【微积分/高等数学】无穷级数 之 和函数的快速求法（九阴真经）
本笔记资料中的方法是考研数学王谱老师的“九阴真经”，对于求和函数的题可快速解决. 现将笔记分享出来，也方便自己翻阅笔记.
前言
此类题目的出题方式一般为给出无穷级数，要求写出和函数及收敛域. 本笔记中的方法是先记住常用的九个无穷级数（不妨称其为“标准型”），对于具体题目，可先将原级数进行因式分解等操作，然后化作九种标准型的和、差即可快速写出和函数.
对于收敛域的求法，则可根据阿贝尔判别法求出收敛区间，再对区间端点单独做验证即可求出收敛域.
九个无穷级数的标准型可分为三大类，即整式型、分式型、阶乘型，值得注意的是本笔记中给出的顺序具有一定规律（可自行体会），根据规律进行记忆更加快速.
一、整式型
{ ( a ) ∑ n = 0 ∞ x n = 1 1 − x ( 收敛域 ( − 1 , 1 ) ) ( b ) ∑ n = 0 ∞ ( n + 1 ) x n = 1 ( 1 − x ) 2 ( 收敛域 ( − 1 , 1 ) ) ( c ) ∑ n = 0 ∞ ( n + 2 ) ( n + 1 ) x n = 2 ( 1 − x ) 3 ( 收敛域 ( − 1 , 1 ) ) \begin{cases} \left( \mathrm{a} \right) \sum_{n=0}^{\infty}{x^n}=\frac{1}{1-x}\,\,\left( \text{收敛域}\left( -1,1 \right) \right)\\ \\ \left( \mathrm{b} \right) \sum_{n=0}^{\infty}{\left( n+1 \right) x^n}=\frac{1}{\left( 1-x \right) ^2}\,\,\left( \text{收敛域}\left( -1,1 \right) \right)\\ \\ \left( \mathrm{c} \right) \sum_{n=0}^{\infty}{\left( n+2 \right) \left( n+1 \right) x^n}=\frac{2}{\left( 1-x \right) ^3}\,\,\left( \text{收敛域}\left( -1,1 \right) \right)\\ \end{cases} ⎩ ⎨ ⎧(a)∑n=0∞xn=1−x1(收敛域(−1,1))(b)∑n=0∞(n+1)xn=(1−x)21(收敛域(−1,1))(c)∑n=0∞(n+2)(n+1)xn=(1−x)32(收敛域(−1,1))
【例1】(2014·数学三)
求幂级数 ∑ n = 0 ∞ ( n + 1 ) ( n + 3 ) x n \sum_{n=0}^{\infty}{\left( n+1 \right) \left( n+3 \right) x^n} ∑n=0∞(n+1)(n+3)xn的收敛域及和函数.
解：
收敛域 ( − 1 , 1 ) (-1,1) (−1,1).
注意需将级数先改写为 n = 0 ， x n n=0，x^n n=0，xn的标准型.
S ( x ) = ∑ n = 0 ∞ ( n + 1 ) ( n + 3 ) x n = ∑ n = 0 ∞ ( n + 1 ) ( n + 2 ) x n + ∑ n = 0 ∞ ( n + 1 ) x n = 2 ( 1 − x ) 3 + 1 ( 1 − x ) 2 . \begin{aligned} S(x)&= \sum_{n=0}^{\infty}{\left( n+1 \right) \left( n+3 \right) x^n} \\ &=\sum_{n=0}^{\infty}{\left( n+1 \right) \left( n+2 \right) x^n}+\sum_{n=0}^{\infty}{\left( n+1 \right) x^n} \\ &=\frac{2}{\left( 1-x \right) ^3}+\frac{1}{\left( 1-x \right) ^2}. \end{aligned} S(x)=n=0∑∞(n+1)(n+3)xn=n=0∑∞(n+1)(n+2)xn+n=0∑∞(n+1)xn=(1−x)32+(1−x)21.
