DLMF:4.9.E1 (Q1609)

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DLMF:4.9.E1
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    test
    ln ⁡ ( 1 + z ) = z 1 + z 2 + z 3 + 4 ⁢ z 4 + 4 ⁢ z 5 + 9 ⁢ z 6 + 9 ⁢ z 7 + ⁢ ⋯ , 1 𝑧 continued-fraction 𝑧 1 continued-fraction 𝑧 2 continued-fraction 𝑧 3 continued-fraction 4 𝑧 4 continued-fraction 4 𝑧 5 continued-fraction 9 𝑧 6 continued-fraction 9 𝑧 limit-from 7 ⋯ {\displaystyle{\displaystyle\ln\left(1+z\right)=\cfrac{z}{1+\cfrac{z}{2+\cfrac% {z}{3+\cfrac{4z}{4+\cfrac{4z}{5+\cfrac{9z}{6+\cfrac{9z}{7+}}}}}}}\cdots,}}
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    DLMF id
    DLMF:4.9.E1
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