DLMF:1.12.E18 (Q448)

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DLMF:1.12.E18
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    test
    p 0 + ∑ k = 1 n p 1 ⁢ p 2 ⁢ ⋯ ⁢ p k = p 0 + p 1 1 - p 2 1 + p 2 - p 3 1 + p 3 - ⋯ p n 1 + p n , subscript 𝑝 0 subscript superscript 𝑛 𝑘 1 subscript 𝑝 1 subscript 𝑝 2 ⋯ subscript 𝑝 𝑘 subscript 𝑝 0 continued-fraction subscript 𝑝 1 1 continued-fraction subscript 𝑝 2 1 subscript 𝑝 2 continued-fraction subscript 𝑝 3 1 subscript 𝑝 3 ⋯ continued-fraction subscript 𝑝 𝑛 1 subscript 𝑝 𝑛 {\displaystyle{\displaystyle p_{0}+\sum^{n}_{k=1}p_{1}p_{2}\cdots p_{k}=p_{0}+% \cfrac{p_{1}}{1-\cfrac{p_{2}}{1+p_{2}-\cfrac{p_{3}}{1+p_{3}-\cdots\cfrac{p_{n}% }{1+p_{n}}}}},}}
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    DLMF id
    DLMF:1.12.E18
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    comment
    Warning: Falling back to standard tex;
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