\((-x+\sum_1^n\frac{p^m}{m!})=o(x^n)\;(p=\sum_1^n\frac{-(-x)^{k-1}}{k})\)

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题：设  \(p={\small\displaystyle\sum_{k=1}^n\frac{(-1)^{k-1}}{k}}x^k\), 试证\(\,{\small\displaystyle\sum_{m=1}^n\frac{p^m}{m!}}=x+o(x^n).\)
证：\(\because\;{\frac{d}{dx}}p={\frac{1-(-x)^n}{1+x}},\)
\(\quad\small\displaystyle\lim_{x\to 0}\frac{-x+\sum_{m=1}^n\frac{p^m}{m!}}{x^n}\;\overset{L'Hospital}{=\hspace{-2px}=\hspace{-2px}=}\lim_{x\to0}\frac{-1+\frac{1-(-x)^n}{1+x}(1+\sum_{m=1}^{n-1}\frac{p^m}{m!})}{nx^{n-1}}\)
\(\small=\displaystyle\lim_{x\to 0}\frac{-x+\sum_{m=1}^{n-1}\frac{p^m}{m!}}{nx^{n-1}}=\cdots=\lim_{x\to 0}\frac{-x+p}{n!x}=0.\quad\square\)

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这个结果在构造可逆矩阵\(\,A\,\)的对数矩阵\(\,\log A\,\)时有用。

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注记： 主贴极限运算的细节：
\(\small\displaystyle\frac{-1+\frac{1-(-x)^n}{1+x}(1+\sum_{m=1}^{n-1}\frac{p^m}{m!})}{nx^{n-1}}=\frac{-1-x+1+\sum_{m=1}^{n-1}\frac{p^m}{m!}+o(x^n)}{n(1+x)x^{n-1}}\)
\(\therefore\small\displaystyle\frac{-x+\sum_{m=1}^n\frac{p^m}{m!}}{x^n}\sim\frac{-x+\sum_{m=1}^{n-1}\frac{p^m}{m!}}{nx^{n-1}}\)