Section02_幂级数

定义

  1. {Un(x)}\{U_{n}(x)\} 为函数列，∑n=1∞Un(x)\displaystyle \sum_{n=1}^{\infty}U_{n}(x) 为函数项级数
    • x=x0x=x_{0} 时，若 ∑n=1∞Un(x0)\displaystyle \sum_{n=1}^{\infty}U_{n}(x_{0}) 收敛，则称 x0x_{0} 为收敛点
    • x=x1x=x_{1} 时，若 ∑n=1∞Un(x1)\displaystyle \sum_{n=1}^{\infty}U_{n}(x_{1}) 发散，则称 x1x_{1} 为发散点
    • 一切收敛点组成的集合称为收敛域
    • 一切发散点组成的集合称为发散域
  2. 幂级数 ∑n=0∞anxn=a0+a1x+a2x2+⋯+anxn+⋯OR∑n=0∞an(x−x0)n=a0+a1(x−x0)+a2(x−x0)2+⋯+an(x−x0)n+⋯ \begin{array}{cc} \displaystyle \sum_{n=0}^{\infty}a_{n}x^{n} = a_{0} + a_{1}x + a_{2}x^{2} + \cdots + a_{n}x^{n} + \cdots \\ \text{OR} \\ \displaystyle \sum_{n=0}^{\infty}a_{n}(x-x_{0})^{n} = a_{0} + a_{1}(x-x_{0}) + a_{2}(x-x_{0})^{2} + \cdots + a_{n}(x-x_{0})^{n} + \cdots \\ \end{array}
  3. Abel 判别法: 对于 ∑n=0∞anxn\displaystyle \sum_{n=0}^{\infty}a_{n}x^{n}:
    • ∣x∣<R\vert x \vert < R：∑n=0∞anxn\displaystyle \sum_{n=0}^{\infty}a_{n}x^{n} 绝对收敛
    • ∣x∣>R\vert x \vert > R：∑n=0∞anxn\displaystyle \sum_{n=0}^{\infty}a_{n}x^{n} 发散
    • ∣x∣=R\vert x \vert = R：未知

Notes

收敛半径，收敛区间，收敛域

Th 1 对 ∑n=0∞anxn,lim⁡n→∞∣an+1an∣=ρ\displaystyle \sum_{n=0}^{\infty}a_{n}x^{n}, \lim_{n\to \infty}\left\vert \frac{a_{n+1}}{a_{n}}\right\vert = \rho 或 lim⁡n→∞∣an∣n=ρ\displaystyle \lim_{n\to \infty}\sqrt[n]{\vert a_{n} \vert} = \rho

