已知无穷级数∑(n=1~∞)an=a,则级数∑(n=1~∞)(an-a(n1))的和等于则级数∑(n=1~∞)(an-a(n+1))的和等于

Question

已知无穷级数∑(n=1~∞)an=a,则级数∑(n=1~∞)(an-a(n 1))的和等于
则级数∑(n=1~∞)(an-a(n+1))的和等于

Answer 1

因为无穷级数∑(n=1~∞)an=a
所以原级数收敛,所以n→无穷时,an=0
∑(n=1~∞)(an-a(n+1))=（a1-a2）+(a2-a3)+……+(an-a(n+1))=a1-a(n+1)=a1-0=a1