求这个级数的和函数x^n / n(n+1)(n+2) n从1到无穷
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 16:25:00
![求这个级数的和函数x^n / n(n+1)(n+2) n从1到无穷](/uploads/image/z/13341522-66-2.jpg?t=%E6%B1%82%E8%BF%99%E4%B8%AA%E7%BA%A7%E6%95%B0%E7%9A%84%E5%92%8C%E5%87%BD%E6%95%B0x%5En+%2F+n%28n%2B1%29%28n%2B2%29+n%E4%BB%8E1%E5%88%B0%E6%97%A0%E7%A9%B7)
求这个级数的和函数x^n / n(n+1)(n+2) n从1到无穷
求这个级数的和函数
x^n / n(n+1)(n+2) n从1到无穷
求这个级数的和函数x^n / n(n+1)(n+2) n从1到无穷
记
f(x) = Σ(n=1~inf.)[(x^n)/n(n+1)(n+2)],-1
1、由题可知
an=x^n / n(n+1)(n+2) =1/2* x^n *{[(1/n-1/(n+1)]-[(1/(n+1)-1/(n+2)]}
同理
an-1=1/2*x^(n-1) *{[(1/(n-1)-1/n]-[(1/n-1/(n+1)]}
an-2=1/2*x^(n-2) *{[(1/(n-2)-1/(n-1)]-[(1/(n-1)-1/n]}
全部展开
1、由题可知
an=x^n / n(n+1)(n+2) =1/2* x^n *{[(1/n-1/(n+1)]-[(1/(n+1)-1/(n+2)]}
同理
an-1=1/2*x^(n-1) *{[(1/(n-1)-1/n]-[(1/n-1/(n+1)]}
an-2=1/2*x^(n-2) *{[(1/(n-2)-1/(n-1)]-[(1/(n-1)-1/n]}
......
a2=1/2*x^2 *{[(1/2-1/3]-[(1/3-1/4]}
a1=1/2*x^1 *{[(1-1/2]-[(1/2-1/3]}
2、那么和函数
Sn=a1+a2+...1+an =1/2*x^1 *{[(1-1/2]-[(1/2-1/3]}+1/2*x^2 *{[(1/2-1/3]-[(1/3-1/4]}+...+1/2*x^n *{[(1/n-1/(n+1)]-[(1/(n+1)-1/(n+2)]}
左右各乘以x
x*Sn= =1/2*x^2 *{[(1-1/2]-[(1/2-1/3]}+1/2*x^3 *{[(1/2-1/3]-[(1/3-1/4]}+...+1/2*x^(n+1) *{[(1/n-1/(n+1)]-[(1/(n+1)-1/(n+2)]}
因此Sn-x*Sn=(1-x)*Sn=1/2*x^1 *{[(1-1/2]-[(1/2-1/3]}+1/2*x^2 *(1-1/2)-1/2*x^n*[(1/n-1/(n+1)]+1/2*x^(n+1) *{[(1/n-1/(n+1)]-[(1/(n+1)-1/(n+2)]}
所以Sn=1/2*{x/3+x^2/2+x^n*[(1/n-1/(n+1)]+x^(n+1) *{[(1/n-1/(n+1)]-[(1/(n+1)-1/(n+2)]} }/(1-x)
收起