Σειρά με ζήτα

Tolaso J Kos · #1

Ας δηλώσουμε με $\zeta$ τη συνάρτηση ζήτα του Riemann. Δειχθήτω:
$\displaystyle{\frac{1}{2\pi} {\rm Li}_2 \left ( e^{-2\pi} \right ) = \log 2\pi - 1 -\frac{5 \pi}{12} - \sum_{n=1}^{\infty} \frac{(-1)^n \zeta(2n)}{n \left ( 2n+1 \right )}}$ όπου ${\rm Li}_2$ είναι η διλογαριθμική συνάρτηση.

#2

Tolaso J Kos έγραψε:Δειχθήτω: $\displaystyle{\frac{1}{2\pi} {\rm Li}_2 \left ( e^{-2\pi} \right ) = \log 2\pi - 1 -\frac{5 \pi}{12} - \sum_{n=1}^{\infty} \frac{(-1)^n \zeta(2n)}{n \left ( 2n+1 \right )}}$

Σχόλια:

$\displaystyle{\sum\limits_{n = 1}^\infty {\left( {\frac{{{{\left( { - 1} \right)}^n}}}{{n \cdot {m^{2n}}}}} \right)} = - \log \left( {\frac{{1 + {m^2}}}{{{m^2}}}} \right){\rm{ }}\left( * \right)}$ και $\displaystyle{\sum\limits_{n = 1}^\infty {\left( {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right) \cdot {m^{2n}}}}} \right)} = 1 - m \cdot \arctan \left( {\frac{1}{m}} \right){\rm{ }}\left( {2*} \right)}$ (γνωστές σειρές).

$\displaystyle{\Gamma \left( i \right)\Gamma \left( { - i} \right) = \frac{{2\pi }}{{{e^\pi } - {e^{ - \pi }}}}{\rm{ }}\left( {3*} \right)}$ διότι $\displaystyle{\Gamma \left( i \right)\Gamma \left( { - i} \right) = \frac{{\Gamma \left( i \right)\left( { - i} \right)\Gamma \left( { - i} \right)}}{{ - i}} = \frac{{\Gamma \left( i \right)\Gamma \left( {1 - i} \right)}}{{ - i}} = \frac{\pi }{{ - i \cdot \sin \left( {\pi \cdot i} \right)}} = \frac{{2\pi }}{{{e^\pi } - {e^{ - \pi }}}}}$

$\displaystyle{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{z \cdot \left( {1 + z} \right) \cdot .. \cdot \left( {m + z} \right)}}{{{{\left( {m + 1} \right)}^z}m!}}} \right) = \frac{1}{{\Gamma \left( z \right)}}{\rm{ }}\left( {4*} \right)}$ (εναλλακτικός ορισμός της συνάρτησης Gamma function)

$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{1}{{{x^2} + {n^2}}}} = - \frac{1}{{2{x^2}}} + \frac{\pi }{{2x}} \cdot \frac{{{e^{\pi x}} + {e^{ - \pi x}}}}{{{e^{\pi x}} - {e^{ - \pi x}}}}{\rm{ }}\left( {5*} \right)}$ (έχει αποδειχθεί αρκετές φορές στην σελίδα)

Στο θέμα μας

$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}\zeta \left( {2n} \right)}}{{n\left( {2n + 1} \right)}}} = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{n\left( {2n + 1} \right)}}\sum\limits_{m = 1}^\infty {\frac{1}{{{m^{2n}}}}} } = 2\sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\left( {\frac{1}{{2n \cdot {m^{2n}}}} - \frac{1}{{\left( {2n + 1} \right) \cdot {m^{2n}}}}} \right)} } = }$

$\displaystyle{ = 2\sum\limits_{m = 1}^\infty {\left( {\sum\limits_{n = 1}^\infty {\left( {\frac{{{{\left( { - 1} \right)}^n}}}{{2n \cdot {m^{2n}}}}} \right) - \sum\limits_{n = 1}^\infty {\left( {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right) \cdot {m^{2n}}}}} \right)} } } \right)} \mathop { = = = }\limits_{\left( {2*} \right)}^{\left( * \right)} }$ $\displaystyle{\sum\limits_{n = 1}^\infty {\left( { - \log \left( {\frac{{1 + {m^2}}}{{{m^2}}}} \right) + 2\left( {1 - m \cdot \arctan \left( {\frac{1}{m}} \right)} \right)} \right)} }$

