答疑问题

求 $$\lim _{x \rightarrow 0} \frac{e^{(1+x)^{1/x}}-(1+x)^{e / x}}{x^{2}}$$

由泰勒公式可知
$$\begin{align*} (1+x)^{1 / x} &=\exp \left\{\frac{\ln (1+x)}{x}\right\}=\exp \left\{\frac{x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+o\left(x^{4}\right)}{x}\right\} \\ &=e \cdot \exp \left\{-\frac{x}{2}+\frac{x^{2}}{3}+o\left(x^{3}\right)\right\}\\ &=e\left[1+\left(-\frac{x}{2}+\frac{x^{2}}{3}+o\left(x^{3}\right)\right)+\frac{\left(-\frac{x}{2}+\frac{x^{2}}{3}+o\left(x^{3}\right)\right)^{2}}{2}+o\left(x^{3}\right)\right] \\ &=e-\frac{e}{2} x+\frac{11 e}{24} x^{2}+o\left(x^{3}\right). \end{align*}$$
于是
$$e^{(1+x)^{1/{x}}}=e^{e}-\frac{1}{2} e^{e+1} x+\frac{1}{24} e^{e+1}(3 e+11) x^{2}+o\left(x^{2}\right).$$
类似地有
$$(1+x)^{e / x}=e^{e}-\frac{1}{2} e^{e+1} x+\frac{1}{24} e^{e+1}(3 e+8) x^{2}+o\left(x^{2}\right).$$
于是所求极限为
$$\lim _{x \rightarrow 0} \frac{e^{(1+x)^{l x}}-(1+x)^{e / x}}{x^{2}}=\frac{1}{8} e^{e+1}.$$

若$\lim_{n\to\infty}a_n=b>0$,判断$\sum_ {n=1}^{\infty}\left(\frac{b}{a_n}\right)^n$的敛散性.

取$a_n=b\sqrt[n]{n^2}\to b$,则$\sum_ {n=1}^{\infty}\left(\frac{b}{a_n}\right)^n=\sum_ {n=1}^{\infty}\frac{1}{n^2}$收敛.

取$a_n=b\sqrt[n]{n}\to b$,则$\sum_ {n=1}^{\infty}\left(\frac{b}{a_n}\right)^n=\sum_ {n=1}^{\infty}\frac{1}{n}$发散.

求极限$\lim_{n\to\infty}n^2\left(\arctan\frac{a}{n}-\arctan{a}{n+1}\right)$.