  1. ρ=0⇒R=+∞\rho = 0 \Rightarrow R = +\infty
  2. ρ=+∞⇒R=0\rho = +\infty \Rightarrow R = 0
  3. 0<ρ<+∞⇒R=1ρ0 < \rho < +\infty \Rightarrow R = \frac{1}{\rho}
  1. 求 ∑n=0∞n!xn\displaystyle \sum_{n=0}^{\infty}n!x^{n} 的收敛域 ∵lim⁡n→∞∣an+1an∣=lim⁡n→∞(n+1)!n!=+∞∴R=0⇒收敛域为{0} \begin{array}{ll} \because & \lim\limits_{n\to \infty} \left\vert \frac{a_{n+1}}{a_{n}}\right\vert = \lim\limits_{n\to \infty}\frac{(n+1)!}{n!} = +\infty \\ \therefore & R = 0 \Rightarrow \text{收敛域为} \{0\} \\ \end{array}
  2. 求 ∑n=1∞xnn!\displaystyle \sum_{n=1}^{\infty}\frac{x^{n}}{n!} 的收敛域 ∵lim⁡n→∞∣an+1an∣=lim⁡n→∞1n+1=0∴R=+∞⇒收敛域为(−∞,+∞) \begin{array}{ll} \because & \lim\limits_{n\to \infty} \left\vert \frac{a_{n+1}}{a_{n}}\right\vert = \lim\limits_{n\to \infty} \frac{1}{n+1} = 0 \\ \therefore & R = +\infty \Rightarrow \text{收敛域为} (-\infty, +\infty) \\ \end{array}
  3. 求 ∑n=1∞(x−1)nn⋅2n\displaystyle \sum_{n=1}^{\infty}\frac{(x-1)^{n}}{n\cdot 2^{n}} 的收敛域 ∵lim⁡n→∞∣an+1an∣=lim⁡n→∞n⋅2n(n+1)⋅2n+1=12lim⁡n→∞nn+1=12∴R=2当x−1=2时, 原式=∑n=1∞1n发散当x−1=−2时, 原式=∑n=1∞(−1)nn收敛∴−2≤x−1<2∴收敛域为[−1,3) \begin{array}{ll} \because & \lim\limits_{n\to \infty} \left\vert \frac{a_{n+1}}{a_{n}}\right\vert = \lim\limits_{n\to \infty} \frac{n\cdot 2^{n}}{(n+1)\cdot 2^{n+1}} = \frac{1}{2}\lim\limits_{n\to \infty} \frac{n}{n+1} = \frac{1}{2}\\ \therefore & R = 2 \\ & \text{当} x-1 = 2 \text{时, 原式} = \sum_{n=1}^{\infty}\frac{1}{n} \text{发散}\\ & \text{当} x-1 = -2 \text{时, 原式} = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} \text{收敛}\\ \therefore & -2 \le x-1 < 2 \\ \therefore & \text{收敛域为} [-1,3) \\ \end{array}

幂级数的分析性质

  • 和函数 当 −R<x<R-R<x<R 时，S(x)=∑n=0∞anxnS(x) = \displaystyle \sum_{n=0}^{\infty}a_{n}x^{n} 被称为和函数

Th 1 逐项可导性

  • 对于 ∑n=0∞anxn,x∈(−R,R)\displaystyle \sum_{n=0}^{\infty}a_{n}x^{n}, x\in (-R,R) 则 (∑n=0∞anxn)′=∑n=0∞(anxn)′=∑n=0∞nanxn−1\displaystyle \bigg(\sum_{n=0}^{\infty}a_{n}x^{n} \bigg)' = \sum_{n=0}^{\infty}(a_{n}x^{n})' = \sum_{n=0}^{\infty} na_{n}x^{n-1}，且收敛半径依然为 RR

Th 2 逐项可积性

  • 对于 ∑n=0∞anxn,x∈(−R,R)\displaystyle \sum_{n=0}^{\infty}a_{n}x^{n}, x\in (-R,R) 则 ∫0x(∑n=0∞anxn)⋅dx=∑n=0∞∫0x(anxn)⋅dx=∑n=0∞ann+1xn+1\displaystyle \int_{0}^{x}\bigg(\sum_{n=0}^{\infty}a_{n}x^{n} \bigg)\cdot dx = \sum_{n=0}^{\infty}\int_{0}^{x}(a_{n}x^{n})\cdot dx = \sum_{n=0}^{\infty} \frac{a_{n}}{n+1}x^{n+1}，且收敛半径依然为 RR

任务一: 展开-f(x)⇒∑n=0∞an(x−x0)n\displaystyle f(x)\Rightarrow \sum_{n=0}^{\infty}a_{n}(x-x_{0})^{n}

直接法(公式法)

泰勒级数

  • 若f(x)f(x)在x=x0x=x_{0}邻域内任意阶可导，则 f(x)=∑n=0∞an(x−x0)n,an=f(n)(x0)n! f(x) = \sum_{n=0}^{\infty}a_{n}(x-x_{0})^{n}, a_{n} = \frac{f^{(n)}(x_{0})}{n!}

