对∑(2^n)/n!则an=(2^n)/n!因为a(n+1)/an=[(2^(n+1))/(n+1)!]/[(2^n)/n!]=2/(n+1)所以lim[a(n+1)/an]=lim[(2^(n+1))/(n+1)!]/[(2^n)/n!]=lim[2/(n+1)]=0<1由比值审敛法知∑(2^n)/n!收敛lim(n/(n+1))^n=lim[1/(1+1/n)^n]=1/e<1
