(1)
S1=a1=1+ka1,
a1=1/(1-k);
Sn-1=1+kan-1,
Sn=1+kan,
an=Sn-Sn-1=k(an-an-1)
an=(an-1)k/(k-1),即得出遞推關(guān)系,
所以a2=-k/(k-1)^2,a3=-k^2/(k-1)^3,...
an=-k^(n-1)/(k-1)^n;
（2)
Sn=1+kan=1-(k/k-1)^n,
因?yàn)閘imn→∞Sn=1,
所以limn→∞(k/k-1)^n=0,
即要求-1