Υπολογισμός αθροίσματος (10)

#1

Ας υπολογισθεί το άθροισμα $\displaystyle{\sum_{n=1}^{+\infty}\frac{H_{n}\left(\zeta(2)-\sum_{k=1}^{n}\frac{1}{k^2}\right)}{n}}$ .

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ

#2

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ

#3

Ήταν πολύ δύσκολο, όντως !!

Λήμμα : $\displaystyle{\sum\limits_{n = 1}^\infty {{a_n}\sum\limits_{k = n + 1}^\infty {{b_k}} } = \sum\limits_{n = 1}^\infty {{b_n}\sum\limits_{k = 1}^n {{a_k}} } - \sum\limits_{n = 1}^\infty {\left( {{a_n}{b_n}} \right)} }$ διότι $\displaystyle{\sum\limits_{n = 1}^\infty {{a_n}\sum\limits_{k = n + 1}^\infty {{b_k}} } = {a_1}\left( {{b_2} + {b_3} + {b_4} + ..} \right) + {a_2}\left( {{b_3} + {b_4} + {b_5} + ..} \right) + .. = }$
$\displaystyle{ = {b_2}{a_1} + {b_3}\left( {{a_1} + {a_2}} \right) + .. = \sum\limits_{n = 2}^\infty {{b_n}\sum\limits_{k = 1}^{n - 1} {{a_k}} } = \sum\limits_{n = 2}^\infty {{b_n}\left( { - {a_n} + \sum\limits_{k = 1}^n {{a_k}} } \right)} = .. = \sum\limits_{n = 1}^\infty {{b_n}\sum\limits_{k = 1}^n {{a_k}} } - \sum\limits_{n = 1}^\infty {{b_n}{a_n}} }$
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Τότε $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{n}\left( {\zeta \left( 2 \right) - \sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)} = \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{n}\left( {\sum\limits_{k = 1}^\infty {\frac{1}{{{k^2}}}} - \sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)} = \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{n}\left( {\sum\limits_{k = n + 1}^\infty {\frac{1}{{{k^2}}}} } \right)} \mathop = \limits^{\left\lfloor {lemma} \right\rfloor } \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{{{H_k}}}{k}} } \right)} - \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^3}}}} }$
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Όμως ισχύει $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^3}}}} = \frac{1}{{72}}{\pi ^4}}$ και γενικότερα (Euler) $\displaystyle{2\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^m}}}} = \left( {m + 2} \right)\zeta \left( {m + 1} \right) - \sum\limits_{n = 1}^{m - 2} {\zeta \left( {m - n} \right)\zeta \left( {n + 1} \right)} }$
αναφέρεται και εδώ http://mathworld.wolfram.com/HarmonicNumber.html αποδεικνύεται εδώ http://projecteuclid.org/DPubS/Reposito ... 1102636166
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Επίσης $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{{{H_k}}}{k}} } \right)} \mathop = \limits^{\left\lfloor 1 \right\rfloor } - \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\int\limits_0^1 {{x^{k - 1}}\ln \left( {1 - x} \right)dx} } } \right)} = - \int\limits_0^1 {\ln \left( {1 - x} \right)\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \left( {\sum\limits_{k = 1}^n {{x^{k - 1}}} } \right)dx} = }$

$\displaystyle{ = - \int\limits_0^1 {\frac{{\ln \left( {1 - x} \right)}}{{1 - x}}\left( {\sum\limits_{n = 1}^\infty {\frac{{1 - {x^n}}}{{{n^2}}}} } \right)dx} = \frac{1}{2}\int\limits_0^1 {{{\left( {{{\ln }^2}\left( {1 - x} \right)} \right)}{'}}\left( {\sum\limits_{n = 1}^\infty {\frac{{1 - {x^n}}}{{{n^2}}}} } \right)dx} \mathop = \limits^{\left\lfloor 2 \right\rfloor } \frac{1}{2}\int\limits_0^1 {{{\ln }^2}\left( {1 - x} \right)\left( {\sum\limits_{n = 1}^\infty {\frac{{{x^{n - 1}}}}{n}} } \right)dx} = }$

