( 1 \right)} \right) = \ln \left( 1 \right) = 0$$$ $TEX: $$0 = - \gamma + \sum\limits_{j = 2}^\infty {\frac{{\left( { - 1} \right)^j \zeta \left( j \right)}}{j}}$$$ $TEX: $$\therefore \gamma = \sum\limits_{j = 2}^\infty {\frac{{\left( { - 1} \right)^j \zeta \left( j \right)}}{j}}$$$ Mensaje modificado por neo shykerex el Nov 14 2008, 07:59 PM -------------------- Miembro de Anime No Seishin Doukokai, podrías ser el próximo.

PaulRS Ver Perfil del Miembro Ver los mensajes de este miembro	Nov 16 2008, 07:57 AM Publicado: #3
Doctor en Matemáticas Grupo: Usuario FMAT Mensajes: 156 Registrado: 11-June 07 Miembro Nº: 6.615 Nacionalidad: Sexo:	Excelente! Otra solución: Notemos que: $TEX: $$<br />A_n = \sum\limits_{k = 1}^n {\left( {\tfrac{1}<br />{k}} \right)} - \ln \left( {n + 1} \right) = \sum\limits_{k = 1}^n {\left[ {\tfrac{1}<br />{k} - \ln \left( {1 + \tfrac{1}<br />{k}} \right)} \right]} <br />$$$ Por definición: $TEX: $$<br />\mathop {\lim }\limits_{n \to + \infty } A_n = \gamma <br />$$$ Informalmente tenemos: $TEX: $$ <br />\ln \left( {1 + \tfrac{1}<br />{k}} \right) = \sum\limits_{j = 1}^\infty {\tfrac{{\left( { - 1} \right)^{j + 1} }}<br />{{j \cdot k^j }}} <br />$$$ y luego: $TEX: $$<br />\gamma = \mathop {\lim }\limits_{n \to + \infty } A_n = \sum\limits_{k = 1}^\infty {\left[ {\tfrac{1}<br />{k} - \ln \left( {1 + \tfrac{1}<br />{k}} \right)} \right]} = \sum\limits_{k = 1}^\infty {\sum\limits_{j = 2}^\infty {\tfrac{{\left( { - 1} \right)^j }}<br />{{j \cdot k^j }}} } = \sum\limits_{j = 2}^\infty {\tfrac{{\left( { - 1} \right)^j }}<br />{j} \cdot \sum\limits_{k = 1}^\infty {\tfrac{1}<br />{{k^j }}} } = \sum\limits_{j = 2}^\infty {\tfrac{{\left( { - 1} \right)^j }}<br />{j} \cdot \zeta \left( j \right)} <br />$$$ Para más detalles ver el spoiler: Por Taylor: $TEX: $$<br />\ln \left( {1 + \tfrac{1}<br />{k}} \right) = \sum\limits_{j = 1}^n {\tfrac{{\left( { - 1} \right)^{j + 1} }}<br />{{j \cdot k^j }}} + R\left( {k,n} \right)<br />$$$ donde: $TEX: $$<br />R\left( {k,n} \right) = \left( { - 1} \right)^n \cdot \int_0^{\tfrac{1}<br />{k}} {\tfrac{{t^n }}<br />{{1 + t}}dt} <br />$$$ Así: $TEX: $$<br />A_n = \sum\limits_{k = 1}^n {\sum\limits_{j = 2}^n {\tfrac{{\left( { - 1} \right)^j }}<br />{{j \cdot k^j }}} } - \sum\limits_{k = 1}^n {R\left( {k,n} \right)} <br />$$$ Revirtiendo el orden en las sumas (son finitas): $TEX: $$<br />A_n = \sum\limits_{j = 2}^n {\left[ {\tfrac{{\left( { - 1} \right)^{j } }}<br />{j} \cdot \sum\limits_{k = 1}^n {\left( {\tfrac{1}<br />{{k^j }}} \right)} } \right]} - \sum\limits_{k = 1}^n {R\left( {k,n} \right)} <br />$$$ Ahora: $TEX: $$<br />\zeta \left( j \right) - \sum\limits_{k = n + 1}^\infty {\left( {\tfrac{1}<br />{{k^j }}} \right)} = \sum\limits_{k = 1}^n {\left( {\tfrac{1}<br />{{k^j }}} \right)} <br />$$$ por lo tanto: $TEX: $$<br />0 \leqslant \sum\limits_{k = n + 1}^\infty {\left( {\tfrac{1}<br />{{k^j }}} \right)} = \int_{n + 1}^\infty {\tfrac{{dx}}<br />{{\left\lfloor x \right\rfloor ^j }}} \leqslant \int_{n + 1}^\infty {\tfrac{{dx}}<br />{{\left( {x - 1} \right)^j }}} = \int_n^\infty {\tfrac{{dx}}<br />{{x^j }}} = \tfrac{1}<br />{{\left( {j - 1} \right) \cdot n^{j - 1} }}\therefore \zeta \left( j \right) + O\left( {\tfrac{1}<br />{{n^{j - 1} }}} \right) = \sum\limits_{k = 1}^n {\left( {\tfrac{1}<br />{{k^j }}} \right)} <br />$$$ Así: $TEX: $$<br />A_n = \sum\limits_{j = 2}^n {\left[ {\tfrac{{\left( { - 1} \right)^{j + 1} }}<br />{j} \cdot \zeta \left( j \right)} \right]} + O\left( {\sum\limits_{j = 2}^n {\tfrac{1}<br />{{j \cdot n^{j - 1} }}} } \right) - \sum\limits_{k = 1}^n {R\left( {k,n} \right)} <br />$$<br />$ Y dado que: $TEX: $$<br />0 \leqslant \sum\limits_{j = 2}^n {\tfrac{1}<br />{{j \cdot n^{j - 1} }}} \leqslant \sum\limits_{j = 1}^n {\tfrac{1}<br />{{n^j }}} = \tfrac{1}<br />{{n - 1}}<br />$$$ Se sigue: $TEX: $$<br />A_n = \sum\limits_{j = 2}^n {\left[ {\tfrac{{\left( { - 1} \right)^{j + 1} }}<br />{j} \cdot \zeta \left( j \right)} \right]} + O\left( {\tfrac{1}<br />{n}} \right) - \sum\limits_{k = 1}^n {R\left( {k,n} \right)} <br />$$$ $TEX: $$<br />\left\| {R\left( {k,n} \right)} \right\| = \int_0^{\tfrac{1}<br />{k}} {\tfrac{{t^n }}<br />{{1 + t}}dt} \leqslant \int_0^{\tfrac{1}<br />{k}} {t^{n - 1} dt} = \tfrac{1}<br />{{n \cdot k^n }} \Rightarrow \left\| {\sum\limits_{k = 1}^n {R({k,n}) } } \right\| \leqslant \tfrac{{\sum\nolimits_{k = 1}^n {\left( {\tfrac{1}<br />{{k^n }}} \right)} }}<br />{n} \leqslant \tfrac{{\sum\nolimits_{k = 1}^\infty {\left( {\tfrac{1}<br />{{k^n }}} \right)} }}<br />{n} = \tfrac{{\zeta \left( n \right)}}<br />{n}<br />$$$ Y finalmente: $TEX: $$<br />A_n = \sum\limits_{j = 2}^n {\left[ {\tfrac{{\left( { - 1} \right)^{j + 1} }}<br />{j} \cdot \zeta \left( j \right)} \right]} + O\left( {\tfrac{1}<br />{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to + \infty } \sum\limits_{j = 2}^n {\left[ {\tfrac{{\left( { - 1} \right)^{j + 1} }}<br />{j} \cdot \zeta \left( j \right)} \right]} = \gamma <br />$$$ Saludos Mensaje modificado por PaulRS el Nov 16 2008, 08:05 AM -------------------- $TEX: $\sqrt[3]{\displaystyle\sum_{i=1}^n{i^{3\cdot{\sqrt[]{3}}-1}}}\approx{\displaystyle\sum_{i=1}^n{i^{\sqrt[]{3}-1}}}$$

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