I: limn→∞1+3+6+….+n(n+1)2n3=16II:limn→∞1.1!+2.2!+3.3!+……+n.n!(n+1)=1

Ilimn→∞∑n2+n2n3==limn→∞12n(n+1)(2n+1)6+n(n+1)2n3=limx→∞12nnn1+1n2+1n+n·n1n+136n3=   12  limx→∞1+1n2+1n6+1n+136n=12(1+0)(2+0)6+0 II  limn→∞(n+1)!-1(n+1)!=limn→∞(n+1)!(n+1)!-1(n+1)!=limn→∞1-1(n+1)!1-0=1