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一个级数求解 - Eufisky - The lost book
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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 , 1275 阅读
n=1lnnn2=π26(12lnA+γ+ln2+lnπ).
证明:(Glaisher–Kinkelin constant)lnA=lim
(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).
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)}}.
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).
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