Notes 上式仅在 f(x)f(x) 的定义域的和幂级数收敛域的公共部分上相等

  • 特别地，若 x0=0x_{0} = 0，f(x)=∑n=0∞f(n)(0)n!xnf(x) = \displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}

常见展开(记！)

  1. ex=1+x+12x2+13!x3+⋯=∑n=0∞1n!xn(−∞<x<+∞)\displaystyle e^{x} = 1+x + \frac{1}{2}x^{2} + \frac{1}{3!}x^{3} + \cdots = \sum_{n=0}^{\infty}\frac{1}{n!} x^{n} \quad (-\infty<x< +\infty)
  2. sin⁡x=x−x33!x3+x55!x5−⋯=∑n=0∞(−1)n(2n+1)!x2n+1(−∞<x<+∞)\displaystyle \sin x = x - \frac{x^{3}}{3!}x^{3} + \frac{x^{5}}{5!}x^{5}- \cdots = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}x^{2n+1}\quad (-\infty < x < +\infty)
  3. cos⁡x=1−x22!x2+x44!x4−⋯=∑n=0∞(−1)n(2n)!x2n(−∞<x<+∞)\displaystyle \cos x = 1 - \frac{x^{2}}{2!}x^{2} + \frac{x^{4}}{4!}x^{4}- \cdots = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{2n}\quad (-\infty < x < +\infty)
  4. 11−x=1+x+x2+x3+⋯=∑n=0∞xn(−1<x<1)\displaystyle \frac{1}{1-x} = 1 + x + x^{2} + x^{3} + \cdots = \sum_{n=0}^{\infty}x^{n} \quad (-1 < x < 1)
  5. 11+x=1−x+x2−x3+⋯=∑n=0∞(−1)nxn(−1<x<1)\displaystyle \frac{1}{1+x} = 1 - x +x^{2} - x^{3} + \cdots = \sum_{n=0}^{\infty} (-1)^{n}x^{n}\quad (-1 < x < 1)
  6. ln⁡(1+x)=x−12x2+13x3+⋯=∑n=1∞(−1)n−1nxn(−1<x≤1)\displaystyle \ln(1+x) = x - \frac{1}{2}x^{2} + \frac{1}{3}x^{3}+ \cdots = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} x^{n} \quad (-1 < x \le 1)
    • 1−12+13−14+15−⋯=ln⁡2\displaystyle 1 - \frac{1}{2} + \frac{1}{3}-\frac{1}{4} + \frac{1}{5} - \cdots = \ln 2
  7. −ln⁡(1−x)=x+12x2+13x3+⋯=∑n=1∞xnn(−1≤x<1)\displaystyle -\ln(1-x) = x+\frac{1}{2}x^{2} + \frac{1}{3}x^{3} + \cdots = \sum_{n=1}^{\infty}\frac{x^{n}}{n}\quad (-1\le x < 1)

