首项为1的无穷递减等比数列各项合为S,SN表示该数列的前n项的和,求lim(s1+s2+...+sn-ns)的值

首项1>0,若公比q0,又数列为递减数列,00 1-qⁿ->1 S=a1/(1-q)
S1+S2+...+Sn-nS
=[a1/(1-q)][(1-q)+(1-q²)+...+(1-qⁿ)] -na1/(1-q)
=[a1/(1-q)][n-(q+q²+...+qⁿ)] -na1/(1-q)
=[a1/(1-q)][n-q(1-qⁿ)/(1-q)] -na1/(1-q)
=na1/(1-q) -a1q(1-qⁿ)/(1-q)² -na1/(1-q)
=-a1q(1-qⁿ)/(1-q)²
=-[a1q/(q-1)²](1-qⁿ)
n->+∞,qⁿ->0 1-qⁿ->1
S1+S2+...+Sn-nS->-a1q/(q-1)²
lim(S1+S2+...+Sn-nS)=-a1q/(q-1)²
n->+∞