高等数学（第七版）同济大学 习题12-3 个人解答

高等数学第七版同济大学 习题12-3

$\begin{array}{rlr}& 1.求下列幂级数的收敛区间& \end{array}$

$\begin{array}{rlr}& (1)x+2x^2+3x^3+\cdot \cdot \cdot +n{x}^n+\cdot \cdot \cdot & \\ & & \\ & (2)1-x+\frac{x^2}{2^2}+\cdot \cdot \cdot +(-1{)}^n\frac{x^n}{n^2}+\cdot \cdot \cdot & \\ & & \\ & (3)\frac{x}{2}+\frac{x^2}{2\cdot 4}+\frac{x^3}{2\cdot 4\cdot 6}+\cdot \cdot \cdot +\frac{x^n}{2\cdot 4\cdot \cdot \cdot \cdot \cdot (2n)}+\cdot \cdot \cdot & \\ & & \\ & (4)\frac{x}{1\cdot 3}+\frac{x^2}{2\cdot 3^2}+\frac{x^3}{3\cdot 3^3}+\cdot \cdot \cdot +\frac{x^n}{n\cdot 3^n}+\cdot \cdot \cdot & \\ & & \\ & (5)\frac{2}{2}x+\frac{2^2}{5}{x}^2+\frac{2^3}{10}{x}^3+\cdot \cdot \cdot +\frac{2^n}{n^2+1}{x}^n+\cdot \cdot \cdot & \\ & & \\ & (6)\sum _{n=1}^{\infty }(-1{)}^n\frac{x^{2n+1}}{2n+1}& \\ & & \\ & (7)\sum _{n=1}^{\infty }\frac{2n-1}{2^n}{x}^{2n-2}& \\ & & \\ & (8)\sum _{n=1}^{\infty }\frac{(x-5{)}^n}{\sqrt{n}}.& \end{array}$

解

$\begin{array}{rlr}& (1)因为\underset{n\to \infty }{\lim }\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{\lim }\frac{n+1}{n}=1所以收敛半径为1收敛区间为(-1,1).& \\ & & \\ & (2)当n\ge 1时\frac{|a_{n+1}|}{|a_n|}=\frac{\frac{1}{(n+1{)}^2}}{\frac{1}{n^2}}={\left(\frac{n}{n+1}\right)}^2\underset{n\to \infty }{\lim }\frac{|a_{n+1}|}{|a_n|}=1所以收敛半径为1收敛区间为(-1,1).& \\ & & \\ & (3)因为\underset{n\to \infty }{\lim }\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{\lim }\frac{1}{2(n+1)}=0所以收敛半径为+\infty 收敛区间为(-\infty ,+\infty ).& \\ & & \\ & (4)因为\underset{n\to \infty }{\lim }\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{\lim }\frac{1}{3}\cdot \frac{n}{n+1}=\frac{1}{3}所以收敛半径为3收敛区间为(-3,3).& \\ & & \\ & (5)因为\underset{n\to \infty }{\lim }\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{\lim }2\cdot \frac{n^2+1}{(n+1{)}^2+1}=2所以收敛半径为\frac{1}{2}收敛区间为\left(-\frac{1}{2},\frac{1}{2}\right).& \\ & & \\ & (6)将(-1{)}^n\frac{x^{2n+1}}{2n+1}看作数项级数的一般项u_n因为\underset{n\to \infty }{\lim }\frac{|u_{n+1}|}{|u_n|}=\underset{n\to \infty }{\lim }\frac{2n+1}{2n+3}|x{|}^2=|x{|}^2& \\ & & \\ & 当|x|<1时级数绝对收敛当|x|>1时由于一般项u_n\nrightarrow 0(n\to \infty )级数发散& \\ & & \\ & 所以该级数收敛半径为1收敛区间为(-1,1).& \\ & & \\ & (7)令t=x^2原式=\sum _{n=1}^{\infty }\frac{2n-1}{2^n}{t}^{n-1}因为\underset{n\to \infty }{\lim }\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{\lim }\frac{1}{2}\cdot \frac{2n+1}{2n-1}=\frac{1}{2}所以该级数的收敛半径为2& \\ & & \\ & 因此原级数的收敛半径为\sqrt{2}收敛区间为(-\sqrt{2},\sqrt{2}).& \\ & & \\ & (8)因为\underset{n\to \infty }{\lim }\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{\lim }\frac{\sqrt{n}}{\sqrt{n+1}}=1所以该级数收敛半径为1当|x-5|<1时级数收敛& \\ & & \\ & 当|x-5|>1时级数发散因此该级数收敛区间为(4,6).& \end{array}$