间接法

方法

  1. 分析性质
  2. 常见展开公式

例题

  1. 将 f(x)=1x2−1\displaystyle f(x) = \frac{1}{x^{2}-1} 展开为 (x−2)(x-2) 的幂级数 f(x)=1x2−1=12(1x−1−1x+1)1x−1=1(x−2)+1=∑n=0∞(−1)n(x−2)n1<x<31x+1=1(x−2)+3=131(x−2)3+1=13∑n=0∞(−1)n3n(x−2)n=∑n=0∞(−1)n3n+1(x−2)n−1<x<5∴f(x)=∑n=0∞(−1)n2(1−13n+1)(x−2)n1<x<3 \begin{array}{ll} &f(x) = \frac{1}{x^{2} - 1} = \frac{1}{2}(\frac{1}{x-1} - \frac{1}{x+1})\\ & \frac{1}{x-1} = \frac{1}{(x-2) + 1} = \sum_{n=0}^{\infty} (-1)^{n}(x-2)^{n}\quad 1<x<3 \\ & \frac{1}{x+1} = \frac{1}{(x-2) + 3} = \frac{1}{3}\frac{1}{\frac{(x-2)}{3} + 1} = \frac{1}{3}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{3^{n}} (x-2)^{n} \\ & = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{3^{n+1}}(x-2)^{n} \quad -1 < x < 5\\ \therefore & f(x) = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{2} (1- \frac{1}{3^{n+1}}) (x-2)^{n}\quad 1<x<3 \\ \end{array}
  2. 将 f(x)=5x−1x2−x−2\displaystyle f(x)=\frac{5x - 1}{x^{2} - x -2} 展开为 (x−1)(x-1) 的幂级数 5x−1x2−x−2=Ax−2+Bx+1∴A(x+1)+B(x−2)=5x−1⇒{A+B=5A−2B=−1∴5x−1x2−x−2=3x−2+2x+11x−2=1(x−1)−1=−11−(x−1)=−∑n=0∞(x−1)n0<x<21x+1=1(x−1)+2=121(x−1)2+1=∑n=0∞(−1)n2n+1(x−1)n−1<x<3∴5x−1x2−x−2=∑n=0∞((−1)n2n−3)(x−1)n0<x<2 \begin{array}{ll} & \frac{5x-1}{x^{2}- x - 2} = \frac{A}{x-2} + \frac{B}{x+1} \\ \therefore & A(x+1) + B(x-2) = 5x -1 \Rightarrow \begin{cases} A+B = 5 \\ A - 2B = -1 \end{cases}\\ \therefore & \frac{5x - 1}{x^{2}- x -2} = \frac{3}{x-2} + \frac{2}{x+1} \\ & \frac{1}{x-2} = \frac{1}{(x-1)- 1} = - \frac{1}{1- (x-1)} = -\sum_{n=0}^{\infty}(x-1)^{n} \quad 0<x<2 \\ & \frac{1}{x+1} = \frac{1}{(x-1) + 2} = \frac{1}{2} \frac{1}{\frac{(x-1)}{2} + 1} = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{2^{n+1}} (x-1)^{n} \quad -1<x<3 \\ \therefore & \frac{5x - 1}{x^{2} - x -2} = \sum_{n=0}^{\infty}(\frac{(-1)^{n}}{2^{n}}-3)(x-1)^{n} \quad 0<x<2\\ \end{array}
  3. f(x)=arctan⁡xf(x) = \arctan x 展开为 xx 的幂级数 ∵f′(x)=11+x2=∑n=0∞(−1)nx2n−1<x<1∴f(x)=∫0xf′(x)⋅dx=∑n=0∞∫0x(−1)nx2n⋅dx=∑n=0∞(−1)n2n+1x2n+1当x=1时，∑n=0∞(−1)n2n+1收敛当x=−1时，∑n=0∞(−1)3n+12n+1收敛∴f(x)=∑n=0∞(−1)n2n+1x2n+1−1≤x≤1 \begin{array}{ll} \because & f'(x) = \frac{1}{1+x^{2}} = \sum_{n=0}^{\infty}(-1)^{n}x^{2n} \quad -1<x<1 \\ \therefore & f(x) = \int_{0}^{x}f'(x) \cdot dx = \sum_{n=0}^{\infty}\int_{0}^{x}(-1)^{n}x^{2n}\cdot dx \\ & = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1}x^{2n+1}\\ & \text{当}x = 1 \text{时，}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1} \text{收敛}\\ & \text{当}x = -1 \text{时，}\sum_{n=0}^{\infty}\frac{(-1)^{3n+1}}{2n+1} \text{收敛}\\ \therefore & f(x) = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}x^{2n+1}\quad -1\le x\le1 \\ \end{array}

任务二: 求和-S(x)S(x)

