Question

已知数列﹛an﹜中,a1=½,Sn为数列的前n项和,且Sn与1/an的一个等比中项为n,则Sn（n趋于∞）的极

Answer 1

Sn=an*n^2

S(n+1)=a(n+1)*(n+1)^2

a(n+1)= a(n+1)*(n+1)^2- an*n^2

an*n^2= a(n+1)*(n^2+2n)

a(n+1)=n/(n+2)an=n/(n+2)*(n-1)/(n+1)*(n-2)/n*…*1/3*a1

=n!/(n+2)!*2*1/2=n!/(n+2)!=1/(n+2)(n+1)

对n=0成立a1=1/2

于是an=1/n(n+1),n=1,2…

则Sn=n^2/n(n+1)

于是Sn->1,n->∞