∑[n=2:∞]1/(n²-1)=∑[n=2:∞]1/(n-1)(n+1)=½ ∑[n=2:∞][1/(n-1)-1/(n+1)]部分和Sn=½[1-1/3+1/2-1/4+1/3-1/5+……+1/(n-2)-1/n+1/(n-1)-1/(n+1)]=½ [1+1/2-1/n-1/(n+1)]故和S=lim[n→∞]Sn=½(1+1/2)=3/4