Case1 ∑p(n)xn\sum_{}^{}p(n)x^{n}

相关公式

  1. ∑n=0∞xn=11−x−1<x<1\displaystyle \sum_{n=0}^{\infty}x^{n} = \frac{1}{1-x}\quad -1<x<1
  2. ∑n=0∞(−1)nxn=11+x−1<x<1\displaystyle \sum_{n=0}^{\infty}(-1)^{n}x^{n} = \frac{1}{1+x}\quad -1 <x<1
  1. ∑n=0∞(n+2)xn\displaystyle \sum_{n=0}^{\infty}(n+2)x^{n} 求 S(x)S(x) ∵lim⁡n→∞∣an+1an∣=lim⁡n→∞n+3n+2=1∴R=1当x=±1时，lim⁡n→∞(n+2)(±1)n≠0∴收敛域为(−1,1)∑n=0∞(n+2)xn=21−x+x∑n=1∞nxn−1=21−x+x∑n=1∞(xn)′=21−x+x(∑n=1∞xn)′=21−x+x(x1−x)′=21−x+x1(1−x)2=2−x(1−x)2∴S(x)=2−x(1−x)2x∈(−1,1) \begin{array}{ll} \because & \lim\limits_{n\to \infty} \left\vert \frac{a_{n+1}}{a_{n}} \right\vert = \lim\limits_{n\to \infty} \frac{n+3}{n+2} = 1 \\ \therefore & R = 1 \\ & \text{当} x=\pm 1 \text{时，} \lim\limits_{n\to \infty} (n+2)(\pm 1)^{n} \ne 0 \\ \therefore & \text{收敛域为} (-1, 1) \\ \\ & \sum_{n=0}^{\infty}(n+2)x^{n} = \frac{2}{1-x} + x \sum_{n=1}^{\infty}n x^{n-1} \\ & = \frac{2}{1-x} + x \sum_{n=1}^{\infty}(x^{n})' = \frac{2}{1-x} + x (\sum_{n=1}^{\infty}x^{n})' \\ & = \frac{2}{1-x} + x (\frac{x}{1-x})'= \frac{2}{1-x} + x \frac{1}{(1-x)^{2}} \\ & = \frac{2 - x}{(1-x)^{2}}\\ \therefore & S(x) = \frac{2-x}{(1-x)^{2}} \quad x\in (-1,1) \end{array}
  2. ∑n=0∞n2xn\displaystyle \sum_{n=0}^{\infty}n^{2}x^{n} 求 S(x)S(x) ∵lim⁡n→∞∣an+1an∣=lim⁡n→∞(n+1n)2=1∴R=1当x=±1时，lim⁡n→∞n2(±1)n≠0∴收敛域为(−1,1)∑n=0∞n2xn=∑n=2∞n(n−1)xn+∑n=1∞nxn=x2∑n=2∞n(n−1)xn−2+x∑n=1∞nxn−1=x2(∑n=2∞xn)′′+x(∑n=1∞xn)′=x2(x21−x)′′+x(x1−x)′=x22(1−x)3+x1(1−x)2=2x2(1−x)3+x(1−x)2=x2+x(1−x)3∴S(x)=x2+x(1−x)3x∈(−1,1) \begin{array}{ll} \because & \lim\limits_{n\to \infty} \left\vert \frac{a_{n+1}}{a_{n}} \right\vert = \lim\limits_{n\to \infty} (\frac{n+1}{n})^{2} = 1 \\ \therefore & R = 1 \\ & \text{当}x=\pm 1 \text{时，} \lim\limits_{n\to \infty}n^{2} (\pm 1)^{n} \ne 0\\ \therefore & \text{收敛域为} (-1, 1) \\ \\ & \sum_{n=0}^{\infty}n^{2}x^{n} = \sum_{n=2}^{\infty} n(n-1) x^{n} + \sum_{n=1}^{\infty}nx^{n} \\ & = x^{2}\sum_{n=2}^{\infty}n(n-1)x^{n-2} + x \sum_{n=1}^{\infty}n x^{n-1} \\ & = x^{2}(\sum_{n=2}^{\infty}x^{n})'' + x (\sum_{n=1}^{\infty}x^{n})' \\ & = x^{2}(\frac{x^{2}}{1-x})'' + x(\frac{x}{1-x})' \\ & = x^{2}\frac{2}{(1-x)^{3}} + x \frac{1}{(1-x)^{2}} \\ & = \frac{2x^{2}}{(1-x)^{3}} + \frac{x}{(1-x)^{2}} = \frac{x^{2} + x}{(1-x)^{3}} \\ \therefore & S(x) = \frac{x^{2} + x}{(1-x)^{3}} \quad x\in (-1,1) \\ \end{array}

