在数列an中,其前N项和Sn=1/3n(n+1)(n+2).记Tn为数列(1/an)的前N项和.求lim(n→∞)Tn

an=Sn-Sn-1 =1/3 n(n+1)(n+2) - 1/3 n(n+1)(n-1) = n(n+1)
所以 1/an = 1/ n(n+1) = 1/n -1/ n+1
数列(1/an)的前n项和 = 1-1/2 + 1/2-1/3 +...+ 1/n -1/ n+1 =1- 1/n+1
所以lim(n→∞)Tn
=lim(n→∞)1- 1/n+1
=1