$\displaystyle{ = \frac{1}{2}\int\limits_0^1 {\frac{{{{\ln }^2}\left( {1 - x} \right)}}{x}\left( {\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{n}} } \right)dx} = - \frac{1}{2}\int\limits_0^1 {\frac{{{{\ln }^3}\left( {1 - x} \right)}}{x}dx} = - \frac{1}{2}\int\limits_0^1 {\frac{{{{\ln }^3}\left( x \right)}}{{1 - x}}dx} = - \frac{1}{2}\int\limits_0^1 {{{\ln }^3}\left( x \right)\sum\limits_{n = 0}^\infty {{x^n}} dx} = }$

$\displaystyle{ = - \frac{1}{2}\sum\limits_{n = 0}^\infty {\left( {\int\limits_0^1 {{{\ln }^3}\left( x \right){x^n}dx} } \right)} \mathop = \limits^{\left\lfloor 3 \right\rfloor } 3\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^4}}}} = 3\zeta \left( 4 \right) = \frac{{{\pi ^4}}}{{30}}}$

Και τελικά $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{n}\left( {\zeta \left( 2 \right) - \sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{{{H_k}}}{k}} } \right)} - \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^3}}}} = \frac{{{\pi ^4}}}{{30}} - \frac{{{\pi ^4}}}{{72}} \Rightarrow \boxed{\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{n}\left( {\zeta \left( 2 \right) - \sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)} = \frac{{7{\pi ^4}}}{{360}}}}$

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$\displaystyle{\left\lfloor 1 \right\rfloor :}$ Διότι $\displaystyle{{H_k} = 1 + \frac{1}{2} + \frac{1}{3} + .. + \frac{1}{k} = \int\limits_0^1 {\left( {1 + x + {x^2} + .. + {x^{k - 1}}} \right)dx} = \int\limits_0^1 {\frac{{1 - {x^k}}}{{1 - x}}dx} = - \int\limits_0^1 {\left( {1 - {x^k}} \right){{\left( {\ln \left( {1 - x} \right)} \right)}{'}}dx} = }$
$\displaystyle{ = - k\int\limits_0^1 {{x^{k - 1}}\ln \left( {1 - x} \right)dx} \Rightarrow \boxed{\frac{{{H_k}}}{k} = - \int\limits_0^1 {{x^{k - 1}}\ln \left( {1 - x} \right)dx} }}$

$\displaystyle{\left\lfloor 2 \right\rfloor :}$ Παραγοντική ολοκλήρωση με σχετικά απλά όρια.

$\displaystyle{\left\lfloor 3 \right\rfloor :}$ Διότι $\displaystyle{\int\limits_0^1 {{{\ln }^3}\left( x \right){x^n}dx} = \frac{1}{{n + 1}}\int\limits_0^1 {{{\ln }^3}\left( x \right){{\left( {{x^{n + 1}}} \right)}{'}}dx} = \frac{{ - 3}}{{n + 1}}\int\limits_0^1 {{{\ln }^2}\left( x \right){x^n}dx} = .. \Rightarrow \boxed{\int\limits_0^1 {{{\ln }^3}\left( x \right){x^n}dx} = \frac{{ - 6}}{{{{\left( {n + 1} \right)}^4}}}}}$ .

#4

Σεραφείμ πολύ ωραία λύση!

#5

Παρεμπιπτόντως ο Wolfran το υπολογίζει απελπιστικά λάθος. Δίδει αποτέλεσμα $\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{n}\left( {\zeta \left( 2 \right) - \sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)} = \frac{\gamma }{2}\left( {{\pi ^2}\ln 2 - 4\zeta \left( 3 \right)} \right)}$

Είναι αλήθεια πως με παρέσυρε ..

parmenides51 · #6

Wolfram έγραψε:Σεραφείμ

Υπολογισμός αθροίσματος (10)

Υπολογισμός αθροίσματος (10)

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