DLMF:4.6.E1 (Q1570)

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DLMF:4.6.E1
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    test
    ln ⁡ ( 1 + z ) = z - 1 2 ⁢ z 2 + 1 3 ⁢ z 3 - ⋯ , 1 𝑧 𝑧 1 2 superscript 𝑧 2 1 3 superscript 𝑧 3 ⋯ {\displaystyle{\displaystyle\ln\left(1+z\right)=z-\tfrac{1}{2}z^{2}+\tfrac{1}{% 3}z^{3}-\cdots,}}
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    DLMF:4.6.E1
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