Case2 ∑xnp(n)\sum_{}^{}\frac{x^{n}}{p(n)}

相关公式

  1. ∑n=1∞xnn=−ln⁡(1−x)−1≤x<1\displaystyle \sum_{n=1}^{\infty}\frac{x^{n}}{n} = -\ln(1-x) \quad -1\le x<1
  2. ∑n=1∞(−1)nxnn=ln⁡(1+x)−1<x≤1\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}x^{n}}{n} = \ln(1+x)\quad -1 <x\le 1
  1. ∑n=1∞xnn(n+1)\displaystyle \sum_{n=1}^{\infty} \frac{x^{n}}{n(n+1)} ∵lim⁡n→∞∣an+1an∣=1∴R=1∵当x=±1时，∑n=1∞(±1)nn(n+1)收敛∴收敛域为[−1,1]x=0时，∑n=1∞0nn(n+1)=0x∈[−1,0)∪(0,1)时: ∑n=1∞xnn(n+1)=∑n=1∞xnn−∑n=1∞xnn+1=−ln⁡(1−x)−1x∑n=1∞xn+1n+1=−ln⁡(1−x)−1x∑n=2∞xnn=−ln⁡(1−x)−1x(∑n=1∞xnn−x)=(1x−1)ln⁡(1−x)+1x=1时，∑n=1∞1n(n+1)=1∴S(x)={0,x=0(1x−1)ln⁡(1−x)+1,x∈[−1,0)∪(0,1)1,x=1 \begin{array}{ll} \because & \lim\limits_{n\to \infty} \left\vert \frac{a_{n+1}}{a_{n}}\right\vert = 1\\ \therefore & R = 1 \\ \because & \text{当}x = \pm 1 \text{时，} \sum_{n=1}^{\infty} \frac{(\pm 1)^{n}}{n(n+1)} \text{收敛}\\ \therefore & \text{收敛域为} [-1,1] \\ \\ & x=0 \text{时，} \sum_{n=1}^{\infty}\frac{0^{n}}{n(n+1)} = 0 \\ & x\in[-1,0)\cup(0,1) \text{时: } \\ & \sum_{n=1}^{\infty}\frac{x^{n}}{n(n+1)} = \sum_{n=1}^{\infty}\frac{x^{n}}{n} - \sum_{n=1}^{\infty} \frac{x^{n}}{n+1} \\ & = -\ln(1-x) - \frac{1}{x}\sum_{n=1}^{\infty}\frac{x^{n+1}}{n+1} \\ & = -\ln(1-x) - \frac{1}{x} \sum_{n=2}^{\infty} \frac{x^{n}}{n}\\ & = -\ln(1-x) - \frac{1}{x}(\sum_{n=1}^{\infty}\frac{x^{n}}{n} - x) \\ & = (\frac{1}{x}-1)\ln(1-x) +1 \\ & x= 1 \text{时，} \sum_{n=1}^{\infty}\frac{1}{n(n+1)} = 1\\ \therefore & S(x) = \begin{cases} 0,& x = 0 \\ (\frac{1}{x} - 1)\ln(1-x) + 1, & x\in [-1,0)\cup (0,1)\\ 1, & x=1 \end{cases} \\ \end{array}
