DLMF:24.4.E8 (Q7421)

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DLMF:24.4.E8
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    test
    ∑ k = 1 m ( - 1 ) m - k ⁢ k n = E n ⁡ ( m + 1 ) + ( - 1 ) m ⁢ E n ⁡ ( 0 ) 2 . superscript subscript 𝑘 1 𝑚 superscript 1 𝑚 𝑘 superscript 𝑘 𝑛 Euler-polynomial-E 𝑛 𝑚 1 superscript 1 𝑚 Euler-polynomial-E 𝑛 0 2 {\displaystyle{\displaystyle\sum_{k=1}^{m}(-1)^{m-k}k^{n}=\frac{E_{n}\left(m+1% \right)+(-1)^{m}E_{n}\left(0\right)}{2}.}}
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    DLMF:24.4.E8
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