问题及解答

判断级数 $\sum\limits_{n=1}^{\infty}n^n\sin^n(\frac{2}{n})$ 的敛散性.

Posted by haifeng on 2021-01-11 09:59:47 last update 2021-01-11 10:00:14 | Edit | Answers (1)

1

Posted by haifeng on 2021-01-11 10:18:10

$\lim_{n\rightarrow\infty}n^n\sin^n(\frac{2}{n})=\lim_{n\rightarrow\infty}\dfrac{\sin^n(\frac{2}{n})}{(\frac{1}{n})^n}=\lim_{n\rightarrow\infty}\dfrac{\sin^n(\frac{2}{n})}{(\frac{2}{n})^n\cdot\frac{1}{2^n}}=\lim_{n\rightarrow\infty}\Biggl[\biggl(\dfrac{\sin\frac{2}{n}}{\frac{2}{n}}\biggr)^n\cdot 2^n\Biggr]=\infty$

$\lim_{x\rightarrow 0}\biggl(\frac{\sin x}{x}\biggr)^{\frac{1}{x}}=\lim_{x\rightarrow 0}e^{\frac{1}{x}\ln\frac{\sin x}{x}}=e^{\lim\limits_{x\rightarrow 0}\frac{1}{x}\ln\frac{\sin x}{x}}$

$\lim\limits_{x\rightarrow 0}\frac{1}{x}\ln\frac{\sin x}{x}=\lim_{x\rightarrow 0}\frac{1}{x}\ln(1-\frac{1}{3!}x^2+o(x^3))=\lim_{x\rightarrow 0}\frac{-\frac{1}{3!}x^2+o(x^3)}{x}=\lim_{x\rightarrow 0}(-\frac{1}{3!}x+o(x^2))=0,$

$\lim_{x\rightarrow 0}\biggl(\frac{\sin x}{x}\biggr)^{\frac{1}{x}}=e^0=1.$

$\lim_{n\rightarrow\infty}\biggl(\dfrac{\sin\frac{2}{n}}{\frac{2}{n}}\biggr)^n=\lim_{n\rightarrow\infty}\Biggl[\biggl(\dfrac{\sin\frac{2}{n}}{\frac{2}{n}}\biggr)^{n/2}\Biggr]^2=1^2=1.$