  2. ∑n=1∞x2nn(2n−1)\displaystyle \sum_{n=1}^{\infty}\frac{x^{2n}}{n(2n-1)} 求 S(x)S(x) ∵lim⁡n→∞∣an+1an∣=1∴R=1∵当x=±1时，∑n=1∞(±1)2nn(2n−1)收敛∴收敛域为[−1,1]∑n=1∞x2nn(2n−1)=2(x∑n=1∞x2n−12n−1−∑n=1∞x2n2n)=2[xS1(x)−S2(x)]S1(x)=∑n=1∞∫0xx2n−2⋅dx=∫0x∑n=1∞x2n−2⋅dx=∫0x11−x2⋅dx=−12∫0x(1x−1−1x+1)⋅dx=−12ln⁡x−1x+1−1<x<1S2(x)=12∑n=1∞(x2)nn=−12ln⁡(1−x2)−1≤x<1∴S(x)=ln⁡(1−x2)−xln⁡x−1x+1−1<x<1而 S(±1)=∑n=1∞1n(2n−1)=2∑n=1∞(12n−1−12n)=2(1−12+13−14+⋯ )=2ln⁡2∴S(x)={ln⁡(1−x2)−xln⁡x−1x+1x∈(−1,1)2ln⁡2x=±1 \begin{array}{ll} \because & \lim\limits_{n\to \infty} \left\vert \frac{a_{n+1}}{a_{n}}\right\vert = 1\\ \therefore & R = 1 \\ \because & \text{当}x = \pm 1 \text{时，} \sum_{n=1}^{\infty} \frac{(\pm 1)^{2n}}{n(2n-1)} \text{收敛}\\ \therefore & \text{收敛域为} [-1,1] \\ \\ & \sum_{n=1}^{\infty}\frac{x^{2n}}{n(2n-1)} = 2 (x\sum_{n=1}^{\infty}\frac{x^{2n-1}}{2n-1} - \sum_{n=1}^{\infty}\frac{x^{2n}}{2n}) = 2[xS_{1}(x) - S_{2}(x)] \\\\ &S_{1}(x) = \sum_{n=1}^{\infty}\int_{0}^{x} x^{2n-2}\cdot dx = \int_{0}^{x}\sum_{n=1}^{\infty}x^{2n-2}\cdot dx = \int_{0}^{x} \frac{1}{1-x^{2}}\cdot dx \\ & = -\frac{1}{2}\int_{0}^{x}(\frac{1}{x-1} - \frac{1}{x+1})\cdot dx = -\frac{1}{2} \ln \frac{x-1}{x+1} \quad -1<x<1 \\ & S_{2}(x) = \frac{1}{2}\sum_{n=1}^{\infty}\frac{(x^{2})^{n}}{n} = -\frac{1}{2}\ln (1-x^{2}) \quad -1 \le x <1 \\ \therefore & S(x) = \ln(1-x^{2}) -x\ln \frac{x-1}{x+1} \quad -1<x<1 \\ & \text{而 } S(\pm1) = \sum_{n=1}^{\infty}\frac{1}{n(2n-1)} = 2 \sum_{n=1}^{\infty} (\frac{1}{2n-1} - \frac{1}{2n}) = 2(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots) = 2\ln 2 \\ \therefore & S(x) = \begin{cases} \ln(1-x^{2}) -x\ln \frac{x-1}{x+1} & x\in (-1,1) \\ 2\ln2 & x =\pm 1 \end{cases} \\ \end{array}
  3. ∑n=0∞(−1)n4n2+4n+22n+1x2n\displaystyle \sum_{n=0}^{\infty}(-1)^{n}\frac{4n^{2}+4n + 2}{2n + 1}x^{2n} 求 