Όμως $\displaystyle{\sum\limits_{n = 1}^\infty {\log \left( {\frac{{1 + {m^2}}}{{{m^2}}}} \right) = \log \left( {\prod\limits_{n = 1}^\infty {\frac{{\left( {m + i} \right)\left( {m - i} \right)}}{{{m^2}}}} } \right)} = }$ $\displaystyle{\log \left( {\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{i \cdot \left( {1 + i} \right) \cdot .. \cdot \left( {m + i} \right)}}{{m!}} \cdot \frac{{ - i \cdot \left( {1 - i} \right) \cdot .. \cdot \left( {m - i} \right)}}{{m!}}} \right)} \right) = }$

$\displaystyle{ = \log \left( {\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{i \cdot \left( {1 + i} \right) \cdot .. \cdot \left( {m + i} \right)}}{{{{\left( {m + 1} \right)}^i}m!}} \cdot \frac{{ - i \cdot \left( {1 - i} \right) \cdot .. \cdot \left( {m - i} \right)}}{{{{\left( {m + 1} \right)}^{ - i}}m!}}} \right)} \right)\mathop { = = }\limits^{\left( {4*} \right)} }$ $\displaystyle{ - \log \left( {\Gamma \left( i \right)\Gamma \left( { - i} \right)} \right)\mathop { = = }\limits^{\left( {3*} \right)} - \log \frac{{2\pi }}{{{e^\pi } - {e^{ - \pi }}}}}$

Επομένως $\displaystyle{{\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}\zeta \left( {2n} \right)}}{{n\left( {2n + 1} \right)}}} = \log \frac{{2\pi }}{{{e^\pi } - {e^{ - \pi }}}} + 2\sum\limits_{n = 1}^\infty {\left( {1 - m \cdot \arctan \left( {\frac{1}{m}} \right)} \right)} }}$

Επίσης $\displaystyle{\sum\limits_{n = 1}^\infty {\left( {1 - n \cdot \arctan \left( {\frac{1}{n}} \right)} \right)} = \sum\limits_{n = 1}^\infty {n\left( {\frac{1}{n} - \arctan \left( {\frac{1}{n}} \right)} \right)} = }$ $\displaystyle{\sum\limits_{n = 1}^\infty {n\left( {\frac{1}{n} - \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{\left( {2k + 1} \right) \cdot {n^{2k + 1}}}}} } \right)} = \sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{\left( {2k + 1} \right) \cdot {n^{2k}}}}} } = }$

$\displaystyle{ = \sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{n^{2k}}}}\int\limits_0^1 {{x^{2k}}dx} } } = \sum\limits_{n = 1}^\infty {\int\limits_0^1 {\sum\limits_{k = 1}^\infty {\left( {{{\left( { - 1} \right)}^{k - 1}}{{\left( {\frac{{{x^2}}}{{{n^2}}}} \right)}^k}} \right)dx} } } = }$ $\displaystyle{\sum\limits_{n = 1}^\infty {\int\limits_0^1 {\frac{{{x^2}}}{{{x^2} + {n^2}}}dx} } = \int\limits_0^1 {{x^2}\left( {\sum\limits_{n = 1}^\infty {\frac{1}{{{x^2} + {n^2}}}} } \right)dx} \mathop { = = = }\limits^{\left( {5*} \right)} }$

$\displaystyle{ = \frac{1}{2}\int\limits_0^1 {\left( { - 1 + \pi x\frac{{{e^{\pi x}} + {e^{ - \pi x}}}}{{{e^{\pi x}} - {e^{ - \pi x}}}}} \right)dx} = - \frac{1}{2} + \pi \int\limits_0^1 {x \cdot \frac{{{e^{\pi x}} + {e^{ - \pi x}}}}{{{e^{\pi x}} - {e^{ - \pi x}}}}dx} = }$ $\displaystyle{ - \frac{1}{2} + \frac{\pi }{2}\int\limits_0^1 {x \cdot \frac{{1 + {e^{ - 2\pi x}}}}{{1 - {e^{ - 2\pi x}}}}dx} = }$

