一个级数求解

Eufisky posted @ 2014年6月24日 23:28 in 数学分析 with tags 级数 , 1191 阅读

\[\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^2}}}} = - \frac{{{\pi ^2}}}{6}\left( { - 12\ln A + \gamma + \ln 2 + \ln \pi } \right).\]

证明：(Glaisher–Kinkelin constant)\[\boxed{\ln A = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^2}}}{4} - \left( {\frac{{{n^2}}}{2} + \frac{n}{2} + \frac{1}{{12}}} \right)\ln n + \sum\limits_{k = 1}^n {k\ln k} } \right].}\]

(Riemann zeta函数的导函数)\[\boxed{\zeta '\left( s \right) = - \sum\limits_{k = 1}^\infty {\frac{{\ln k}}{{{k^s}}}} .}\]

(Gamma 函数)

\[\Gamma \left( s \right) = \int_0^\infty {{x^{s - 1}}{e^{ - x}}dx}. \]

首先，证明$\zeta'(-1)=\frac{1}{12}-\ln A$.

再证明$\Gamma'(2)=1-\gamma.$

\[\Gamma '\left( 2 \right) = \int_0^\infty {x{e^{ - x}}\ln xdx} = \int_0^\infty {\left( {{e^{ - x}} + {e^{ - x}}\ln x} \right)dx} = 1 - \gamma .\]

利用

\[\zeta \left( s \right) = {2^s}{\pi ^{s - 1}}\sin \frac{{\pi s}}{2}\Gamma \left( {1 - s} \right)\zeta \left( {1 - s} \right).\]

令$s=-1$，我们有$\zeta{-1}=-\frac{1}{12}.$

两边同取对数得

\[\ln \zeta \left( s \right) = s\ln 2 + \left( {s - 1} \right)\ln \pi + \ln \sin \frac{{\pi s}}{2} + \ln \Gamma \left( {1 - s} \right) + \ln \zeta \left( {1 - s} \right).\]

求导，得到\[\frac{{\zeta '\left( s \right)}}{{\zeta \left( s \right)}} = \ln \left( {2\pi } \right) + \frac{\pi }{{2\tan \frac{{\pi s}}{2}}} - \frac{{\Gamma '\left( {1 - s} \right)}}{{\Gamma \left( {1 - s} \right)}} - \frac{{\zeta '\left( {1 - s} \right)}}{{\zeta \left( {1 - s} \right)}}.\]

令$s=-1$，我们有

\[\frac{{\zeta '\left( { - 1} \right)}}{{\zeta \left( { - 1} \right)}} = 12\ln A - 1 = \ln \left( {2\pi } \right) - 1 + \gamma - \frac{{\zeta '\left( 2 \right)}}{{\zeta \left( 2 \right)}}.\]

\[ \Rightarrow \zeta '\left( 2 \right) = \frac{{{\pi ^2}}}{6}\left( {\ln \left( {2\pi } \right) - 12\ln A + \gamma } \right).\]

因此我们得到\[\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^2}}}} = - \frac{{{\pi ^2}}}{6}\left( { - 12\ln A + \gamma + \ln 2 + \ln \pi } \right).\]