1/n²-1=1/(n(-1)(n+1)=(1/2)*[1/(n-1)-1/(n+1)]所以 1/2²-1+1/3²-1+1/4²-1+。。。+1/n²-1=(1/2)*[1-1/3+1/2-1/4+1/3-1/5+....+1/(n-1)+1/(n+1)]=(1/2)[1+1/2-1/n-1/(n+1)]=(1/2)[3/2-(2n+1)/n(n+1)]=3/4-(2n+1)/[2n(n+1)]
楼主呢？
