S.{x|x=1-1/2^n,n∈N+}证明supS=1,infS=1/2
问题描述：
S.{x|x=1-1/2^n,n∈N+}证明supS=1,infS=1/2
答
证明：①对于任意x∈S,有x=1-1/2^n0,存在x=1-1/2^([log2(1/e)]+1) [x]是求整函数使得x-1-e=-1/2^([log2(1/e)]+1)-e>-1/2^(log2(1/e))-e=e-e=0即x>1-e综上所述,supS=1同理,①对于任意x∈S,有x=1-1/2^n>=1-1/2=1/2②对...