Σειρά (34)

china university · #1

Να βρεθεί το
$\displaystyle{\sum_{n=0}^{\infty}\dfrac{1}{(6n+1)^5}}$ .
Έχω βρει ότι ισούται με
$\displaystyle{\dfrac{1}{2}\left(\dfrac{2^5-1}{2^5}\cdot\dfrac{3^5-1}{3^5}\zeta(5)+\dfrac{11\pi^5}{8\cdot 3^5\sqrt{3}}\right)}$

από την αντίστοιχη συζήτηση εδώ

#2

china university έγραψε:Να βρεθεί το $\displaystyle{\sum_{n=0}^{\infty}\dfrac{1}{(6n+1)^5}}$ . Έχω βρει ότι ισούται με $\displaystyle{\dfrac{1}{2}\left(\dfrac{2^5-1}{2^5}\cdot\dfrac{3^5-1}{3^5}\zeta(5)+\dfrac{11\pi^5}{8\cdot 3^5\sqrt{3}}\right)}$ από την αντίστοιχη συζήτηση εδώ

Μια λύση καθαρά με polygamma functions ..

$\displaystyle{S = \sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {6k + 1} \right)}^5}}}} = \frac{1}{{{6^5}}}\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {k + 1/6} \right)}^5}}}} }$ . Όμως $\displaystyle{\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {k + z} \right)}^{n + 1}}}}} = {\left( { - 1} \right)^{n + 1}}\frac{{{\psi _n}\left( z \right)}}{{n!}}}$ . Τότε $\displaystyle{\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {6k + 1} \right)}^5}}}} = \frac{1}{{{6^5}}}\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {k + 1/6} \right)}^5}}}} = - \frac{1}{{{6^5} \cdot 4!}} \cdot {\psi _4}\left( {\frac{1}{6}} \right)}$

Επίσης $\displaystyle{{\psi _n}\left( {mz} \right) = {\delta _{n,0}}\log m + \frac{1}{{{m^{n + 1}}}}\sum\limits_{k = 0}^{m - 1} {{\psi _n}\left( {z + \frac{k}{m}} \right)} }$ . Τότε $\displaystyle{{\psi _4}\left( {2 \cdot \frac{1}{6}} \right) = \frac{1}{{{2^5}}}\sum\limits_{k = 0}^1 {{\psi _4}\left( {\frac{1}{6} + \frac{k}{2}} \right)} \Rightarrow {\psi _4}\left( {\frac{1}{6}} \right) + {\psi _4}\left( {\frac{2}{3}} \right) = 32 \cdot {\psi _4}\left( {\frac{1}{3}} \right)}$

Επίσης $\displaystyle{{\psi _n}\left( {1 - z} \right) + {\left( { - 1} \right)^{n + 1}}{\psi _n}\left( z \right) = {\left( { - 1} \right)^n}\pi \frac{{{d^n}}}{{d{z^n}}}\cot \left( {\pi z} \right)}$ . Τότε $\displaystyle{{\psi _4}\left( {1 - \frac{1}{3}} \right) - {\psi _4}\left( {\frac{1}{3}} \right) = \pi {\left. {\frac{{{d^4}}}{{d{z^4}}}\cot \left( {\pi z} \right)} \right|_{z = 1/3}} = }$

$\displaystyle{ = 8{\pi ^5} \cdot \frac{{{{\cos }^3}\left( {\pi /3} \right) + 2\cos \left( {\pi /3} \right)}}{{{{\sin }^5}\left( {\pi /3} \right)}} = \frac{{32 \cdot {\pi ^5}}}{{\sqrt 3 }} \Rightarrow }$ \displaystyle{\displaystyle{{\psi _4}\left( {\frac{2}{3}} \right) = {\psi _4}\left( {\frac{1}{3}} \right) + \frac{{32 \cdot {\pi ^5}}}{{\sqrt 3 }}} Οπότε

\displaystyle{{\psi _4}\left( {\frac{1}{6}} \right) = 32 \cdot {\psi _4}\left( {\frac{1}{3}} \right) - {\psi _4}\left( {\frac{2}{3}} \right) = 31 \cdot {\psi _4}\left( {\frac{1}{3}} \right) - \frac{{32 \cdot {\pi ^5}}}{{\sqrt 3 }}} . Άρα

\displaystyle{\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {6k + 1} \right)}^5}}}} = \frac{1}{{{6^5} \cdot 4!}} \cdot \frac{{32 \cdot {\pi ^5}}}{{\sqrt 3 }} - \frac{1}{{{6^5} \cdot 4!}} \cdot 31 \cdot {\psi _4}\left( {\frac{1}{3}} \right)} Επίσης

\displaystyle{{\psi _n}\left( {mz} \right) = {\delta _{n,0}}\log m + \frac{1}{{{m^{n + 1}}}}\sum\limits_{k = 0}^{m - 1} {{\psi _n}\left( {z + \frac{k}{m}} \right)} \mathop {\mathop \Rightarrow \limits_{z = 1/3} }\limits^{m = 3} }} $\displaystyle{{\psi _4}\left( 1 \right) = \frac{1}{{{3^5}}}\left( {{\psi _4}\left( {\frac{1}{3}} \right) + {\psi _4}\left( {\frac{2}{3}} \right) + {\psi _4}\left( 1 \right)} \right) = }$

$\displaystyle{ = \frac{1}{{{3^5}}}\left( {2 \cdot {\psi _4}\left( {\frac{1}{3}} \right) + \frac{{32 \cdot {\pi ^5}}}{{\sqrt 3 }} + {\psi _4}\left( 1 \right)} \right) \Rightarrow {\psi _4}\left( {\frac{1}{3}} \right) = \frac{{{3^5} - 1}}{2}{\psi _4}\left( 1 \right) - \frac{{16 \cdot {\pi ^5}}}{{\sqrt 3 }}}$

Τότε $\displaystyle{\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {6k + 1} \right)}^5}}}} = \frac{{33 \cdot {\pi ^5}}}{{48 \cdot {3^5} \cdot \sqrt 3 }} - \frac{{31\left( {{3^5} - 1} \right)}}{{48 \cdot {6^5}}}{\psi _4}\left( 1 \right)}$ και $\displaystyle{\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {k + z} \right)}^{n + 1}}}}} = {\left( { - 1} \right)^{n + 1}}\frac{{{\psi _n}\left( z \right)}}{{n!}} \Rightarrow }$

$\displaystyle{\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {k + 1} \right)}^5}}}} = - \frac{{{\psi _4}\left( 1 \right)}}{{4!}} \Rightarrow {\psi _4}\left( 1 \right) = - 4! \cdot \zeta \left( 5 \right)}$ . Τελικά $\displaystyle{\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {6k + 1} \right)}^5}}}} = \frac{{33 \cdot {\pi ^5}}}{{48 \cdot {3^5} \cdot \sqrt 3 }} + \frac{{31\left( {{3^5} - 1} \right)}}{{2 \cdot {6^5}}}\zeta \left( 5 \right)}$

Όλες οι ιδιότητες της Polygamma function που χρησιμοποιήθηκαν βρίσκονται εδώ http://mathworld.wolfram.com/PolygammaFunction.html

Σειρά (34)

Σειρά (34)

Re: Σειρά (34)

Μέλη σε σύνδεση