二、分式型
{ ( d ) ∑ n = 0 ∞ x n + 1 n + 1 = − ln ( 1 − x ) ( 收敛域 [ − 1 , 1 ) ) ( e ) ∑ n = 0 ∞ x 2 n + 1 2 n + 1 = 1 2 ln 1 + x 1 − x = tanh − 1 x ( 收敛域 ( − 1 , 1 ) ) ( f ) ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 2 n + 1 = arctan x ( 收敛域 [ − 1 , 1 ] ) \begin{cases} \left( \mathrm{d} \right) \sum_{n=0}^{\infty}{\frac{x^{n+1}}{n+1}}=-\ln \left( 1-x \right) \,\, \left( \text{收敛域}\left[ -1,1 \right) \right)\\ \\ \left( \mathrm{e} \right) \sum_{n=0}^{\infty}{\frac{x^{2n+1}}{2n+1}}=\frac{1}{2}\ln \frac{1+x}{1-x}=\tanh ^{-1}x\,\, \left( \text{收敛域}\left( -1,1 \right) \right)\\ \\ \left( \mathrm{f} \right) \sum_{n=0}^{\infty}{\left( -1 \right) ^n\frac{x^{2n+1}}{2n+1}}=\arctan x\,\, \left( \text{收敛域}\left[ -1,1 \right] \right)\\ \end{cases} ⎩ ⎨ ⎧(d)∑n=0∞n+1xn+1=−ln(1−x)(收敛域[−1,1))(e)∑n=0∞2n+1x2n+1=21ln1−x1+x=tanh−1x(收敛域(−1,1))(f)∑n=0∞(−1)n2n+1x2n+1=arctanx(收敛域[−1,1])
【例2】(1987·数学一)
求幂级数 ∑ n = 1 ∞ 1 n ⋅ 2 n x n − 1 \sum_{n=1}^{\infty}{\frac{1}{n\cdot 2^n}x^{n-1}} ∑n=1∞n⋅2n1xn−1的和函数.
解：
注意先凑分母
x ≠ 0 时， S ( x ) = ∑ n = 1 ∞ 1 n ⋅ 2 n x n − 1 = ∑ n = 0 ∞ 1 n + 1 ⋅ x n 2 n + 1 = 1 x ∑ n = 0 ∞ 1 n + 1 ⋅ ( x 2 ) n + 1 = − 1 x ln ( 1 − x 2 ) \begin{aligned} x\ne0时， S\left( x \right) &=\sum_{n=1}^{\infty}{\frac{1}{n\cdot 2^n}x^{n-1}} \\ &=\sum_{n=0}^{\infty}{\frac{1}{n+1}\cdot \frac{x^n}{2^{n+1}}} \\ &=\frac{1}{x}\sum_{n=0}^{\infty}{\frac{1}{n+1}\cdot \left( \frac{x}{2} \right) ^{n+1}} \\ &=-\frac{1}{x}\ln \left( 1-\frac{x}{2} \right) \end{aligned} x=0时，S(x)=n=1∑∞n⋅2n1xn−1=n=0∑∞n+11⋅2n+1xn=x1n=0∑∞n+11⋅(2x)n+1=−x1ln(1−2x)
因此和函数
S ( x ) = { − 1 x ln ( 1 − x 2 ) , x ≠ 0 1 2 , x = 0 S\left( x \right) = \begin{cases} -\frac{1}{x}\ln \left( 1-\frac{x}{2} \right) ,x\ne 0\\ \frac{1}{2},x=0\\ \end{cases} S(x)={−x1ln(1−2x),x=021,x=0
【例3】(2009·数学一)
设 a n a_n an为曲线 y = x n y=x^n y=xn与 y = x n + 1 ( n = 1 , 2 , . . . ) y=x^{n+1}(n=1,2,...) y=xn+1(n=1,2,...)所围成区域的面积，记 S 1 = ∑ n = 1 ∞ a n S_1=\sum_{n=1}^{\infty}a_n S1=∑n=1∞an, S 2 = ∑ n = 1 ∞ a 2 n − 1 S_2=\sum_{n=1}^{\infty}a_{2n-1} S2=∑n=1∞a2n−1，求 S 1 , S 2 S_1,S_2 S1,S2的值.
解：
a n = ∫ 0 1 ( x n − x n + 1 ) d x = 1 n + 1 − 1 n + 2 a_n=\int_0^1{\left( x^n-x^{n+1} \right) \mathrm{d}x}=\frac{1}{n+1}-\frac{1}{n+2} an=∫01(xn−xn+1)dx=n+11−n+21
S 1 = ∑ n = 1 ∞ ( 1 n + 1 − 1 n + 2 ) = 1 2 S_1=\sum_{n=1}^{\infty}{\left( \frac{1}{n+1}-\frac{1}{n+2} \right)}=\frac{1}{2} S1=n=1∑∞(n+11−n+21)=21
S 2 = ∑ n = 1 ∞ ( 1 2 n − 1 2 n + 1 ) = ∑ n = 2 ∞ ( − 1 ) n n = ∑ n = 0 ∞ ( − 1 ) n + 1 n + 1 + 1 = 1 − ln 2 \begin{aligned} S_2&=\sum_{n=1}^{\infty}{\left( \frac{1}{2n}-\frac{1}{2n+1} \right)} \\ &=\sum_{n=2}^{\infty}{\frac{\left( -1 \right) ^n}{n}} \\ &=\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^{n+1}}{n+1}}+1 \\ &=1-\ln 2 \end{aligned} S2=n=1∑∞(2n1−2n+11)=n=2∑∞n(−1)n=n=0∑∞n+1(−1)n+1+1=1−ln2
