已知等差数列{bn},公差d>0,b1>0,Sn=1/b1b2+1/b2b3+1/b3b4+...+1/bnb(n+1),则lim(n→∞)Sn=

Question

已知等差数列{bn},公差d>0,b1>0,Sn=1/b1b2+1/b2b3+1/b3b4+...+1/bnb(n+1),则lim(n→∞)Sn=_______

Answer 1

Sn=1/b1b2+1/b2b3+1/b3b4+...+1/bnb(n+1),
Sn=d(1/b1-1/b2+1/b2-1/b3+...+1/bn-1/b(n+1))
得Sn=d(1/b1-1/b(n+1))因为bn为等差数列
b(n+1)=b1+nd
Sn=nd^2/(b1^2+nb1d)
lim(n→∞)Sn=lim(n→∞)nd^2/(b1^2+nb1d)=
lim(n→∞)d^2/(b1^2/n+b1d)=d/b1