User:Zelmerszoetrop/Test1: Difference between revisions

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

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

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