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PolyLog[z,-k]在1的留数

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hbghlyj Posted at 2023-1-2 05:38:12 |Read mode
Last edited by hbghlyj at 2023-1-11 13:02:00
  1. Residue[Sum[n^2 z^n,{n,1,Infinity}],{z,1}]
  2. Residue[Sum[n^3 z^n,{n,1,Infinity}],{z,1}]
  3. Residue[Sum[n^4 z^n,{n,1,Infinity}],{z,1}]
  4. Residue[Sum[n^5 z^n,{n,1,Infinity}],{z,1}]
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输出$-1,1,-1,1$
猜测$k\in\Bbb N,\sum_{n=1}^\infty n^k z^n$在1的留数为$(-1)^{k-1}$

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 Author| hbghlyj Posted at 2023-1-11 20:14:35
根据这帖7#
$M(t)=\frac{z \exp (t)}{1-z \exp (t)}$, 则$\sum_{n=1}^\infty n^kz^n=M^{(k)}(0)$
如何求在1的留数呢

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2025-4-21 14:24 GMT+8

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