级数[无穷,n=1] n(n+1)x^n
答案:2 悬赏:70
解决时间 2021-04-22 21:06
- 提问者网友:刀枪不入
- 2021-04-22 03:50
求级数的收敛域中,并求和函数。(请给出详细过程)
最佳答案
- 二级知识专家网友:都不是誰的誰
- 2021-04-22 04:27
lim(n+1)(n+2)/n(n+1)=1,收敛半径为1
当x=1时,limn(n+1)=无穷大,级数发散
当x=-1时,级数n(n+1)(-1)^n发散
所以收敛域为(-1,1)
设和函数S(x)=∑n(n+1)x^n
则∫S(x)dx=∑n∫(n+1)x^ndx=∑nx^(n+1)=∑(n+2)x^(n+1)-2∑x^(n+1)
=[∑∫(n+2)x^(n+1)dx]'-2x^2/(1-x)
=[∑x^(n+2)]'-2x^2/(1-x)
=[x^3/(1-x)]'-2x^2/(1-x)
=(3x^2-2x^3)/(1-x)^2-2x^2/(1-x)
=x^2/(1-x)^2
s(x)=[x^2/(1-x)^2]'=2x/(1-x)^3,-1
当x=1时,limn(n+1)=无穷大,级数发散
当x=-1时,级数n(n+1)(-1)^n发散
所以收敛域为(-1,1)
设和函数S(x)=∑n(n+1)x^n
则∫S(x)dx=∑n∫(n+1)x^ndx=∑nx^(n+1)=∑(n+2)x^(n+1)-2∑x^(n+1)
=[∑∫(n+2)x^(n+1)dx]'-2x^2/(1-x)
=[∑x^(n+2)]'-2x^2/(1-x)
=[x^3/(1-x)]'-2x^2/(1-x)
=(3x^2-2x^3)/(1-x)^2-2x^2/(1-x)
=x^2/(1-x)^2
s(x)=[x^2/(1-x)^2]'=2x/(1-x)^3,-1
全部回答
- 1楼网友:留下所有热言
- 2021-04-22 04:47
收敛区间我就不求了,你自己算下
收敛区间为(-1,1)
设s(x)=∑ nx^n/(n+1)
两边同乘x
xs(x)=∑ nx^(n+1)/(n+1)
令f(x)=xs(x)=∑ nx^(n+1)/(n+1)
两边同时求导
f'(x)=∑ nx^n
两边同除x
f'(x)/x=∑ nx^(n-1)
令g(x)=f'(x)/x=∑ nx^(n-1)
两边同时积分
∫g(x)dx=∑ x^n=x+x²+x³+……
无穷等比级数
=x/(1-x)
所以g(x)=[x/(1-x)]'=(1-x+x)/(1-x)²=1/(1-x)²
f'(x)=x/(1-x)²=x/(x-1)²=(x-1+1)/(x-1)²=1/(x-1) +1/(x-1)²
积分得
f(x)=ln(1-x) - 1/(x-1) +c
因为f(0)=0s(0)=0
得c=-1
则f(x)=ln(1-x) + 1/(1-x) -1=ln(1-x) + x/(1-x)
所以s(x)=(1/x) [ln(1-x) + x/(1-x)]
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