1/aka(k+1)=[a(k+1)-ak]/[d*ak*a(k+1)]=(1/d)*[1/ak-1/a(k+1)] 所以Sn=(1/d)*[1/a1-1/a2+1/a2-1/a3+...+1/an-1/a(n+1)] =(1/d)*[1/a1-1/a(n+1)]