$(1+1/z)^{nz}$在0和∞的级数

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WolframAlpha
Series expansion at z=0
$$1 + n z \log(1/z) + \frac{1}{2} z^2 \left(n^2 \log ^2\left(\frac{1}{z}\right)+2 n\right)+O(z^3)$$
(Puiseux series)

Series expansion at z=∞
\[e^n-\frac{e^n n}{2 z}+\frac{e^n n (8+3 n)}{24 z^2}+O\left(\left(\frac{1}{z}\right)^3\right)\]
(Laurent series)

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$(1+\frac1z)^{nz}=\exp\left(nz\log(1+\frac1z)\right)$
用$\log(1+\frac1z)=\log \left(\frac{1}{z}\right)+z-\frac{z^2}{2}+\frac{z^3}{3}+O\left(z^4\right)$就得到在0的级数
用$\log(1+\frac1z)=\frac1z-\frac1{2z^2}+\frac1{3z^3}+O(\frac1{z^4})$就得到在∞的级数

分析$(1+1/z)^{nz}$的奇点:
$$\lim_{z\to0}(1+\frac1z)^{nz}=1\;\forall n\inC$$
0是removable singularity
对于$\log(1+\frac1z)$, $\frac1z\le-1$是branch cut, 即$z\in[-1,0]$

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contour $\gamma$ encloses the branch cut $[-1,0]$
interior of $\gamma$ contains $\infty$

1. NIntegrate[(1 + 1/z)^z, {z, 0, -2 + I, -2 - I, 0}]

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$$\int_\gamma(1+\frac1z)^z=-i\pi e$$

$$\int_\gamma(1+\frac1z)^{nz}=-i\pi n e^n$$