$\begin{array}{rlr}& 2.利用逐项求导或逐项积分求下列级数的和函数& \end{array}$

$\begin{array}{rlr}& (1)\sum _{n=1}^{\infty }n{x}^{n-1}& \\ & & \\ & (2)\sum _{n=1}^{\infty }\frac{x^{4n+1}}{4n+1}& \\ & & \\ & (3)x+\frac{x^3}{3}+\frac{x^5}{5}+\cdot \cdot \cdot +\frac{x^{2n-1}}{2n-1}+\cdot \cdot \cdot & \\ & & \\ & (4)\sum _{n=1}^{\infty }(n+2)x^{n+3}.& \end{array}$

解

$\begin{array}{rlr}& (1)该级数的收敛半径为1当-1<x<1时{\int }_0^x(\sum _{n=1}^{\infty }n{x}^{n-1}))dx=\sum _{n=1}^{\infty }\left({\int }_0^{x}n{x}^{n-1}dx\right)=\sum _{n=1}^{\infty }{x}^n=\frac{x}{1-x}& \\ & & \\ & 上式两端对x求导得\sum _{n=1}^{\infty }n{x}^{n-1}=\frac{1}{(1-x{)}^2}又因原级数在x=\pm 1处发散所以和函数& \\ & & \\ & s(x)=\frac{1}{(1-x{)}^2}(-1<x<1).& \\ & & \\ & (2)该级数的收敛半径为1当-1<x<1时{\left(\sum _{n=1}^{\infty }\frac{x^{4n+1}}{4n+1}\right)}^{\prime }=\sum _{n=1}^{\infty }{\left(\frac{x^{4n+1}}{4n+1}\right)}^{\prime }=\sum _{n=1}^{\infty }{x}^{4n}=\frac{x^4}{1-x^4}& \\ & & \\ & 上式两端分别从0到x积分由于\sum _{n=1}^{\infty }\frac{x^{4n+1}}{4n+1}在x=0处收敛于0所以得\sum _{n=1}^{\infty }\frac{x^{4n+1}}{4n+1}=& \\ & & \\ & {\int }_0^x\frac{x^4}{1-x^4}dx={\int }_0^x\left(-1+\frac{1}{2}\cdot \frac{1}{1+x^2}+\frac{1}{2}\cdot \frac{1}{1-x^2}\right)dx=\frac{1}{4}ln\frac{1+x}{1-x}+\frac{1}{2}arctanx-x& \\ & & \\ & 又因原级数在x=\pm 1处发散所以和函数s(x)=\frac{1}{4}ln\frac{1+x}{1-x}+\frac{1}{2}arctanx-x(-1<x<1).& \\ & & \\ & (3)级数为\sum _{n=1}^{\infty }\frac{x^{2n-1}}{2n-1}收敛半径为1当-1<x<1时{\left(\sum _{n=1}^{\infty }\frac{x^{2n-1}}{2n-1}\right)}^{\prime }=\sum _{n=1}^{\infty }{x}^{2n-2}=\frac{1}{1-x^2}& \\ & & \\ & 上式两端分别从0到x积分且\sum _{n=1}^{\infty }\frac{x^{2n-1}}{2n-1}在x=0处收敛于0所以得\sum _{n=1}^{\infty }\frac{x^{2n-1}}{2n-1}={\int }_0^x\frac{1}{1-x^2}dx=& \\ & & \\ & \frac{1}{2}ln\frac{1+x}{1-x}又因原级数在x=\pm 1处发散所以和函数s(x)=\frac{1}{2}ln\frac{1+x}{1-x}(-1<x<1).& \\ & & \\ & (4)该级数收敛半径为1收敛域为(-1,1)当x\in (-1,1)时\sum _{n=1}^{\infty }(n+2)x^{n+3}=x^2\sum _{n=1}^{\infty }(n+2)x^{n+1}=& \\ & & \\ & x^2{\left(\sum _{n=1}^{\infty }{x}^{n+2}\right)}^{\prime }其中\sum _{n=1}^{\infty }{x}^{n+2}=x^3\sum _{n=0}^{\infty }{x}^n=\frac{x^3}{1-x}又因{\left(\sum _{n=1}^{\infty }{x}^{n+2}\right)}^{\prime }={\left(\frac{x^3}{1-x}\right)}^{\prime }=\frac{3x^2-2x^3}{(1-x{)}^2}& \\ & & \\ & 所以和函数s(x)=x^2\cdot \frac{3x^2-2x^3}{(1-x{)}^2}=\frac{3x^4-2x^5}{(1-x{)}^2}(-1<x<1).& \end{array}$