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Δημοσιεύτηκε: **Παρ Φεβ 28, 2014 8:19 pm**

Να υπολογιστεί το άθροισμα $\displaystyle{{S = \sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} }}$

Δημοσιεύτηκε: **Κυρ Αύγ 23, 2015 7:56 am**

$\displaystyle{\sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} = \frac{{\log \left( {{{\left( {\frac{1}{3}} \right)}^2}} \right)}}{{\frac{1}{3}}} + \sum\limits_{n = 1}^\infty {\left( {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}} + \frac{{\log \left( {{{\left( { - n + \frac{1}{3}} \right)}^2}} \right)}}{{ - n + \frac{1}{3}}}} \right)} = }$

$\displaystyle{ = - 3\log \left( 9 \right) + \sum\limits_{n = 1}^\infty {\left( {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}} - \frac{{\log \left( {{{\left( {n - \frac{1}{3}} \right)}^2}} \right)}}{{n - \frac{1}{3}}}} \right)} = - 3\log \left( 9 \right) + 6\sum\limits_{n = 1}^\infty {\left( {\frac{{\log \left( {\frac{{3n + 1}}{3}} \right)}}{{3n + 1}} - \frac{{\log \left( {\frac{{3n - 1}}{3}} \right)}}{{3n - 1}}} \right)} = }$

$\displaystyle{ = - 6\log \left( 3 \right) + 6\sum\limits_{n = 1}^\infty {\left( {\frac{{\log \left( {3n + 1} \right) - \log 3}}{{3n + 1}} - \frac{{\log \left( {3n - 1} \right) - \log 3}}{{3n - 1}}} \right)} = }$

$\displaystyle{ = - 6\log \left( 3 \right) + 6\sum\limits_{n = 1}^\infty {\left( {\frac{{\log 3}}{{3n - 1}} - \frac{{\log 3}}{{3n + 1}}} \right)} + 6\sum\limits_{n = 1}^\infty {\left( {\frac{{\log \left( {3n + 1} \right)}}{{3n + 1}} - \frac{{\log \left( {3n - 1} \right)}}{{3n - 1}}} \right)} }$

Όμως $\displaystyle{ - 6\log \left( 3 \right) + 6\sum\limits_{n = 1}^\infty {\left( {\frac{{\log 3}}{{3n - 1}} - \frac{{\log 3}}{{3n + 1}}} \right)} = - 6\log \left( 3 \right)\left( {1 + \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{3n + 1}} - \frac{1}{{3n - 1}}} \right)} } \right) = }$

$\displaystyle{ = - 6\log \left( 3 \right)\frac{2}{{\sqrt 3 }}\sum\limits_{n = 1}^\infty {\frac{{\sin \frac{{2n\pi }}{3}}}{n}} = - \frac{{4\sqrt 3 \log \left( 3 \right)}}{{2i}}\sum\limits_{n = 1}^\infty {\frac{{{e^{2i\pi n/3}} - {e^{ - 2i\pi n/3}}}}{n}} = - \frac{{4\sqrt 3 \log \left( 3 \right)}}{{2i}}\log \frac{{1 - {e^{ - 2i\pi /3}}}}{{1 - {e^{2i\pi /3}}}} = }$

$\displaystyle{ = - \frac{{4\sqrt 3 \log \left( 3 \right)}}{{2i}}\log \left( { - {e^{{e^{ - 2i\pi /3}}}}} \right) = - \frac{{4\sqrt 3 \log \left( 3 \right)}}{{2i}}\log \left( {{e^{i\pi /3}}} \right) = - \frac{{4\sqrt 3 \log \left( 3 \right)}}{{2i}}\left( {i\frac{\pi }{3}} \right) = - \frac{{2\pi \log 3}}{{\sqrt 3 }}}$

Οπότε $\displaystyle{\sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} = - \frac{{2\pi \log 3}}{{\sqrt 3 }} + 6\sum\limits_{n = 1}^\infty {\left( {\frac{{\log \left( {3n + 1} \right)}}{{3n + 1}} - \frac{{\log \left( {3n - 1} \right)}}{{3n - 1}}} \right)} = - \frac{{2\pi \log 3}}{{\sqrt 3 }} + 4\sqrt 3 \sum\limits_{n = 1}^\infty {\frac{{\log n \cdot \sin \frac{{2n\pi }}{3}}}{n}} }$

Από την φόρμουλα του Kummer http://www.math.titech.ac.jp/~tosho/Pre ... df/128.pdf έχουμε

$\displaystyle{\log \Gamma \left( x \right) = \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\log n \cdot \sin \left( {2\pi nx} \right)}}{n}} + \frac{{\log 2\pi + \gamma }}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {2\pi nx} \right)}}{n}} + \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {2\pi nx} \right)}}{n}} + \frac{1}{2}\log 2\pi }$

Οπότε για $\displaystyle{x = \frac{1}{3}}$ προκύπτει $\displaystyle{\log \Gamma \left( {\frac{1}{3}} \right) = \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\log n \cdot \sin \left( {\frac{{2\pi n}}{3}} \right)}}{n}} + \frac{{\log 2\pi + \gamma }}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\frac{{2\pi n}}{3}} \right)}}{n}} + \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {\frac{{2\pi n}}{3}} \right)}}{n}} + \frac{1}{2}\log 2\pi }$

Ισχύει $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{\sin nx}}{n}} = \frac{{\pi - x}}{2}}$ , επομένως $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\frac{{2\pi n}}{3}} \right)}}{n}} = \frac{{\pi - 2\pi /3}}{2} = \frac{\pi }{6}}$

καθώς και $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{\cos nx}}{n}} = - \log \left( {2\left| {\sin \frac{x}{2}} \right|} \right)}$ , δηλαδή $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {\frac{{2\pi n}}{3}} \right)}}{n}} = - \log \left( {2\sin \frac{\pi }{3}} \right) = - \frac{1}{2}\log 3}$

τα παραπάνω δυο αθροίσματα έχουν αποδειχθεί στο mathematica

οπότε $\displaystyle{\log \Gamma \left( {\frac{1}{3}} \right) = \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\log n \cdot \sin \left( {\frac{{2\pi n}}{3}} \right)}}{n}} + \frac{{\log 2\pi + \gamma }}{\pi }\frac{\pi }{6} - \frac{1}{4}\log 3 + \frac{1}{2}\log 2\pi }$

δηλαδή $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{\log n \cdot \sin \left( {\frac{{2\pi n}}{3}} \right)}}{n}} = \frac{\pi }{{12}}\left( {12\log \Gamma \left( {\frac{1}{3}} \right) - 2\gamma + 3\log 3 - 8\log 2\pi } \right)}$

Τελικά $\displaystyle{\sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} = \frac{\pi }{{\sqrt 3 }}\left( {12\;\log \left( {\Gamma \left( {\frac{1}{3}} \right)} \right) + \log 3 - 2\;\gamma - 8\log 2\pi } \right)}$

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Να υπολογιστεί το άθροισμα.

Να υπολογιστεί το άθροισμα.

Re: Να υπολογιστεί το άθροισμα.