S(x)S(x) ∵lim⁡n→∞∣an+1an∣=1∴R=1当x=±1时，lim⁡n→∞4n2+4n+22n+1=+∞∴收敛域为(−1,1)∑n=0∞(−1)n4n2+4n+22n+1x2n=∑n=0∞(−1)n(2n+1)x2n+∑n=0∞(−1)n12n+1x2n=S1(x)+S2(x)当x=0时，S(0)=0当x∈(−1,0)∪(0,1)时S1(x)=∑n=0∞(−1)n(x2)n+∑n=1∞(−1)n(2n)x2n=11+x2+x∑n=1∞(−1)n(2n)x2n−1=11+x2+x∑n=1∞(−1)n(x2n)′=11+x2+x(∑n=1∞(−1)nx2n)′=11+x2+x(11+x2−1)′=11+x2−2x2(1+x2)2=1−x2(1+x2)2S2(x)=∑n=0∞(−1)n12n+1x2n=1x∑n=0∞(−1)n∫0xx2n⋅dx=1x∫0x∑n=0∞(−1)n(x2)n⋅dx=1x∫0x11+x2⋅dx=1xarctan⁡x∴S(x)=1−x2(1+x2)2+1xarctan⁡xx∈(−1,0)∪(0,1)∴S(x)={0,x=01−x2(1+x2)2+1xarctan⁡xx∈(−1,0)∪(0,1) \begin{array}{ll} \because & \lim\limits_{n\to \infty}\left\vert \frac{a_{n+1}}{a_{n}}\right\vert = 1 \\ \therefore & R = 1 \\ & \text{当} x= \pm 1 \text{时，} \lim\limits_{n\to \infty} \frac{4n^{2} + 4n + 2}{2n+1} = +\infty \\ \therefore & \text{收敛域为} (-1,1) \\\\ & \sum_{n=0}^{\infty}(-1)^{n} \frac{4n^{2} + 4n + 2}{2n + 1}x^{2n} = \sum_{n=0}^{\infty}(-1)^{n}(2n+1)x^{2n} + \sum_{n=0}^{\infty}(-1)^{n}\frac{1}{2n+1}x^{2n} \\ & = S_{1}(x) + S_{2}(x) \\\\ & \text{当} x = 0 \text{时，} S(0) = 0\\ & \text{当} x\in (-1,0)\cup(0,1) \text{时}\\ & S_{1}(x) = \sum_{n=0}^{\infty}(-1)^{n}(x^{2})^{n} + \sum_{n=1}^{\infty}(-1)^{n}(2n)x^{2n} \\ & = \frac{1}{1+x^{2}} + x \sum_{n=1}^{\infty}(-1)^{n} (2n)x^{2n-1} = \frac{1}{1+x^{2}} + x \sum_{n=1}^{\infty}(-1)^{n} (x^{2n})' \\ & = \frac{1}{1+x^{2}} + x(\sum_{n=1}^{\infty}(-1)^{n}x^{2n})' = \frac{1}{1+x^{2}} + x(\frac{1}{1+x^{2}} - 1) ' \\ & = \frac{1}{1+x^{2}} - \frac{2x^{2}}{(1+x^{2})^{2}} = \frac{1-x^{2}}{(1+x^{2})^{2}} \\ &S_{2}(x) = \sum_{n=0}^{\infty}(-1)^{n} \frac{1}{2n+1} x^{2n} = \frac{1}{x}\sum_{n=0}^{\infty}(-1)^{n} \int_{0}^{x}x^{2n}\cdot dx \\ & = \frac{1}{x} \int_{0}^{x}\sum_{n=0}^{\infty} (-1)^{n} (x^{2})^{n}\cdot dx = \frac{1}{x} \int_{0}^{x}\frac{1}{1+x^{2}}\cdot dx \\ & = \frac{1}{x}\arctan x \\ \therefore & S(x) = \frac{1-x^{2}}{(1+x^{2})^{2}} + \frac{1}{x}\arctan x\quad x\in(-1,0)\cup(0,1)\\ \therefore & S(x) = \begin{cases} 0, & x = 0 \\ \frac{1-x^{2}}{(1+x^{2})^{2}}+\frac{1}{x}\arctan x & x\in (-1,0)\cup(0,1) \end{cases}\\ \end{array}

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