$\displaystyle{ = - \frac{1}{2} + \frac{\pi }{2}\int\limits_0^1 {x \cdot \left( {1 + {e^{ - 2\pi x}}} \right)\sum\limits_{n = 0}^\infty {{e^{ - 2n\pi x}}} dx} = - \frac{1}{2} + \frac{\pi }{2}\int\limits_0^1 {x\sum\limits_{n = 0}^\infty {{e^{ - 2n\pi x}}} dx} }$ $\displaystyle{ + \frac{\pi }{2}\int\limits_0^1 {x\sum\limits_{n = 0}^\infty {{e^{ - 2\left( {n + 1} \right)\pi x}}} dx} = }$

$\displaystyle{ = - \frac{1}{2} + \frac{\pi }{2}\sum\limits_{n = 0}^\infty {\int\limits_0^1 {x \cdot {e^{ - 2n\pi x}}dx} } + \frac{\pi }{2}\sum\limits_{n = 0}^\infty {\int\limits_0^1 {x \cdot {e^{ - 2\left( {n + 1} \right)\pi x}}dx} } = }$ $\displaystyle{ - \frac{1}{2} + \frac{\pi }{4} + \frac{\pi }{2}\sum\limits_{n = 1}^\infty {\int\limits_0^1 {x \cdot {e^{ - 2n\pi x}}dx} } + \frac{\pi }{2}\sum\limits_{n = 0}^\infty {\int\limits_0^1 {x \cdot {e^{ - 2\left( {n + 1} \right)\pi x}}dx} } \Rightarrow }$

$\displaystyle{ \Rightarrow \sum\limits_{n = 1}^\infty {\left( {1 - n \cdot \arctan \left( {\frac{1}{n}} \right)} \right)} = - \frac{1}{2} + \frac{\pi }{4} + \pi \sum\limits_{n = 1}^\infty {\left( { - \frac{{{e^{ - 2n\pi }}}}{{4{\pi ^2}{n^2}}} + \frac{1}{{4{\pi ^2}{n^2}}} - \frac{{{e^{ - 2n\pi }}}}{{2\pi n}}} \right)} }$ $\displaystyle{ = - \frac{1}{2} + \frac{\pi }{4} - \frac{1}{{4\pi }}\sum\limits_{n = 1}^\infty {\left( {\frac{{{e^{ - 2n\pi }}}}{{{n^2}}}} \right)} + }$

$\displaystyle{ + \frac{1}{{4\pi }}\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{{n^2}}}} \right)} - \frac{1}{2}\sum\limits_{n = 1}^\infty {\left( {\frac{{{e^{ - 2n\pi }}}}{n}} \right)} = - \frac{1}{2} + \frac{{7\pi }}{{24}} - \frac{1}{{4\pi }}L{i_2}\left( {{e^{ - 2\pi }}} \right) + \frac{1}{2}\log \left( {1 - {e^{ - 2\pi }}} \right)}$

Τότε $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}\zeta \left( {2n} \right)}}{{n\left( {2n + 1} \right)}}} = \log \frac{{2\pi }}{{{e^\pi } - {e^{ - \pi }}}} + 2\left( { - \frac{1}{2} + \frac{{7\pi }}{{24}} - \frac{1}{{4\pi }}L{i_2}\left( {{e^{ - 2\pi }}} \right) + \frac{1}{2}\log \left( {1 - {e^{ - 2\pi }}} \right)} \right)}$ $\displaystyle{ = \log 2\pi - 1 - \frac{{5\pi }}{{12}} - \frac{1}{{2\pi }}L{i_2}\left( {{e^{ - 2\pi }}} \right)}$

και τελικά $\displaystyle{\frac{1}{{2\pi }}L{i_2}\left( {{e^{ - 2\pi }}} \right) = \log \left( {2\pi } \right) - 1 - \frac{{5\pi }}{{12}} - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}\zeta \left( {2n} \right)}}{{n\left( {2n + 1} \right)}}} }$

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