$f(z)^{1+1/z}$的渐近阶

hbghlyj

$$\left(1+\frac{1}{z}+O(\frac{1}{z^5})\right)^{1+\frac{1}{z}}=1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{2 z^3}+\frac{1}{3 z^4}+O(\frac{1}{z^5})$$这是怎么得到的
来源: asymptotic Powers, Expontials and Logarithms例子

Czhang271828

\begin{align*}
&\,e^{(1+z^{-1})\log(1+z^{-1}+\mathcal O(z^{-5}))}\\
=&\,e^{(1+z^{-1})(z^{-1}-z^{-2}/2+z^{-3}/3-z^{-4}/4+\mathcal O(z^{-5}))}\\
=&\,e^{z^{-1}+z^{-2}/2-z^{-3}/6+z^{-4}/12+\mathcal O(z^{-5})}\\
=&\,1+z^{-1}+z^{-2}+z^{-3}/2+z^{-4}/3+\mathcal O(z^{-5})
\end{align*}

hbghlyj

MSE$$(1+t^{-1})^t=\exp\left(t\ln\left(1+\frac1t\right)\right)=\exp\left(1-\frac{1}{2t}+O(t^{-2})\right)
=e\left(1-\frac1{2t}+O(t^{-2})\right).$$
Integrate this from $1$ to $x$.