三、阶乘型
{ ( g ) ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! = sin x ( 收敛域 ( − ∞ , + ∞ ) ) ( h ) ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! = cos x ( 收敛域 ( − ∞ , + ∞ ) ) ( i ) ∑ n = 0 ∞ x n n ! = e x ( 收敛域 ( − ∞ , + ∞ ) ) \begin{cases} \left( \mathrm{g} \right) \sum_{n=0}^{\infty}{\left( -1 \right) ^n\frac{x^{2n+1}}{\left( 2n+1 \right) !}}=\sin x\,\,\left( \text{收敛域}\left( -\infty ,+\infty \right) \right)\\ \\ \left( \mathrm{h} \right) \sum_{n=0}^{\infty}{\left( -1 \right) ^n\frac{x^{2n}}{\left( 2n \right) !}}=\cos x\,\,\left( \text{收敛域}\left( -\infty ,+\infty \right) \right)\\ \\ \left( \mathrm{i} \right) \sum_{n=0}^{\infty}{\frac{x^n}{n!}}=\mathrm{e}^x\,\,\left( \text{收敛域}\left( -\infty ,+\infty \right) \right)\\ \end{cases} ⎩ ⎨ ⎧(g)∑n=0∞(−1)n(2n+1)!x2n+1=sinx(收敛域(−∞,+∞))(h)∑n=0∞(−1)n(2n)!x2n=cosx(收敛域(−∞,+∞))(i)∑n=0∞n!xn=ex(收敛域(−∞,+∞))
【例4】
求
∑ n = 0 ∞ n 2 + 1 2 n ⋅ n ! . \sum_{n=0}^{\infty}{\frac{n^2+1}{2^n\cdot n!}}. n=0∑∞2n⋅n!n2+1.
解：
∑ n = 0 ∞ n 2 + 1 2 n ⋅ n ! = ∑ n = 0 ∞ n 2 + 1 n ! ( 1 2 ) n = ∑ n = 0 ∞ n ( n − 1 ) + n + 1 n ! ( 1 2 ) n = ∑ n = 2 ∞ 1 ( n − 2 ) ! ( 1 2 ) n + ∑ n = 1 ∞ 1 ( n − 1 ) ! ( 1 2 ) n + ∑ n = 0 ∞ 1 n ! ( 1 2 ) n = 1 4 ∑ n = 0 ∞ 1 n ! ( 1 2 ) n + 1 2 ∑ n = 0 ∞ 1 n ! ( 1 2 ) n + ∑ n = 0 ∞ 1 n ! ( 1 2 ) n = ( 1 4 + 1 2 + 1 ) e = 7 4 e \begin{aligned} \sum_{n=0}^{\infty}{\frac{n^2+1}{2^n\cdot n!}}&=\sum_{n=0}^{\infty}{\frac{n^2+1}{n!}}\left( \frac{1}{2} \right) ^n \\ &=\sum_{n=0}^{\infty}{\frac{n\left( n-1 \right) +n+1}{n!}}\left( \frac{1}{2} \right) ^n \\ &=\sum_{n=2}^{\infty}{\frac{1}{\left( n-2 \right) !}}\left( \frac{1}{2} \right) ^n+\sum_{n=1}^{\infty}{\frac{1}{\left( n-1 \right) !}}\left( \frac{1}{2} \right) ^n+\sum_{n=0}^{\infty}{\frac{1}{n!}}\left( \frac{1}{2} \right) ^n \\ &=\frac{1}{4}\sum_{n=0}^{\infty}{\frac{1}{n!}}\left( \frac{1}{2} \right) ^n+\frac{1}{2}\sum_{n=0}^{\infty}{\frac{1}{n!}}\left( \frac{1}{2} \right) ^n+\sum_{n=0}^{\infty}{\frac{1}{n!}}\left( \frac{1}{2} \right) ^n \\ &=\left( \frac{1}{4}+\frac{1}{2}+1 \right) \sqrt{\mathrm{e}} \\ &=\frac{7}{4}\sqrt{\mathrm{e}} \end{aligned} n=0∑∞2n⋅n!n2+1=n=0∑∞n!n2+1(21)n=n=0∑∞n!n(n−1)+n+1(21)n=n=2∑∞(n−2)!1(21)n+n=1∑∞(n−1)!1(21)n+n=0∑∞n!1(21)n=41n=0∑∞n!1(21)n+21n=0∑∞n!1(21)n+n=0∑∞n!1(21)n=(41+21+1)e=47e
【拓展】
对于 “二、分式型” ，可将结论推广为
∑ n = 0 ∞ ( n + k − 1 ) . . . ( n + 1 ) x n = ( k − 1 ) ! ( 1 − x ) k ( 收敛域 ( − 1 , 1 ) ) . \sum_{n=0}^{\infty}{\left( n+k-1 \right) ...\left( n+1 \right) x^n}=\frac{\left( k-1 \right) !}{\left( 1-x \right) ^k}\,\,\left( \text{收敛域}\left( -1,1 \right) \right) . n=0∑∞(n+k−1)...(n+1)xn=(1−x)k(k−1)!(收敛域(−1,1)).