无穷级数收敛计算级数-1+1/3-1/5+1/7-1/9+.+1/（2m-1）*(-1)的m次方收敛吗?要是收敛,收敛于多少呢?

Question

无穷级数收敛计算
级数-1+1/3-1/5+1/7-1/9+.+1/（2m-1）*(-1)的m次方收敛吗?要是收敛,收敛于多少呢?

Answer 1

由Leibniz判别法,交错级数∑(-1)^m*am收敛当且仅当am->0.现在am=1/(2m-1)->0,所以级数收敛.
为求其值,注意到arctan(x)的Taylor展开式为
arctan(x)
=∑{n>=0} (-1)^n*x^(2n+1)/(2n+1)
(换元n=m-1)
=∑{m>=1} (-1)^(m-1)*x^(2m-1)/(2m-1)
=-∑{m>=1} (-1)^m*x^(2m-1)/(2m-1),
代入x=1,得
arctan(1)
=-∑{m>=1} (-1)^m/(2m-1),
所以待求级数
∑{m>=1} (-1)^m/(2m-1)
=-arctan(1)
=-π/4