幂级数Σ(n=1到∞)x^(2n)=?

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/16 08:54:44

幂级数Σ(n=1到∞)x^(2n)=?

∑(n从1到正无穷)[(n² 1)/n ] x^(2n)
=∑(n从1到正无穷)nx^(2n) ∑(n从1到正无穷)(1/n)x^(2n)
=x/2∑(n从1到正无穷)2nx^(2n-1) 2∑(n从1到正无穷)[x^(2n)]/2n
=x/2∑(n从1到正无穷)[x^(2n)]′ 2∑(n从1到正无穷)∫x^(2n-1)dx(积分区间为0到x)
=x/...

全部展开

∑(n从1到正无穷)[(n² 1)/n ] x^(2n)
=∑(n从1到正无穷)nx^(2n) ∑(n从1到正无穷)(1/n)x^(2n)
=x/2∑(n从1到正无穷)2nx^(2n-1) 2∑(n从1到正无穷)[x^(2n)]/2n
=x/2∑(n从1到正无穷)[x^(2n)]′ 2∑(n从1到正无穷)∫x^(2n-1)dx(积分区间为0到x)
=x/2[∑(n从1到正无穷)x^(2n)]′ 2∫[∑(n从1到正无穷)x^(2n-1)]dx
=x/2[x²/(1-x²)]′ 2∫[x/(1-x²)]dx
=x²/(1-x²)²-ln|1-x²|

收起