User:Zelmerszoetrop/Test1: Difference between revisions

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$\ln x=\sum _{k=1}^{\infty }(1-1/x)^{k}/k=\sum _{k=1}^{\infty }(1/k)\sum _{j=0}^{k}{n \choose j}(-x)^{-j},$

$\sum _{k=1}^{\infty }(1/k)\sum _{j=1}^{k}{n \choose j}-{\frac {\Gamma (1-j)}{\Gamma (1-j-{\tfrac {1}{N}})}}x^{-j-{\tfrac {1}{N}}}$

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		<math>\ln x = \sum_{k=1}^\infty (1-1/x)^k /k = \sum_{k=1}^\infty (1/k) \sum_{j=0}^k {n \choose j}(-x)^{-j},</math>



		<math>\sum_{k=1}^\infty (1/k) \sum_{j=1}^k {n \choose j} -\frac{\Gamma(1-j)}{\Gamma(1-j-\tfrac{1}{N})} x^{-j-\tfrac{1}{N}}</math>