DLMF:4.9.E2 (Q1610)

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DLMF:4.9.E2
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    ln ( 1 + z 1 - z ) = 2 z 1 - z 2 3 - 4 z 2 5 - 9 z 2 7 - 16 z 2 9 - , 1 𝑧 1 𝑧 continued-fraction 2 𝑧 1 continued-fraction superscript 𝑧 2 3 continued-fraction 4 superscript 𝑧 2 5 continued-fraction 9 superscript 𝑧 2 7 continued-fraction 16 superscript 𝑧 2 limit-from 9 {\displaystyle{\displaystyle\ln\left(\frac{1+z}{1-z}\right)=\cfrac{2z}{1-% \cfrac{z^{2}}{3-\cfrac{4z^{2}}{5-\cfrac{9z^{2}}{7-\cfrac{16z^{2}}{9-}}}}}% \cdots,}}
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    DLMF:4.9.E2
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