# 级数欣赏

$\sum_{n = 1}^\infty {\frac{1}{{{n^3}}}} \left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right) = \frac{7}{4}\zeta \left( 3 \right)\ln 2 - \frac{{{\pi ^4}}}{{288}}$

$\sum_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^3}}}\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)} = \frac{{{\pi ^4}}}{{60}} + \frac{{{\pi ^2}}}{{12}}{\ln ^2}2 - \frac{1}{{12}}{\ln ^4}2 - 2\mathrm{Li}_4\left( {\frac{1}{2}} \right)$

$\sum_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)} = - \frac{{17{\pi ^4}}}{{480}} - \frac{{{\pi ^2}}}{6}{\ln ^2}2 + \frac{1}{6}{\ln ^4}2 + 4\mathrm{Li}_4\left( {\frac{1}{2}} \right) + \frac{7}{2}\zeta \left( 3 \right)\ln 2$

${\sum_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)} ^2} = - \frac{{61{\pi ^4}}}{{1440}} + \frac{{{\pi ^2}}}{3}{\ln ^2}2 + \frac{1}{6}{\ln ^4}2 + 4\mathrm{Li}_4\left( {\frac{1}{2}} \right) + \frac{7}{4}\zeta \left( 3 \right)\ln 2$

$\sum_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)} = \frac{{29{\pi ^4}}}{{1440}} + \frac{{{\pi ^2}}}{8}{\ln ^2}2 - \frac{1}{8}{\ln ^4}2 - 3\mathrm{Li}_4\left( {\frac{1}{2}} \right)$

\begin{align*} \sum_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^3}}}{{\left( {\sum_{k = 1}^n {\frac{1}{k}} } \right)}^2}}&=- \frac{1}{9}{\pi ^2}{\ln ^3}2 + \frac{2}{{15}}{\ln ^5}2 + 4\mathrm{Li}_4\left( {\frac{1}{2}} \right)\ln 2 + 4\mathrm{Li}_5\left( {\frac{1}{2}} \right) \\ &~~~- \frac{{11}}{{48}}{\pi ^2}\zeta \left( 3 \right) + \frac{7}{2}\zeta \left( 3 \right){\ln ^2}2 - \frac{{19}}{{32}}\zeta \left( 5 \right) \end{align*}

\begin{align*} \sum_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)}&= \frac{1}{9}{\pi ^2}{\ln ^3}2 - \frac{2}{{15}}{\ln ^5}2 - 4\mathrm{Li}_4\left( {\frac{1}{2}} \right)\ln 2 - 4\mathrm{Li}_5\left( {\frac{1}{2}} \right) \\ &~~~+ \frac{5}{{32}}{\pi ^2}\zeta \left( 3 \right) - \frac{7}{4}\zeta \left( 3 \right){\ln ^2}2 + \frac{{23}}{8}\zeta \left( 5 \right) \end{align*}

$\sum_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{k^2}}}} } \right)}= - \frac{{13}}{{48}}{\pi ^2}\zeta \left( 3 \right) + \frac{{125}}{{32}}\zeta \left( 5 \right)$

\begin{align*} \sum_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^3}}}\left( {\sum_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)}&= \frac{2}{{45}}{\pi ^4}\ln 2 + \frac{1}{{36}}{\pi ^2}{\ln ^3}2 - \frac{1}{{60}}{\ln ^5}2 + 2\mathrm{Li}_5\left( {\frac{1}{2}} \right) \\ &~~~- \frac{1}{{48}}{\pi ^2}\zeta \left( 3 \right) - \frac{7}{8}\zeta \left( 3 \right){\ln ^2}2 - \frac{{37}}{{16}}\zeta \left( 5 \right) \end{align*}

\begin{align*} \sum_{n = 1}^\infty {\frac{1}{{{n^3}}}\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)} &= \frac{2}{{45}}{\pi ^4}\ln 2 + \frac{1}{{36}}{\pi ^2}{\ln ^3}2 - \frac{1}{{60}}{\ln ^5}2 + 2\mathrm{Li}_5\left( {\frac{1}{2}} \right) \\ &~~~+ \frac{1}{{16}}{\pi ^2}\zeta \left( 3 \right) - \frac{7}{8}\zeta \left( 3 \right){\ln ^2}2 - \frac{{193}}{{64}}\zeta \left( 5 \right) \end{align*}

\begin{align*} \sum_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}{{\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)}^3}} &= - \frac{1}{6}{\pi ^2}{\ln ^3}2 + \frac{1}{5}{\ln ^5}2 + 6\mathrm{Li}_4\left( {\frac{1}{2}} \right)\ln 2 + 6\mathrm{Li}_5\left( {\frac{1}{2}} \right) \\ &~~~- \frac{9}{{32}}{\pi ^2}\zeta \left( 3 \right) + \frac{{21}}{8}\zeta \left( 3 \right){\ln ^2}2 - \frac{9}{4}\zeta \left( 5 \right) \end{align*}

\begin{align*} \sum_{n = 1}^\infty {\frac{1}{{{n^2}}}{{\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)}^3}} &= - \frac{{29}}{{160}}{\pi ^4}\ln 2 + \frac{{11}}{{12}}{\pi ^2}{\ln ^3}2 - \frac{1}{{20}}{\ln ^5}2 - 6\mathrm{Li}_4\left( {\frac{1}{2}} \right)\ln 2 \\ &~~~- 24\mathrm{Li}_5\left( {\frac{1}{2}} \right) + \frac{1}{{12}}{\pi ^2}\zeta \left( 3 \right) + \frac{{367}}{{16}}\zeta \left( 5 \right) \end{align*}

\begin{align*} &\sum_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{n^2}}}\left( {\sum_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)\left( {\sum_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{k^2}}}} } \right)} \\&= \frac{{49}}{{720}}{\pi ^4}\ln 2 - \frac{1}{{18}}{\pi ^2}{\ln ^3}2 + \frac{1}{{10}}{\ln ^5}2 + 4\mathrm{Li}_4\left( {\frac{1}{2}} \right)\ln 2+ 8\mathrm{Li}_5\left( {\frac{1}{2}} \right) + \frac{1}{{96}}{\pi ^2}\zeta \left( 3 \right) - \frac{{35}}{4}\zeta \left( 5 \right) \end{align*}

$\sum_{n = 1}^\infty {\frac{1}{{{n^3}}}\left( {\sum_{k = 1}^n {\frac{1}{k}} } \right)\left( {\sum_{k = 1}^n {\frac{1}{{{k^2}}}} } \right)} = - \frac{{101}}{{45360}}{\pi ^6} + \frac{5}{2}{\zeta ^2}\left( 3 \right)$

$\sum_{n = 1}^\infty {\frac{1}{{{n^3}}}{{\left( {\sum\limits_{k = 1}^n {\frac{1}{k}} } \right)}^3} = } \frac{{31}}{{5040}}{\pi ^6} - \frac{5}{2}{\zeta ^2}\left( 3 \right)$

posted @ 2016-05-12 15:40  Renascence_5  阅读(322)  评论(0编辑  收藏