f(1)=1;f(2)=6*(2-1)+1;f(3)=6*(3-1)+f(2)……
可知f(n)=6*(n-1)+f(n-1)=6*(n-1)+6*(n-2)+f(n-2)=……=6*[(n-1)+(n-2)+……+(2-1)]+f(1)=6n(n-1)/2+1=3n(n-1)+1
所以,1/f(1)+1/f(2)+1/f(3)…+1/f(n)=1+1/7+1/9+……+1/[3n(n-1)+1]＜1+1/7+1/9+……+1/[3n(n-1)]【∵1/[3n(n-1)+1]＜1/[3n(n-1)]】=1/3{3+1/(2*1)+1/(3*2)+……+1/[n(n-1)]}=1/3[3+1-1/2+1/2-1/3+……+1/(n-1)-1/n]=1/3(3+1-1/n)=4/3-1/3n＜4/3
证毕