CVS2

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Find the Maclaurin series for \tan ^{{-1}}z\,.

{\frac  {d}{dz}}\tan ^{{-1}}z={\frac  {1}{1+z^{2}}}=\sum _{{n=0}}^{\infty }(-z^{2})^{n}=\sum _{{n=0}}^{\infty }(-1)^{n}z^{{2n}},|z|<1\,.

Integrating termwise,

\int _{0}^{z}\sum _{{n=0}}^{\infty }(-1)^{n}\xi ^{{2n}}d\xi =\sum _{{n=0}}^{\infty }(-1)^{n}\int _{0}^{z}\xi ^{{2n}}d\xi =\sum _{{n=0}}^{\infty }(-1)^{n}{\frac  {z^{{2n+1}}}{2n+1}},|z|<1\,



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