《数学分析解题指南》（林源渠，方企勤）里边的 4.1.17 题

luyuanhong

随便112

=sigma(-1)^(n+1)*1/(n+1).n=0到无穷
令a(n)=(-1)^(n+1)/(n+1),则c(n)=sigma(a(n)a(n-k),k=0,1,2***，n
=sigma(-1)^(k+1+n-k+1)*(1/(k+1)*1/(n-k+1)
=(-1)^nsigma(1/(k+1)(n-k+1)=2(-1)^n/(n+2)[1+1/2+***+1/（n+1)]
因为b(n)=1+1/2+1/3+***+1/n/(n+1)递减趋于0，所以c(n)收敛

luyuanhong

此题证明如下：