chapter 9 series

chapter 9 series

Positive series

9.1
$$\begin{gathered} f(x)={\int }_0^{sinx}sin(t^2)dt,g(x)=\sum _{n=1}^{\infty }\frac{x^{2n+1}}{n^n+2},则x\to 0时，f(x)是g(x)的（）无穷小 \\ 解：f(x)={\int }_0^{sinx}sin(t^2)dt\sim {\int }_0^x{t}^{2}dt=\frac{2}{3}{x}^3 \\ x\to 0,g(x)=\frac{1}{3}{x}^3+o(x^3)\sim \frac{1}{3}{x}^3,所以f(x)是g(x)的等价无穷小 \end{gathered}$$
9.2
$$\begin{gathered} 正向级数\sum _{n=1}^{\infty }{a}_n收敛，正向级数\sum _{n=1}^{\infty }{b}_n发散，则 \\ \sum _{n=1}^{\infty }{a}_n{b}_n必发散 \\ \sum _{n=1}^{\infty }{a}_n^2必收敛 \\ \sum _{n=1}^{\infty }{a}_n{b}_n必收敛 \\ \sum _{n=1}^{\infty }{b}_n^2必发散 \end{gathered}$$
9.3
$$当\sum _{n=1}^{\infty }{a}_n^2、\sum _{n=1}^{\infty }{b}_n^2都收敛则\sum _{n=1}^{\infty }{a}_n{b}_n绝对收敛$$
9.4
$$\begin{gathered} (1)\sum _{n=1}^{\infty }(\frac{n}{3n+2}{)}^n \\ 解：\underset{n\to \infty }{\lim }{\sqrt[n]{u}}_n=\underset{n\to \infty }{\lim }\frac{n}{3n+2}=\frac{1}{3}<1,收敛 \\ (2)\sum _{n=1}^{\infty }{\int }_0^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^2}dx \\ 解：{\int }_0^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^2}dx<{\int }_0^{\frac{1}{n}}\sqrt{x}dx=\frac{2}{3}\frac{1}{n^{\frac{3}{2}}},p=\frac{3}{2}>1,收敛 \\ (3)\sum _{n=1}^{\infty }\sqrt[3]{n+1}-\sqrt[3]{n} \\ 解：\sqrt[3]{n}(\sqrt[3]{n+1}-1)\sim \sqrt[3]{n}\cdot \frac{1}{3}\cdot \frac{1}{n},p<1,发散 \\ 另解，由拉格朗日中值定理，\sqrt[3]{n+1}-\sqrt[3]{n}=\frac{1}{3}{\xi }^{-\frac{2}{3}}\cdot 1\sim \frac{1}{3}{n}^{-\frac{2}{3}},p=\frac{2}{3}<1，发散 \end{gathered}$$
9.5
$$\begin{gathered} \sum _{n=1}^{\infty }{\int }_n^{\frac{\pi }{n}}\frac{\sin x}{1+x}dx \\ 解：\sum _{n=1}^{\infty }{\int }_n^{\frac{\pi }{n}}\frac{\sin x}{1+x}dx\le \sum _{n=1}^{\infty }{\int }_n^{\frac{\pi }{n}}\frac{1}{1+x}dx=\ln (1+\frac{\pi }{n}),发散，但说明不了问题 \\ 因为大的发散，小的不知道 \\ \sum _{n=1}^{\infty }{\int }_n^{\frac{\pi }{n}}\frac{\sin x}{1+x}dx\le \sum _{n=1}^{\infty }{\int }_n^{\frac{\pi }{n}}\frac{x}{1+x}dx\le \sum _{n=1}^{\infty }{\int }_n^{\frac{\pi }{n}}xdx \\ {\int }_n^{\frac{\pi }{n}}xdx=\frac{{\pi }^2}{2n^2},p=2>1，收敛，原级数收敛 \end{gathered}$$